chapter 11 systems of equations and inequalitiesshubbard/hpcanswers/pr_9e_ism_11all.pdf · 45 3 4...
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 11
Systems of Equations and InequalitiesSection 11.1
1. 3 4 8
4 4
1
x x
x
x
+ = −==
The solution set is { }1 .
2. a. 3 4 12x y+ =
x-intercept: ( )3 4 0 12
3 12
4
x
x
x
+ ===
y-intercept: ( )3 0 4 12
4 12
3
y
y
y
+ ===
b. 3 4 12
4 3 12
33
4
x y
y x
y x
+ == − +
= − +
A parallel line would have slope 3
4− .
3. inconsistent
4. consistent; independent
5. (3, 2)−
6. consistent; dependent
7. 2 5
5 2 8
x y
x y
− = + =
Substituting the values of the variables: 2(2) ( 1) 4 1 5
5(2) 2( 1) 10 2 8
− − = + = + − = − =
Each equation is satisfied, so 2, 1x y= = − , or
(2, 1)− , is a solution of the system of equations.
8. 3 2 2
7 30
x y
x y
+ = − = −
Substituting the values of the variables: 3( 2) 2(4) 6 8 2
( 2) 7(4) 2 28 30
− + = − + = − − = − − = −
Each equation is satisfied, so 2, 4x y= − = , or
( 2, 4)− , is a solution of the system of equations.
9. 3 4 4
1 13
2 2
x y
x y
− = − = −
Substituting the values of the variables:
13(2) 4 6 2 4
2
1 1 3 1(2) 3 1
2 2 2 2
− = − =
− = − = −
Each equation is satisfied, so 12, 2x y= = , or
( )12, 2 , is a solution of the system of equations.
10. 12 02
193 4 2
x y
x y
+ =
− = −
Substituting the values of the variables, we obtain:
( )
( )
1 12 2 1 1 0
2 2
1 3 193 4 2 8
2 2 2
− + = − + = − − = − − = −
Each equation is satisfied, so 1 , 22x y= − = , or
( )1 , 22− , is a solution of the system of equations.
11. 3
13
2
x y
x y
− = + =
Substituting the values of the variables, we obtain:
Section 11.1: Systems of Linear Equations: Substitution and Elimination
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4 1 3
1(4) 1 2 1 3
2
− = + = + =
Each equation is satisfied, so 4, 1x y= = , or
(4, 1) , is a solution of the system of equations.
12. 3
3 1
x y
x y
− =− + =
Substituting the values of the variables:
( ) ( )( ) ( )2 5 2 5 3
3 2 5 6 5 1
− − − = − + =
− − + − = − =
Each equation is satisfied, so 2, 5x y= − = − , or
( 2, 5)− − , is a solution of the system of equations.
13.
3 3 2 4
0
2 3 8
x y z
x y z
y z
+ + = − − = − = −
Substituting the values of the variables: 3(1) 3( 1) 2(2) 3 3 4 4
1 ( 1) 2 1 1 2 0
2( 1) 3(2) 2 6 8
+ − + = − + = − − − = + − = − − = − − = −
Each equation is satisfied, so 1, 1, 2x y z= = − = ,
or (1, 1, 2)− , is a solution of the system of
equations.
14.
4 7
8 5 0
5 6
x z
x y z
x y z
− = + − =− − + =
Substituting the values of the variables:
( )( ) ( )
( ) ( )
4 2 1 8 1 7
8 2 5 3 1 16 15 1 0
2 3 5 1 2 3 5 6
− = − =
+ − − = − − =− − − + = − + + =
Each equation is satisfied, so 2x = , 3y = − ,
1z = , or (2, 3,1)− , is a solution of the system of
equations.
15.
3 3 2 4
3 10
5 2 3 8
x y z
x y z
x y z
+ + = − + = − − =
Substituting the values of the variables:
( ) ( ) ( )( )
( ) ( ) ( )
3 2 3 2 2 2 6 6 4 4
2 3 2 2 2 6 2 10
5 2 2 2 3 2 10 4 6 8
+ − + = − + =
− − + = + + = − − − = + − =
Each equation is satisfied, so 2x = , 2y = − ,
2z = , or (2, 2, 2)− is a solution of the system of
equations.
16.
4 5 6
5 17
6 5 24
x z
y z
x y z
− = − = −− − + =
Substituting the values of the variables:
( ) ( )( ) ( )( ) ( ) ( )
4 4 5 2 16 10 6
5 3 2 15 2 17
4 6 3 5 2 4 18 10 24
− = − =
− − = − − = −− − − + = − + + =
Each equation is satisfied, so 4x = , 3y = − ,
2z = , or (4, 3, 2)− , is a solution of the system
of equations.
17. 8
4
x y
x y
+ = − =
Solve the first equation for y, substitute into the second equation and solve:
8
4
y x
x y
= − − =
(8 ) 4
8 4
2 12
6
x x
x x
x
x
− − =− + =
==
Since 6, 8 6 2x y= = − = . The solution of the
system is 6, 2x y= = or using ordered pairs
(6, 2) .
18. 2 7
3
x y
x y
+ = − + = −
Solve the first equation for x, substitute into the second equation and solve:
7 2
3
x y
x y
= − − + = −
( 7 2 ) 3
7 3
4
y y
y
y
− − + = −− − = −
− =
Since 4, 7 2( 4) 1y x= − = − − − = . The solution
of the system is 1, 4x y= = − or using an
ordered pair, (1, 4)− .
Chapter 11: Systems of Equations and Inequalities
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19. 5 21
2 3 12
x y
x y
− = + = −
Multiply each side of the first equation by 3 and add the equations to eliminate y:
15 3 63
2 3 12
x y
x y
− = + = −
17 51
3
x
x
==
Substitute and solve for y: 5(3) 21
15 21
6
6
y
y
y
y
− =− =− =
= −
The solution of the system is 3, 6x y= = − or
using ordered an pair ( )3, 6− .
20. 3 5
2 3 8
x y
x y
+ = − = −
Add the equations: 3 5
2 3 8
x y
x y
+ = − = −
3 3
1
x
x
= −= −
Substitute and solve for y: 1 3 5
3 6
2
y
y
y
− + ===
The solution of the system is 1, 2x y= − = or
using ordered pairs ( 1, 2)− .
21. 3 24
2 0
x
x y
= + =
Solve the first equation for x and substitute into the second equation:
8
2 0
x
x y
= + =
8 2 0
2 8
4
y
y
y
+ == −= −
The solution of the system is 8, 4x y= = − or
using ordered pairs (8, 4)−
22. 4 5 3
2 8
x y
y
+ = − − = −
Solve the second equation for y and substitute into the first equation:
4 5 3
4
x y
y
+ = − =
4 5(4) 3
4 20 3
4 23
23
4
x
x
x
x
+ = −+ = −
= −
= −
The solution of the system is 23
, 44
x y= − = or
using ordered pairs 23
, 44
−
.
23. 3 6 2
5 4 1
x y
x y
− = + =
Multiply each side of the first equation by 2 and each side of the second equation by 3, then add to eliminate y:
6 12 4
15 12 3
x y
x y
− = + =
21 7
1
3
x
x
=
=
Substitute and solve for y:
( )3 1/ 3 6 2
1 6 2
6 1
1
6
y
y
y
y
− =− =− =
= −
The solution of the system is 1 1
,3 6
x y= = − or
using ordered pairs 1 1
,3 6
−
.
Section 11.1: Systems of Linear Equations: Substitution and Elimination
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24. 2
2 43
3 5 10
x y
x y
+ = − = −
Multiply each side of the first equation by 5 and each side of the second equation by 4, then add to eliminate y:
1010 20
312 20 40
x y
x y
+ = − = −
11022
35
3
x
x
=
= −
Substitute and solve for y:
( )3 5 / 3 5 10
5 5 10
5 5
1
y
y
y
y
− − = −− − = −
− = −=
The solution of the system is 5
, 13
x y= − = or
using ordered pairs 5
, 13
−
.
25. 2 1
4 2 3
x y
x y
+ = + =
Solve the first equation for y, substitute into the second equation and solve:
1 2
4 2 3
y x
x y
= − + =
4 2(1 2 ) 3
4 2 4 3
0 1
x x
x x
+ − =+ − =
=
This equation is false, so the system is inconsistent.
26. 5
3 3 2
x y
x y
− =− + =
Solve the first equation for x, substitute into the second equation and solve:
5
3 3 2
x y
x y
= +− + =
3( 5) 3 2
3 15 3 2
0 17
y y
y y
− + + =− − + =
=
This equation is false, so the system is inconsistent.
27. 2 04 2 12
x yx y
− = + =
Solve the first equation for y, substitute into the second equation and solve:
24 2 12y xx y=
+ =
4 2(2 ) 124 4 12
8 123
2
x xx x
x
x
+ =+ =
=
=
Since 3 3
, 2 32 2
x y = = =
The solution of the system is 3
, 32
x y= = or
using ordered pairs 3
,3 .2
28. 3 3 1
84
3
x y
x y
+ = − + =
Solve the second equation for y, substitute into the first equation and solve:
3 3 18
43
x y
y x
+ = − = −
8
3 3 4 13
3 8 12 19 9
1
x x
x xxx
+ − = − + − = −
− = −=
Since 8 8 4
1, 4(1) 4 .3 3 3
x y= = − = − = −
The solution of the system is 4
1,3
x y= = − or
using ordered pairs 4
1,3
−
.
29. 2 4
2 4 8x yx y
+ = + =
Solve the first equation for x, substitute into the second equation and solve:
4 22 4 8x y
x y= −
+ =
Chapter 11: Systems of Equations and Inequalities
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2(4 2 ) 4 88 4 4 8
0 0
y yy y
− + =− + =
=
These equations are dependent. The solution of the system is either 4 2x y= − , where y is any real
number or 4
2
xy
−= , where x is any real number.
Using ordered pairs, we write the solution as
{ }( , ) 4 2 , is any real numberx y x y y= − or as
4( , ) , is any real number
2
xx y y x
− =
.
30. 3 7
9 3 21
x y
x y
− = − =
Solve the first equation for y, substitute into the second equation and solve:
3 7
9 3 21
y x
x y
= − − =
9 3(3 7) 21
9 9 21 21
0 0
x x
x x
− − =− + =
=
These equations are dependent. The solution of the system is either 3 7y x= − , where x is any real
number is 7
3
yx
+= , where y is any real number.
Using ordered pairs, we write the solution as
{ }( , ) 3 7, is any real numberx y y x x= − or as
7( , ) , is any real number
3
yx y x y
+ =
.
31. 2 3 1
10 11
x y
x y
− = − + =
Multiply each side of the first equation by –5, and add the equations to eliminate x:
10 15 5
10 11
x y
x y
− + = + =
16 16
1
y
y
==
Substitute and solve for x: 2 3(1) 1
2 3 1
2 2
1
x
x
x
x
− = −− = −
==
The solution of the system is 1, 1x y= = or
using ordered pairs (1, 1).
32. 3 2 0
5 10 4
x y
x y
− = + =
Multiply each side of the first equation by 5, and add the equations to eliminate y:
15 10 0
5 10 4
x y
x y
− = + =
20 4
1
5
x
x
=
=
Substitute and solve for y:
( )5 1/ 5 10 4
1 10 4
10 3
3
10
y
y
y
y
+ =+ =
=
=
The solution of the system is 1 3
,5 10
x y= = or
using ordered pairs 1 3
,5 10
.
33. 2 3 6
1
2
x y
x y
+ = − =
Solve the second equation for x, substitute into the first equation and solve:
2 3 6
1
2
x y
x y
+ = = +
12 3 6
2
2 1 3 6
5 5
1
y y
y y
y
y
+ + =
+ + ===
Since 1 3
1, 12 2
y x= = + = . The solution of the
system is 3
, 12
x y= = or using ordered pairs
3, 1
2
.
Section 11.1: Systems of Linear Equations: Substitution and Elimination
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34. 1
22
2 8
x y
x y
+ = − − =
Solve the second equation for x, substitute into the first equation and solve:
12
22 8
x y
x y
+ = − = +
1(2 8) 2
24 2
2 6
3
y y
y y
y
y
+ + = −
+ + = −= −= −
Since 3, 2( 3) 8 6 8 2y x= − = − + = − + = . The
solution of the system is 2, 3x y= = − or using
ordered pairs (2, 3)− .
35.
1 13
2 31 2
14 3
x y
x y
+ = − = −
Multiply each side of the first equation by –6 and each side of the second equation by 12, then add to eliminate x:
3 2 18
3 8 12
x y
x y
− − = − − = −
10 30
3
y
y
− = −=
Substitute and solve for x: 1 1
(3) 32 3
11 3
21
22
4
x
x
x
x
+ =
+ =
=
=
The solution of the system is 4, 3x y= = or
using ordered pairs (4, 3).
36.
1 35
3 23 1
114 3
x y
x y
− = − + =
Multiply each side of the first equation by –54 and each side of the second equation by 24, then add to eliminate x:
18 81 270
18 8 264
x y
x y
− + = + =
89 534
6
y
y
==
Substitute and solve for x: 3 1
(6) 114 3
32 11
43
94
12
x
x
x
x
+ =
+ =
=
=
The solution of the system is 12, 6x y= = or
using ordered pairs (12, 6).
37. 3 5 3
15 5 21
x y
x y
− = + =
Add the equations to eliminate y: 3 5 3
15 5 21
x y
x y
− = + =
18 24
4
3
x
x
=
=
Substitute and solve for y:
( )3 4 / 3 5 3
4 5 3
5 1
1
5
y
y
y
y
− =− =− = −
=
The solution of the system is 4 1
,3 5
x y= = or
using ordered pairs 4 1
,3 5
.
Chapter 11: Systems of Equations and Inequalities
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38. 2 1
1 3
2 2
x y
x y
− = − + =
Multiply each side of the second equation by 2, and add the equations to eliminate y:
2 1
2 3
x y
x y
− = − + =
4 2
1
2
x
x
=
=
Substitute and solve for y: 1
2 12
1 1
2
2
y
y
y
y
− = −
− = −− = −
=
The solution of the system is 1
, 22
x y= = or
using ordered pairs 1
, 22
.
39.
1 18
3 50
x y
x y
+ = − =
Rewrite letting 1 1
,u vx y
= = :
8
3 5 0
u v
u v
+ = − =
Solve the first equation for u, substitute into the second equation and solve:
8
3 5 0
u v
u v
= − − =
3(8 ) 5 0
24 3 5 0
8 24
3
v v
v v
v
v
− − =− − =
− = −=
Since 3, 8 3 5v u= = − = . Thus, 1 1
5x
u= = ,
1 1
3y
v= = . The solution of the system is
1 1,
5 3x y= = or using ordered pairs
1 1,
5 3
.
40.
4 30
6 32
2
x y
x y
− = + =
Rewrite letting 1 1
,u vx y
= = :
4 3 0
36 2
2
u v
u v
− = + =
Multiply each side of the second equation by 2, and add the equations to eliminate v:
4 3 0
12 3 4
u v
u v
− = + =
16 4
4 1
16 4
u
u
=
= =
Substitute and solve for v: 1
4 3 04
1 3 0
3 1
1
3
v
v
v
v
− =
− =− = −
=
Thus, 1 1
4, 3x yu v
= = = = . The solution of the
system is 4, 3x y= = or using ordered pairs
(4, 3).
41.
6
2 3 16
2 4
x y
x z
y z
− = − = + =
Multiply each side of the first equation by –2 and add to the second equation to eliminate x:
2 2 12
2 3 16
2 3 4
x y
x z
y z
− + = −− =
− =
Multiply each side of the result by –1 and add to the original third equation to eliminate y:
2 3 4
2 4
4 0
0
y z
y z
z
z
− + = −+ =
==
Substituting and solving for the other variables:
Section 11.1: Systems of Linear Equations: Substitution and Elimination
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2 0 4
2 4
2
y
y
y
+ ===
2 3(0) 16
2 16
8
x
x
x
− ===
The solution is 8, 2, 0x y z= = = or using
ordered triples (8, 2, 0).
42.
2 4
2 4 0
3 2 11
x y
y z
x z
+ = −− + = − = −
Multiply each side of the first equation by 2 and add to the second equation to eliminate y: 4 2 8
2 4 0
4 4 8
x y
y z
x z
+ = −− + =
+ = −
Multiply each side of the result by 1
2 and add to
the original third equation to eliminate z: 2 2 4
3 2 11
5 15
3
x z
x z
x
x
+ = −− = −
= −= −
Substituting and solving for the other variables: 2( 3) 4
6 4
2
y
y
y
− + = −− + = −
=
3( 3) 2 11
9 2 11
2 2
1
z
z
z
z
− − = −− − = −
− = −=
The solution is 3, 2, 1x y z= − = = or using
ordered triples ( 3, 2,1)− .
43. 2 3 7
2 43 2 2 10
x y zx y zx y z
− + = + + =− + − = −
Multiply each side of the first equation by –2 and add to the second equation to eliminate x; and multiply each side of the first equation by 3 and add to the third equation to eliminate x:
2 4 6 14
2 4
5 5 10
x y z
x y z
y z
− + − = −+ + =
− = −
3 6 9 21
3 2 2 10
4 7 11
x y z
x y z
y z
− + =− + − = −
− + =
Multiply each side of the first result by 4
5 and
add to the second result to eliminate y:
4 4 8
4 7 11
y z
y z
− = −− + =
3 3
1
z
z
==
Substituting and solving for the other variables:
1 2
1
y
y
− = −= −
2( 1) 3(1) 7
2 3 7
2
x
x
x
− − + =+ + =
=
The solution is 2, 1, 1x y z= = − = or using
ordered triples (2, 1,1)− .
44.
2 3 0
2 2 7
3 4 3 7
x y z
x y z
x y z
+ − =− + + = − − − =
Multiply each side of the first equation by –2 and add to the second equation to eliminate y; and multiply each side of the first equation by 4 and add to the third equation to eliminate y:
4 2 6 0
2 2 7
6 7 7
x y z
x y z
x z
− − + =− + + = −
− + = −
8 4 12 0
3 4 3 7
11 15 7
x y z
x y z
x z
+ − =− − =
− =
Multiply each side of the first result by 11 and multiply each side of the second result by 6 to eliminate x:
66 77 77
66 90 42
x z
x z
− + = −− =
13 35
35
13
z
z
− = −
=
Substituting and solving for the other variables: 35
6 7 713
2456 7
13336
613
56
13
x
x
x
x
− + = −
− + = −
− = −
=
Chapter 11: Systems of Equations and Inequalities
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56 352 3 0
13 13
112 1050
13 137
13
y
y
y
+ − =
+ − =
= −
The solution is 56 7 35
, ,13 13 13
x y z= = − = or
using ordered triples 56 7 35
, ,13 13 13 −
.
45.
1
2 3 2
3 2 0
x y z
x y z
x y
− − = + + = + =
Add the first and second equations to eliminate z: 1
2 3 2
3 2 3
x y z
x y z
x y
− − =+ + =
+ =
Multiply each side of the result by –1 and add to the original third equation to eliminate y:
3 2 3
3 2 0
0 3
x y
x y
− − = −+ =
= −
This equation is false, so the system is inconsistent.
46.
2 3 0
2 5
3 4 1
x y z
x y z
x y z
− − =− + + = − − =
Add the first and second equations to eliminate z; then add the second and third equations to eliminate z: 2 3 0
2 5
5
x y z
x y z
x y
− − =− + + =
− =
2 5
3 4 1
2 2 6
x y z
x y z
x y
− + + =− − =
− =
Multiply each side of the first result by –2 and add to the second result to eliminate y:
2 2 10
2 2 6
x y
x y
− + = −− =
0 2= −
This equation is false, so the system is inconsistent.
47.
1
2 3 4
3 2 7 0
x y z
x y z
x y z
− − =− + − = − − − =
Add the first and second equations to eliminate x; multiply the first equation by –3 and add to the third equation to eliminate x:
1
2 3 4
4 3
x y z
x y z
y z
− − =− + − = −
− = −
3 3 3 3
3 2 7 0
4 3
x y z
x y z
y z
− + + = −− − =
− = −
Multiply each side of the first result by –1 and add to the second result to eliminate y:
4 3
4 3
0 0
y z
y z
− + =− = −
=
The system is dependent. If z is any real number, then 4 3y z= − .
Solving for x in terms of z in the first equation: (4 3) 1
4 3 1
5 3 1
5 2
x z z
x z z
x z
x z
− − − =− + − =
− + == −
The solution is {( , , )x y z 5 2,x z= − 4 3y z= − ,
z is any real number}.
48.
2 3 0
3 2 2 2
5 3 2
x y z
x y z
x y z
− − = + + = + + =
Multiply the first equation by 2 and add to the second equation to eliminate z; multiply the first equation by 3 and add to the third equation to eliminate z: 4 6 2 0
3 2 2 2
7 4 2
x y z
x y z
x y
− − =+ + =
− =
6 9 3 0
5 3 2
7 4 2
x y z
x y z
x y
− − =+ + =
− =
Multiply each side of the first result by –1 and add to the second result to eliminate y:
Section 11.1: Systems of Linear Equations: Substitution and Elimination
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
7 4 2
7 4 2
0 0
x y
x y
− + = −− =
=
The system is dependent. If y is any real
number, then 4 2
7 7x y= + .
Solving for z in terms of x in the first equation: 2 3
4 22 3
7
8 4 21
713 4
7
z x y
yy
y y
y
= −+ = −
+ −=
− +=
The solution is 4 2
( , , )7 7
x y z x y = +
,
13 4, is any real number
7 7z y y
= − +
.
49.
2 2 3 6
4 3 2 0
2 3 7 1
x y z
x y z
x y z
− + = − + =− + − =
Multiply the first equation by –2 and add to the second equation to eliminate x; add the first and third equations to eliminate x:
4 4 6 12
4 3 2 0
4 12
x y z
x y z
y z
− + − = −− + =
− = −
2 2 3 6
2 3 7 1
4 7
x y z
x y z
y z
− + =− + − =
− =
Multiply each side of the first result by –1 and add to the second result to eliminate y:
4 12
4 7
0 19
y z
y z
− + =− =
=
This result is false, so the system is inconsistent.
50.
3 2 2 6
7 3 2 1
2 3 4 0
x y z
x y z
x y z
− + = − + = − − + =
Multiply the first equation by –1 and add to the second equation to eliminate z; multiply the first equation by –2 and add to the third equation to eliminate z:
3 2 2 6
7 3 2 1
4 7
x y z
x y z
x y
− + − = −− + = −
− = −
6 4 4 12
2 3 4 0
4 12
x y z
x y z
x y
− + − = −− + =
− + = −
Add the first result to the second result to eliminate y:
4 7
4 12
0 19
x y
x y
− = −− + = −
= −
This result is false, so the system is inconsistent.
51.
6
3 2 5
3 2 14
x y z
x y z
x y z
+ − = − + = − + − =
Add the first and second equations to eliminate z; multiply the second equation by 2 and add to the third equation to eliminate z:
6
3 2 5
4 1
x y z
x y z
x y
+ − =− + = −
− =
6 4 2 10
3 2 14
7 4
x y z
x y z
x y
− + = −+ − =
− =
Multiply each side of the first result by –1 and add to the second result to eliminate y:
4 1
7 4
x y
x y
− + = −− =
3 3
1
x
x
==
Substituting and solving for the other variables: 4(1) 1
3
3
y
y
y
− =− = −
=
3(1) 2(3) 5
3 6 5
2
z
z
z
− + = −− + = −
= −
The solution is 1, 3, 2x y z= = = − or using
ordered triplets (1, 3, 2)− .
Chapter 11: Systems of Equations and Inequalities
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
52. 4
2 3 4 155 2 12
x y zx y zx y z
− + = − − + = − + − =
Multiply the first equation by –3 and add to the second equation to eliminate y; add the first and third equations to eliminate y:
3 3 3 122 3 4 15x y zx y z
− + − =− + = −
33
x zz x
− + = −= −
45 2 12
6 8
x y zx y z
x z
− + = −+ − =
− =
Substitute and solve: 6 ( 3) 8
6 3 85 5
1
x xx x
xx
− − =− + =
==
3 1 3 2z x= − = − = −
12 5 2 12 5(1) 2( 2) 3y x z= − + = − + − =
The solution is 1, 3, 2x y z= = = − or using
ordered triplets (1, 3, 2)− .
53. 2 3
2 4 72 2 3 4
x y zx y zx y z
+ − = − − + = −− + − =
Add the first and second equations to eliminate z; multiply the second equation by 3 and add to the third equation to eliminate z: 2 32 4 7
3 2 10
x y zx y z
x y
+ − = −− + = −− = −
6 12 3 212 2 3 4
4 10 17
x y zx y z
x y
− + = −− + − =
− = −
Multiply each side of the first result by –5 and add to the second result to eliminate y:
15 10 504 10 17
x yx y
− + =− = −
11 333
xx
− == −
Substituting and solving for the other variables: 3( 3) 2 10
9 2 102 1
1
2
yyy
y
− − = −− − = −
− = −
=
13 2 3
2
3 1 3
1
1
z
z
z
z
− + − = −
− + − = −− = −
=
The solution is 1
3, , 12
x y z= − = = or using
ordered triplets 1
3, , 12
−
.
54.
4 3 8
3 3 12
6 1
x y z
x y z
x y z
+ − = − − + = + + =
Add the first and second equations to eliminate z; multiply the first equation by 2 and add to the third equation to eliminate z:
4 3 8
3 3 12
4 3 4
x y z
x y z
x y
+ − = −− + =
+ =
2 8 6 16
6 1
3 9 15
x y z
x y z
x y
+ − = −+ + =
+ = −
Multiply each side of the second result by 1/ 3− and add to the first result to eliminate y: 4 3 4
3 5
x y
x y
+ =− − =
3 9
3
x
x
==
Substituting and solving for the other variables: 3 3 5
3 8
8
3
y
y
y
+ = −= −
= −
83 6 1
3
26
31
9
z
z
z
+ − + =
=
=
The solution is 8 1
3, ,3 9
x y z= = − = or using
ordered triplets 8 1
3, ,3 9
−
.
Section 11.1: Systems of Linear Equations: Substitution and Elimination
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
55. Let l be the length of the rectangle and w be the width of the rectangle. Then:
2l w= and 2 2 90l w+ =
Solve by substitution: 2(2 ) 2 90
4 2 90
6 90
15 feet
2(15) 30 feet
w w
w w
w
w
l
+ =+ =
=== =
The floor is 15 feet by 30 feet.
56. Let l be the length of the rectangle and w be the width of the rectangle. Then:
50l w= + and 2 2 3000l w+ =
Solve by substitution: 2( 50) 2 3000
2 100 2 3000
4 2900
725 meters
725 50 775 meters
w w
w w
w
w
l
+ + =+ + =
=== + =
The dimensions of the field are 775 meters by 725 meters.
57. Let x = the number of commercial launches and y = the number of noncommercial launches.
Then: 55x y+ = and 2 1y x= +
Solve by substitution: (2 1) 55
3 54
18
x x
x
x
+ + ===
2(18) 1
36 1
37
y
y
y
= += +=
In 2005 there were 18 commercial launches and 37 noncommercial launches.
58. Let x = the number of adult tickets sold and y = the number of senior tickets sold. Then:
325
9 7 2495
x y
x y
+ = + =
Solve the first equation for y: 325y x= −
Solve by substitution: 9 7(325 ) 2495
9 2275 7 2495
2 220
110
x x
x x
x
x
+ − =+ − =
==
325 110 215y = − =
There were 110 adult tickets sold and 215 senior citizen tickets sold.
59. Let x = the number of pounds of cashews. Let y = is the number of pounds in the mixture. The value of the cashews is 5x . The value of the peanuts is 1.50(30) = 45. The value of the mixture is 3y .
Then 30x y+ = represents the amount of mixture.
5 45 3x y+ = represents the value of the mixture.
Solve by substitution: 5 45 3( 30)
2 45
22.5
x x
x
x
+ = +==
So, 22.5 pounds of cashews should be used in the mixture.
60. Let x = the amount invested in AA bonds. Let y = the amount invested in the Bank
Certificate. a. Then 150,000x y+ = represents the total
investment. 0.10 0.05 12,000x y+ = represents the
earnings on the investment.
Solve by substitution: 0.10(150,000 ) 0.05 12,000
15,000 0.10 0.05 12,000
0.05 3000
60,000
y y
y y
y
y
− + =− + =
− = −=
150,000 60,000 90,000x = − = Thus, $90,000 should be invested in AA Bonds and $60,000 in a Bank Certificate.
b. Then 150,000x y+ = represents the total
investment. 0.10 0.05 14,000x y+ = represents the
earnings on the investment.
Solve by substitution: 0.10(150,000 ) 0.05 14,000
15,000 0.10 0.05 14,000
0.05 1000
20,000
y y
y y
y
y
− + =− + =
− = −=
150,000 20,000 130,000x = − = Thus, $130,000 should be invested in AA Bonds and $20,000 in a Bank Certificate.
Chapter 11: Systems of Equations and Inequalities
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61. Let x = the plane’s average airspeed and y = the average wind speed.
Rate Time Distance
With Wind 3 600
Against 4 600
x y
x y
+−
( )(3) 600
( )(4) 600
x y
x y
+ = − =
Multiply each side of the first equation by 1
3,
multiply each side of the second equation by 1
4,
and add the result to eliminate y 200
150
x y
x y
+ =− =
2 350
175
x
x
==
175 20025
yy
+ ==
The average airspeed of the plane is 175 mph, and the average wind speed is 25 mph.
62. Let x = the average wind speed and y = the
distance.
Rate Time Distance
With Wind 150 2
Against 150 3
x y
x y
+−
(150 )(2)
(150 )(3)
x y
x y
+ = − =
Solve by substitution: (150 )(2) (150 )(3)
300 2 450 3
5 150
30
x x
x x
x
x
+ = −+ = −
==
Thus, the average wind speed is 30 mph.
63. Let x = the number of $25-design. Let y = the number of $45-design.
Then x y+ = the total number of sets of dishes.
25 45x y+ = the cost of the dishes.
Setting up the equations and solving by substitution:
200
25 45 7400
x y
x y
+ = + =
Solve the first equation for y, the solve by substitution: 200y x= −
25 45(200 ) 7400
25 9000 45 7400
20 1600
80
x x
x x
x
x
+ − =+ − =
− = −=
200 80 120y = − =
Thus, 80 sets of the $25 dishes and 120 sets of the $45 dishes should be ordered.
64. Let x = the cost of a hot dog. Let y = the cost of a soft drink.
Setting up the equations and solving by substitution:
10 5 35.00
7 4 25.25
x y
x y
+ = + =
10 5 35.00
2 7
7 2
x y
x y
y x
+ =+ =
= −
7 4(7 2 ) 25.25
7 28 8 25.25
2.75
2.75
x x
x x
x
x
+ − =+ − =
− = −=
7 2(2.75) 1.50y = − =
A single hot dog costs $2.75 and a single soft drink costs $1.50.
65. Let x = the cost per package of bacon. Let y = the cost of a carton of eggs.
Set up a system of equations for the problem: 3 2 13.45
2 3 11.45
x y
x y
+ = + =
Multiply each side of the first equation by 3 and each side of the second equation by –2 and solve by elimination:
9 6 40.35
4 6 22.90
x y
x y
+ =− − = −
5 17.45
3.49
x
x
==
Substitute and solve for y: 3(3.49) 2 13.45
10.47 2 13.45
2 2.98
1.49
y
y
y
y
+ =+ =
==
A package of bacon costs $3.49 and a carton of eggs cost $1.49. The refund for 2 packages of bacon and 2 cartons of eggs will be 2($3.49) + 2($1.49) = $9.96.
Section 11.1: Systems of Linear Equations: Substitution and Elimination
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66. Let x = Pamela’s average speed in still water. Let y = the speed of the current.
Rate Time Distance
Downstream 3 15
Upstream 5 15
x y
x y
+−
Set up a system of equations for the problem:
3( ) 15
5( ) 15
x y
x y
+ = − =
Multiply each side of the first equation by 1
3,
multiply each side of the second equation by 1
5,
and add the result to eliminate y:
5
3
x y
x y
+ =− =
2 8
4
x
x
==
4 5
1
y
y
+ ==
Pamela's average speed is 4 miles per hour and the speed of the current is 1 mile per hour.
67. Let x = the # of mg of compound 1. Let y = the # of mg of compound 2.
Setting up the equations and solving by substitution:
0.2 0.4 40 vitamin C
0.3 0.2 30 vitamin D
x y
x y
+ = + =
Multiplying each equation by 10 yields 2 4 400
6 4 600
x y
x y
+ = + =
Subtracting the bottom equation from the top equation yields
( )2 4 6 4 400 600
2 6 200
4 200
50
x y x y
x x
x
x
+ − + = −− = −− = −
=
( )2 50 4 400
100 4 400
4 300
30075
4
y
y
y
y
+ =+ =
=
= =
So 50 mg of compound 1 should be mixed with 75 mg of compound 2.
68. Let x = the # of units of powder 1. Let y = the # of units of powder 2.
Setting up the equations and solving by substitution:
120.2 0.4 12 vitamin B
0.3 0.2 12 vitamin E
x y
x y
+ = + =
Multiplying each equation by 10 yields
2 4 120
6 4 240
x y
x y
+ = + =
Subtracting the bottom equation from the top equation yields
( )2 4 6 4 120 240
4 120
30
x y x y
x
x
+ − + = −− = −
=
( )2 30 4 120
60 4 120
4 60
6015
4
y
y
y
y
+ =+ =
=
= =
So 30 units of powder 1 should be mixed with 15 units of powder 2.
69. 2y ax bx c= + +
At (–1, 4) the equation becomes: 24 (–1) ( 1)
4
a b c
a b c
= + − += − +
At (2, 3) the equation becomes: 23 (2) (2)
3 4 2
a b c
a b c
= + += + +
At (0, 1) the equation becomes: 21 (0) (0)
1
a b c
c
= + +=
The system of equations is: 4
4 2 3
1
a b c
a b c
c
− + = + + = =
Substitute 1c = into the first and second equations and simplify:
1 4
3
3
a b
a b
a b
− + =− =
= +
4 2 1 3
4 2 2
a b
a b
+ + =+ =
Solve the first result for a, substitute into the second result and solve:
Chapter 11: Systems of Equations and Inequalities
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
4( 3) 2 2
4 12 2 2
6 10
5
3
b b
b b
b
b
+ + =+ + =
= −
= −
5 43
3 3a = − + =
The solution is 4 5
, , 13 3
a b c= = − = . The
equation is 24 51
3 3y x x= − + .
70. 2y ax bx c= + +
At (–1, –2) the equation becomes: 22 ( 1) ( 1)
2
a b c
a b c
− = − + − +− + = −
At (1, –4) the equation becomes: 24 (1) (1)
4
a b c
a b c
− = + ++ + = −
At (2, 4) the equation becomes: 24 (2) (2)
4 2 4
a b c
a b c
= + ++ + =
The system of equations is: 2
– 4
4 2 4
a b c
a b c
a b c
− + = − + + = + + =
Multiply the first equation by –1 and add to the second equation; multiply the first equation by –1 and add to the third equation to eliminate c:
2
– 4
a b c
a b c
− + − = + + =
2
4 2 4
a b c
a b c
− + − =+ + =
2 2
1
b
b
= −= −
3 3 6
2
a b
a b
+ =+ =
Substitute and solve: ( 1) 2
3
a
a
+ − ==
4
3 ( 1) 4
6
c a b= − − −= − − − −= −
The solution is 3, 1, 6a b c= = − = − . The
equation is 23 6y x x= − −
71. 0.06 5000 240
0.06 6000 900
Y r
Y r
− = + =
Multiply the first equation by 1− , the add the result to the second equation to eliminate Y.
0.06 5000 240
0.06 6000 900
Y r
Y r
− + = −+ =
11000 660
0.06
r
r
==
Substitute this result into the first equation to find Y. 0.06 5000(0.06) 240
0.06 300 240
0.06 540
9000
Y
Y
Y
Y
− =− =
==
The equilibrium level of income and interest rates is $9000 million and 6%.
72. 0.05 1000 10
0.05 800 100
Y r
Y r
− = + =
Multiply the first equation by 1− , the add the result to the second equation to eliminate Y.
0.05 1000 10
0.05 800 100
Y r
Y r
− + = −+ =
1800 90
0.05
r
r
==
Substitute this result into the first equation to find Y. 0.05 1000(0.05) 10
0.05 50 10
0.05 60
1200
Y
Y
Y
Y
− =− =
==
The equilibrium level of income and interest rates is $1200 million and 5%.
73. 2 1 3
1 2
2 3
5 3 5 0
10 5 7 0
I I I
I I
I I
= + − − = − − =
Substitute the expression for 2I into the second
and third equations and simplify:
1 1 3
1 3
5 3 5( ) 0
8 5 5
I I I
I I
− − + =− − = −
1 3 3
1 3
10 5( ) 7 0
5 12 10
I I I
I I
− + − =− − = −
Multiply both sides of the first result by 5 and multiply both sides of the second result by –8 to eliminate 1I :
Section 11.1: Systems of Linear Equations: Substitution and Elimination
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
1 3
1 3
40 25 25
40 96 80
I I
I I
− − = −+ =
3
3
71 55
55
71
I
I
=
=
Substituting and solving for the other variables:
1
1
1
1
558 5 5
71275
8 571
808
7110
71
I
I
I
I
− − = −
− − = −
− = −
=
210 55 65
71 71 71I
= + =
The solution is 1 2 310 65 55
, ,71 71 71
I I I= = = .
74. 3 1 2
3 2
1 2
8 4 6
8 4 6
I I I
I I
I I
= + = + = +
Substitute the expression for 3I into the second
equation and simplify:
1 2 2
1 2
1 2
8 4( ) 6
8 4 10
4 10 8
I I I
I I
I I
= + += +
+ =
1 2
1 2
8 4 6
8 6 4
I I
I I
= +− =
Multiply both sides of the first result by –2 and add to the second result to eliminate 1I :
1 2
1 2
8 20 16
8 6 4
I I
I I
− − = −− =
2
2
26 12
12 6
26 13
I
I
− = −−= =−
Substituting and solving for the other variables:
1
1
1
1
64 10 8
13
604 8
1344
41311
13
I
I
I
I
+ =
+ =
=
=
3 1 211 6 17
13 13 13I I I= + = + =
The solution is 1 2 311 6 17
, ,13 13 13
I I I= = = .
75. Let x = the number of orchestra seats. Let y = the number of main seats.
Let z = the number of balcony seats. Since the total number of seats is 500,
500x y z+ + = .
Since the total revenue is $17,100 if all seats are sold, 50 35 25 17,100x y z+ + = .
If only half of the orchestra seats are sold, the revenue is $14,600.
So, 1
50 35 25 14,6002
x y z + + =
.
Thus, we have the following system: 500
50 35 25 17,100
25 35 25 14,600
x y z
x y z
x y z
+ + = + + = + + =
Multiply each side of the first equation by –25 and add to the second equation to eliminate z; multiply each side of the third equation by –1 and add to the second equation to eliminate z:
25 25 25 12,500
50 35 25 17,100
25 10 4600
x y z
x y z
x y
− − − = −+ + =
+ =
50 35 25 17,100
25 35 25 14,600
x y z
x y z
+ + =− − − = −
25 2500
100
x
x
==
Substituting and solving for the other variables: 25(100) 10 4600
2500 10 4600
10 2100
210
y
y
y
y
+ =+ =
==
100 210 500
310 500
190
z
z
z
+ + =+ =
=
There are 100 orchestra seats, 210 main seats, and 190 balcony seats.
Chapter 11: Systems of Equations and Inequalities
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76. Let x = the number of adult tickets. Let y = the number of child tickets.
Let z = the number of senior citizen tickets. Since the total number of tickets is 405,
405x y z+ + = .
Since the total revenue is $2320, 8 4.50 6 2320x y z+ + = .
Twice as many children's tickets as adult tickets are sold. So, 2y x= .
Thus, we have the following system: 405
8 4.50 6 2320
2
x y z
x y z
y x
+ + = + + = =
Substitute for y in the first two equations and simplify:
(2 ) 405
3 405
x x z
x z
+ + =+ =
8 4.50(2 ) 6 2320
17 6 2320
x x z
x z
+ + =+ =
Multiply the first result by –6 and add to the second result to eliminate z:
18 6 2430
17 6 2320
x z
x z
− − = − + =
110
110
x
x
− = −=
2
2(110)
220
y x===
3 405
3(110) 405
330 405
75
x z
z
z
z
+ =+ =+ =
=
There were 110 adults, 220 children, and 75 senior citizens that bought tickets.
77. Let x = the number of servings of chicken. Let y = the number of servings of corn.
Let z = the number of servings of 2% milk.
Protein equation: 30 3 9 66x y z+ + =
Carbohydrate equation: 35 16 13 94.5x y z+ + =
Calcium equation: 200 10 300 910x y z+ + =
Multiply each side of the first equation by –16 and multiply each side of the second equation by 3 and add them to eliminate y; multiply each side of the second equation by –5 and multiply each side of the third equation by 8 and add to eliminate y:
480 48 144 1056
105 48 39 283.5
375 105 772.5
x y z
x y z
x z
− − − = −+ + =
− − = −
175 80 65 472.5
1600 80 2400 7280
1425 2335 6807.5
x y z
x y z
x z
− − − = −+ + =
+ =
Multiply each side of the first result by 19 and multiply each side of the second result by 5 to eliminate x:
7125 1995 14,677.5
7125 11,675 34,037.5
x z
x z
− − = −+ =
9680 19,360
2
z
z
==
Substituting and solving for the other variables: 375 105(2) 772.5
375 210 772.5
375 562.5
1.5
x
x
x
x
− − = −− − = −
− = −=
30(1.5) 3 9(2) 66
45 3 18 66
3 3
1
y
y
y
y
+ + =+ + =
==
The dietitian should serve 1.5 servings of chicken, 1 serving of corn, and 2 servings of 2% milk.
78. Let x = the amount in Treasury bills. Let y = the amount in Treasury bonds.
Let z = the amount in corporate bonds.
Since the total investment is $20,000, 20,000x y z+ + =
Since the total income is to be $1390, 0.05 0.07 0.10 1390x y z+ + =
The investment in Treasury bills is to be $3000 more than the investment in corporate bonds. So, 3000x z= +
Substitute for x in the first two equations and simplify: (3000 ) 20,000
2 17,000
z y z
y z
+ + + =+ =
5(3000 ) 7 10 139,000
7 15 124,000
z y z
y z
+ + + =+ =
Section 11.1: Systems of Linear Equations: Substitution and Elimination
1091
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Multiply each side of the first result by –7 and add to the second result to eliminate y:
7 14 119,000
7 15 124,000
5,000
y z
y z
z
− − = −+ =
=
3000 3000 5000 8000x z= + = + = 2 17,000
2(5000) 17,000
10,000 17,000
7000
y z
y
y
y
+ =+ =+ =
=
Kelly should invest $8000 in Treasury bills, $7000 in Treasury bonds, and $5000 in corporate bonds.
79. Let x = the price of 1 hamburger. Let y = the price of 1 order of fries.
Let z = the price of 1 drink.
We can construct the system 8 6 6 26.10
10 6 8 31.60
x y z
x y z
+ + = + + =
A system involving only 2 equations that contain 3 or more unknowns cannot be solved uniquely.
Multiply the first equation by 1
2− and the
second equation by 1
2, then add to eliminate y:
4 3 3 13.05
5 3 4 15.80
x y z
x y z
− − − = −+ + =
2.75
2.75
x z
x z
+ == −
Substitute and solve for y in terms of z: ( )5 2.75 3 4 15.80
13.75 3 15.80
3 2.05
1 41
3 60
z y z
y z
y z
y z
− + + =+ − =
= +
= +
Solutions of the system are: 2.75x z= − , 1 41
3 60y z= + .
Since we are given that 0.60 0.90z≤ ≤ , we choose values of z that give two-decimal-place values of x and y with 1.75 2.25x≤ ≤ and 0.75 1.00y≤ ≤ .
The possible values of x, y, and z are shown in the table.
x y z
2.13 0.89 0.62
2.10 0.90 0.65
2.07 0.91 0.68
2.04 0.92 0.71
2.01 0.93 0.74
1.98 0.94 0.77
1.95 0.95 0.80
1.92 0.96 0.83
1.89 0.97 0.86
1.86 0.98 0.89
80. Let x = the price of 1 hamburger.
Let y = the price of 1 order of fries.
Let z = the price of 1 drink We can construct the system
8 6 6 26.1010 6 8 31.603 2 4 10.95
x y zx y zx y z
+ + = + + = + + =
Subtract the second equation from the first equation to eliminate y: 8 6 6 26.1010 6 8 31.60
2 2 5.5
x y zx y z
x z
+ + =+ + =− − = −
Multiply the third equation by –3 and add it to the second equation to eliminate y: 10 6 8 31.609 6 12 32.85
4 1.25
x y zx y z
x z
+ + =− − − = −
− = −
Multiply the second result by 2 and add it to the first result to eliminate x:
2 2 5.52 8 2.5x zx z
− − = −− = −
10 80.8
zz
− = −=
Substitute for z to find the other variables: 4(0.8) 1.25
3.2 1.251.95
xx
x
− = −− = −
=
3(1.95) 2 4(0.8) 10.955.85 2 3.2 1.095
2 1.90.95
yy
yy
+ + =+ + =
==
Therefore, one hamburger costs $1.95, one order of fries costs $0.95, and one drink costs $0.80.
Chapter 11: Systems of Equations and Inequalities
1092
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
81. Let x = Beth’s time working alone. Let y = Bill’s time working alone.
Let z = Edie’s time working alone.
We can use the following tables to organize our work:
Beth Bill Edie
Hours to do job
Part of job done 1 1 1
in 1 hour
x y z
x y z
In 10 hours they complete 1 entire job, so
1 1 110 1
1 1 1 1
10
x y z
x y z
+ + =
+ + =
Bill Edie
Hours to do job
Part of job done 1 1
in 1 hour
y z
y z
In 15 hours they complete 1 entire job, so
1 115 1
1 1 1
15
y z
y z
+ =
+ =
.
Beth Bill Edie
Hours to do job
Part of job done 1 1 1
in 1 hour
x y z
x y z
With all 3 working for 4 hours and Beth and Bill working for an additional 8 hours, they complete
1 entire job, so 1 1 1 1 1
4 8 1
12 12 41
x y z x y
x y z
+ + + + =
+ + =
We have the system
1 1 1 1
10
1 1 1
15
12 12 41
x y z
y z
x y z
+ + =
+ =
+ + =
Subtract the second equation from the first equation:
1 1 1 1
10
1 1 1
15
x y z
y z
+ + =
+ =
1 1
3030
xx
=
=
Substitute x = 30 into the third equation: 12 12 4
130
12 4 3
5
y z
y z
+ + =
+ =.
Now consider the system consisting of the last result and the second original equation. Multiply the second original equation by –12 and add it to the last result to eliminate y:
12 12 12
15
12 4 3
5
y z
y z
− − −+ =
+ =
8 3
1540
zz
− = −
=
Plugging z = 40 to find y: 12 4 3
5
12 4 3
40 5
12 1
2
24
y z
y
y
y
+ =
+ =
=
=
Working alone, it would take Beth 30 hours, Bill 24 hours, and Edie 40 hours to complete the job.
82. – 84. Answers will vary.
Section 11.2: Systems of Linear Equations: Matrices
1093
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Section 11.2
1. matrix
2. augmented
3. third; fifth
4. True
5. Writing the augmented matrix for the system of equations:
5 5 5 514 3 6 3 64
x y
x y
− = −→ + =
6. Writing the augmented matrix for the system of equations:
3 4 7 3 744 2 5 54 2
x y
x y
+ = → − = −
7. 2 3 6 0
4 6 2 0
x y
x y
+ − = − + =
Write the system in standard form and then write the augmented matrix for the system of equations:
2 3 6 3 624 6 2 4 6 2
x y
x y
+ = → − = − − −
8. 9 0
3 4 0
x y
x y
− = − − =
Write the system in standard form and then write the augmented matrix for the system of equations:
9 0 9 013 4 3 1 4
x y
x y
− = −→ − = −
9. Writing the augmented matrix for the system of equations:
0.01 0.03 0.06 0.01 0.060.030.13 0.10 0.20 0.13 0.10 0.20
x y
x y
− = −→ + =
10. Writing the augmented matrix for the system of equations:
4 3 3 4 3 3
3 2 4 3 2 41 1 2 1 21
4 3 3 3 34
x y
x y
− = − → − + = −
11. Writing the augmented matrix for the system of equations:
10 1 1 1 10
3 3 5 3 3 0 5
2 2 1 1 2 2
x y z
x y
x y z
− + = − + = → + + =
12. Writing the augmented matrix for the system of equations:
5 0 5 1 1 0
5 1 1 0 5
2 3 2 2 0 3 2
x y z
x y
x z
− − = − − + = → − = −
13. Writing the augmented matrix for the system of equations:
2 1 1 1 2
3 2 2 3 2 0 2
5 3 1 5 3 1 1
x y z
x y
x y z
+ − = − − = → − + − = −
14.
2 3 4 0
5 2 0
2 3 2
x y z
x z
x y z
+ − = − + = + − = −
Write the system in standard form and then write the augmented matrix for the system of equations:
2 3 4 0 2 3 4 0
5 2 1 0 5 2
2 3 2 1 2 3 2
x y z
x z
x y z
+ − = − − = − → − − + − = − − −
15. Writing the augmented matrix for the system of equations:
10 1 1 1 10
2 2 1 2 1 2 1
3 4 5 3 4 0 5
4 5 0 4 5 1 0
x y z
x y z
x y
x y z
− − = − − + + = − − → − + = − − + = −
16. Writing the augmented matrix for the system of equations:
2 5 1 1 2 1 5
3 4 2 2 1 3 4 2 2
3 5 1 3 1 5 1 1
x y z w
x y z w
x y z w
− + − = − − + − + = → − − − − = − − − − −
Chapter 11: Systems of Equations and Inequalities
1094
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
17. 1 3 2 3 2
2 5 5 2 5 5
x y
x y
− − − = −→ − − =
2 1 22R r r= − +
1 3 2 1 3 2
2 5 5 2(1) 2 2( 3) 5 2( 2) 5
1 3 2
0 1 9
− − − − → − − + − − − − +
− − →
18. 1 3 3 3 3
2 5 4 2 5 4
x y
x y
− − − = −→ − − − = −
2 1 22R r r= − +
1 3 3 1 3 3
2 5 4 2(1) 2 2( 3) 5 2( 3) 4
1 3 3
0 1 2
− − − − → − − − + − − − − − −
− − →
19.
1 3 4 3 3 4 3
3 5 6 6 3 5 6 6
5 3 4 6 5 3 4 6
x y z
x y z
x y z
− − + = − → − + = − − + + =
2 1 23R r r= − +
1 3 4 33 5 6 65 3 4 6
1 3 4 33(1) 3 3( 3) 5 3(4) 6 3(3) 6
5 3 4 6
1 3 4 30 4 6 35 3 4 6
− − −
− → − + − − − − + − +
− −
→ − − −
3 1 35R r r= +
1 3 4 30 4 6 35 3 4 6
1 3 4 30 4 6 3
5(1) 5 5( 3) 3 5(4) 4 5(3) 6
1 3 4 30 4 6 30 12 24 21
− − − −
− → − −
− − + + + −
→ − −
−
20.
1 3 3 5 3 3 5
4 5 3 5 4 5 3 5
3 2 4 6 3 2 4 6
x y z
x y z
x y z
− − − + = − − − − − → − − − = − − − − − + =
2 1 24R r r= +
1 3 3 54 5 3 53 2 4 6
1 3 3 54(1) 4 4( 3) 5 4(3) 3 4( 5) 5
3 2 4 6
1 3 3 50 17 9 253 2 4 6
− − − − − − − −
− − → − − − − − −
− − − −
→ − − − −
3 1 33R r r= +
1 3 3 54 5 3 53 2 4 6
1 3 3 50 17 9 25
3(1) 3 3( 3) 2 3(3) 4 3( 5) 6
1 3 3 50 17 9 250 11 13 9
− − − − − − − −
− − → − −
− − − + − + − −
→ − −
− −
21.
1 3 2 6 3 2 6
2 5 3 4 2 5 3 4
3 6 4 6 3 6 4 6
x y z
x y z
x y z
− − − + = − − − → − + = − − − − − + =
2 1 22R r r= − +
1 3 2 62 5 3 43 6 4 6
1 3 2 62(1) 2 2( 3) 5 2(2) 3 2( 6) 4
3 6 4 6
1 3 2 60 1 1 83 6 4 6
− − − − − −
− − → − + − − − − + − − −
− − − −
→ − − −
Section 11.2: Systems of Linear Equations: Matrices
1095
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
3 1 33R r r= +
1 3 2 60 1 1 83 6 4 6
1 3 2 60 1 1 8
3(1) 3 3( 3) 6 3(2) 4 3( 6) 6
1 3 2 60 1 1 80 15 10 12
− − − − −
− − → −
− − − + − + − −
→ −
− −
22.
1 3 4 6 3 4 6
6 5 6 6 6 5 6 6
1 1 4 6 4 6
x y z
x y z
x y z
− − − − − = − − − → − + = − − − + + =
a. 2 1 26R r r= − +
1 3 4 66 5 6 61 1 4 6
1 3 4 66(1) 6 6( 3) 5 6( 4) 6 6( 6) 6
1 1 4 6
1 3 4 60 13 30 301 1 4 6
− − − − − −
− − − → − + − − − − − + − − −
− − − −
→ −
b. 3 1 3R r r= +
1 3 4 6 1 3 4 66 5 6 6 6 5 6 61 1 4 6 1 1 3 1 4 4 6 6
1 3 4 66 5 6 60 2 0 0
− − − − − − − − → − − − − − + − + − +
− − − → − −
−
23.
5 3 1 2 5 3 2
2 5 6 2 2 5 6 2
4 1 4 6 4 4 6
x y z
x y z
x y z
− − − + = − − − → − + = − − − + + =
1 2 12R r r= − +
5 3 1 22 5 6 24 1 4 6
2(2) 5 2( 5) 3 2(6) 1 2( 2) 22 5 6 24 1 4 6
1 7 11 22 5 6 24 1 4 6
− − − − −
− + − − − − + − − − → − −
− −
→ − − −
3 2 32R r r= +
1 7 11 22 5 6 24 1 4 6
1 7 11 22 5 6 2
2(2) ( 4) 2( 5) 1 2(6) 4 2( 2) 6
1 7 11 22 5 6 20 9 16 2
− − − −
− → − −
+ − − + + − + −
→ − −
−
24. 4 3 1 2 4 3 23 5 2 6 3 5 2 63 6 4 6 3 6 4 6
x y zx y zx y z
− − − − = − → − + = − − − − + =
1 2 1R r r= − +
4 3 1 23 5 2 63 6 4 6
(3) 4 ( 5) 3 (2) 1 (6) 23 5 2 63 6 4 6
1 2 3 43 5 2 63 6 4 6
− − − − −
− + − − − − − − + → −
− − − −
→ − − −
3 2 3R r r= +
1 2 3 43 5 2 63 6 4 6
1 2 3 43 5 2 6
3 ( 3) 5 ( 6) 2 4 6 6
1 2 3 43 5 2 60 11 6 12
− − − − −
− − → −
+ − − + − + + − −
→ −
−
25. 5
1
x
y
= = −
Consistent; 5, 1,x y= = − or using ordered pairs
(5, 1)− .
26. 4
0
x
y
= − =
Consistent; 4, 0,x y= − = or using ordered pairs
( 4, 0)− .
Chapter 11: Systems of Equations and Inequalities
1096
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27.
1
2
0 3
x
y
= = =
Inconsistent
28.
0
0
0 2
x
y
= = =
Inconsistent
29.
2 1
4 2
0 0
x z
y z
+ = − − = − =
Consistent; 1 2
2 4
is any real number
x z
y z
z
= − − = − +
or {( , , ) | 1 2 , 2 4 ,x y z x z y z= − − = − + z is any
real number}
30.
4 4
3 2
0 0
x z
y z
+ = + = =
Consistent; 4 4
2 3
is any real number
x z
y z
z
= − = −
or {( , , ) | 4 4 , 2 3 ,x y z x z y z= − = − z is any real
number}
31. 1
2 4
3 4
1
2
2 3
x
x x
x x
= + = + =
Consistent;
1
2 4
3 4
4
1
2
3 2
is any real number
x
x x
x x
x
= = − = −
or 1 2 3 4 1 2 4{( , , , ) | 1, 2 ,x x x x x x x= = −
3 4 43 2 , is any real number}x x x= −
32. 1
2 4
3 4
1
2 2
3 0
x
x x
x x
= + = + =
Consistent;
1
2 4
3 4
4
1
2 2
3
is any real number
x
x x
x x
x
= = − = −
or 1 2 3 4 1 2 4 3 4{( , , , ) | 1, 2 2 , 3 ,x x x x x x x x x= = − = −
4 is any real number}x
33. 1 4
2 3 4
4 2
3 3
0 0
x x
x x x
+ = + + = =
Consistent;
1 4
2 3 4
3 4
2 4
3 3
, are any real numbers
x x
x x x
x x
= − = − −
or 1 2 3 4 1 4 2 3 4{( , , , ) | 2 4 , 3 3 ,x x x x x x x x x= − = − −
3 4and are any real numbers}x x
34. 1
2
3 4
1
2
2 3
x
x
x x
= = + =
Consistent;
1
2
3 4
4
1
2
3 2
is any real number
x
x
x x
x
= = = −
or 1 2 3 4 1 2 3 4{( , , , ) | 1, 2, 3 2 ,x x x x x x x x= = = −
4 is any real number}x
35.
1 4
2 4
3 4
2
2 2
0
0 0
x x
x x
x x
+ = − + = − = =
Consistent;
1 4
2 4
3 4
4
2
2 2
is any real number
x x
x x
x x
x
= − − = − =
or 1 2 3 4 1 4 2 4{( , , , ) | 2 , 2 2 ,x x x x x x x x= − − = −
3 4 4, is any real number}x x x=
Section 11.2: Systems of Linear Equations: Matrices
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36.
1
2
3
4
1
2
3
0
x
x
x
x
= = = =
Consistent;
1
2
3
4
1
2
3
0
x
x
x
x
= = = =
or (1,2,3,0)
37. 8
4
x y
x y
+ = − =
Write the augmented matrix:
( )
( )
( )
2 1 2
12 22
1 2 1
8 81 1 1 1 01 1 4 2 4
81 1 0 1 2
0 61 0 1 2
R r r
R r
R r r
→ = − +
− − −
→ = −
→ = − +
The solution is 6, 2x y= = or using ordered
pairs (6, 2).
38. 2 5
3
x y
x y
+ = + =
Write the augmented matrix:
( )
( )
( )
2 1 2
2 2
1 2 1
5 51 2 1 2 3 01 1 1 2
51 2 0 1 2
01 1 20 1 2
R r r
R r
R r r
→ = − +
− −
→ = −
→ = − +
The solution is 1, 2x y= = or using ordered
pairs (1, 2).
39. 2 4 2
3 2 3
x y
x y
− = − + =
Write the augmented matrix:
( )
( )
( )
( )
11 12
2 1 2
12 283
4
12
1 2 134
2 1 14 2 2 3 3 3 32 2
1 12 3 0 8 6
1 12
0 1
01 2 0 1
R r
R r r
R r
R r r
−− − − =→
−− = − +→ −− =→
= + →
The solution is 1 3
,2 4
x y= = or using ordered pairs
1 3,
2 4
.
40. 3 3 3
84 2
3
x y
x y
+ = + =
Write the augmented matrix:
( )11 18 38
3 3
3 3 3 1 1 1
4 2 4 2R r
=→
( )
( )
( )
2 1 243
12 22
23
13 1 2 123
1 1 1 4 0 2
1 1 1
0 1
01 0 1
R r r
R r
R r r
= − +→ − − = −→
= − + →
The solution is 1 2
,3 3
x y= = or using ordered
pairs 1 2
,3 3
.
41. 2 4
2 4 8
x y
x y
+ = + =
Write the augmented matrix:
( )2 1 21 2 4 1 2 4 2
8 0 0 02 4R r r
= − +→
This is a dependent system. 2 4
4 2
x y
x y
+ == −
The solution is 4 2 , is any real numberx y y= −
or {( , ) | 4 2 , is any real number}x y x y y= −
Chapter 11: Systems of Equations and Inequalities
1098
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42. 3 7
9 3 21
x y
x y
− = − =
Write the augmented matrix:
( )
( )
713
1 13
713 3
2 1 2
113 71 3
9 3 21 9 3 21
-1 90 0 0
R r
R r r
− − → = − −
= − +→
This is a dependent system. 3 7
3 7
x y
x y
− =− =
The solution is 3 7, is any real numbery x x= −
or {( , ) | 3 7, is any real number}x y y x x= −
43. 2 3 6
1
2
x y
x y
+ = − =
Write the augmented matrix:
( )
( )
( )
( )
312
1 121 12 2
32
2 1 25 52 2
322 2 25
332 1 2 12
3 62 31 1 1 1 1
31 0
31 0 1 1
01 0 1 1
R r
R r r
R r
R r r
→ = − − → = − + − −
→ = −
→ = − +
The solution is 3
, 12
x y= = or 3
, 12
.
44. 1
22
2 8
x y
x y
+ = − − =
Write the augmented matrix:
( )
( )
( )
( )
12 1 1
2 1 2
12 24
1 2 1
1 2 41 2 281 281 2
1 2 4 0 124
1 2 4 0 31
01 2 20 31
R r
R r r
R r
R r r
−− → = − −
−→ = − + − −→ = −
−
→ = − + −
The solution is 2, 3x y= = − or (2, 3)− .
45. 3 5 3
15 5 21
x y
x y
− = + =
Write the augmented matrix:
( )
( )
( )
( )
513 1 13
53
2 1 2
53 1
2 23015
453
1 2 1315
3 5 3 1 1 15 5 21 15 5 21
1 1 150 30 6
1 1 0 1
01 0 1
R r
R r r
R r
R r r
− −→ = −
→ = − + − → = → = +
The solution is 4 1
,3 5
x y= = or 4 1
,3 5
.
46. 2 1
1 3
2 2
x y
x y
− = − + =
( )
( )
( )
1 112 2
1 1231 312 2 2 2
1 12 2 2 1 2
112 1 2 12
2 1 1 1 1 1
1 10 1 2
01 0 1 2
R r
R r r
R r r
− − − − → = − −→ = − +
→ = +
The solution is 1
, 22
x y= = or 1
, 22
.
47.
6
2 3 16
2 4
x y
x z
y z
− = − = + =
Write the augmented matrix:
( )
( )
2 1 2
3 12 22 2
0 61 1
0 3 162
0 2 1 4
0 61 1
0 3 22 4
0 2 1 4
0 61 1
0 1 2
0 2 1 4
R r r
R r
−
−
−
→ − = − + −
− → =
Section 11.2: Systems of Linear Equations: Matrices
1099
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( )
32
1 2 132
3 2 3
32
3 13 32 4
31 3 12
32 3 22
0 81
0 1 2 2
0 0 04
0 81
0 1 2
0 0 01
0 0 81
0 0 1 2
0 0 01
R r r
R r r
R r
R r r
R r r
− = + −→ = − + − −→ =
= + → = +
The solution is 8, 2, 0x y z= = = or (8, 2, 0).
48.
2 4
2 4 0
3 2 11
x y
y z
x z
+ = − − + = − = −
Write the augmented matrix:
( )12
11 12
02 1 4
0 042
3 0 112
201
0 0 42
3 0 112
R r
−
− −−
− → =− −−
( )
( )
( )
12
3 1 3
32
12
12 22
32
11 2 12
33 2 32
13 35
201
0 0 342
0 52
201
0 0 1 2
0 52
201 1
0 0 1 2
0 0 5 5
201 1
0 0 1 2
0 0 1 1
R r r
R r
R r r
R r r
R r
− → = − +−
− − − − → = −−
− − − − = − +
→ − = + − − −
→ = −−
1 3 1
2 3 2
0 0 31
0 0 1 22
0 0 1 1
R r r
R r r
−= − +
→ = +
The solution is 3, 2, 1x y z= − = = or ( 3, 2,1)− .
49.
2 3 7
2 4
3 2 2 10
x y z
x y z
x y z
− + = + + =− + − = −
Write the augmented matrix:
( )
2 1 2
3 1 3
12 25
1 2 1
3 2 3
3 71 2
2 1 1 4
3 102 2
3 71 22
0 5 5 10 3
0 7 114
3 71 2
0 1 1 2
0 7 114
0 31 12
0 1 1 24
0 0 3 3
R r r
R r r
R r
R r r
R r r
− − −−
−= − +
→ − − = + − −
→ =− − −
= + → − − = +
( )13 33
0 31 1
0 1 1 2
0 0 1 1
R r
→ =− −
1 3 1
2 3 2
0 01 2
0 0 1 1
0 0 1 1
R r r
R r r
= − +
→ − = +
The solution is 2, 1, 1x y z= = − = or (2, 1,1)− .
50.
2 3 0
2 2 7
3 4 3 7
x y z
x y z
x y z
+ − =− + + = − − − =
Write the augmented matrix:
( )31
2 21
1 12
312 2
2 1 2
3 1 33112 2
2 1 3 0
2 2 1 7
3 4 3 7
1 0
2 2 1 7
3 4 3 7
1 02
0 3 2 7 3
0 7
R r
R r r
R r r
− − − − −
− → − − = − − −
= + → − − = − + −
Chapter 11: Systems of Equations and Inequalities
1100
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
( )31
2 272 1
2 23 3 3311
2 2
1 0
0 1
0 7
R r
−
→ − − = −
( )
7 76 6 1
1 2 12723 3 11
3 2 3213 356 6
7 76 6
7 623 33 3 13
3513
5613 7
1 3 16713 2
2 3 235 313
1 0
0 1
0 0
1 0
0 1
0 0 1
1 0 0
0 1 0
0 0 1
R r r
R r r
R r
R r r
R r r
− = − + → − − = + − −
−
→ − − = −
= + → − = +
The solution is 56 7 35
, ,13 13 13
x y z= = − = or
56 7 35, ,
13 13 13 −
.
51.
2 2 2 2
2 3 2
3 2 0
x y z
x y z
x y
− − = + + = + =
Write the augmented matrix:
( )
( )
11 12
2 1 2
3 1 3
3 2 3
2 22 2
32 1 2
3 0 02
1 1 1 1
32 1 2
3 0 02
1 1 1 12
0 5 3 0 3
0 5 3 3
1 1 1 1
0 5 3 0
0 0 0 3
R r
R r r
R r r
R r r
− −
− −
→ = − −
= − + → = − + −
− −
→ = − + −
There is no solution. The system is inconsistent.
52.
2 3 0
2 5
3 4 1
x y z
x y z
x y z
− − =− + + = − − =
Write the augmented matrix:
( )
1 2
2 1 2
3 1 3
3 2 3
3 02 151 2 1
3 1 14
51 12Interchange
3 02 1 and
3 1 14
51 122
0 10 1 13
0 162 2
51 120 10 21 10 0 0 4
r r
R r r
R r r
R r r
− − − −−
−−− −→ − − −−
−−− = − + → = − + −−− → = − + −
There is no solution. The system is inconsistent.
53.
1
2 3 4
3 2 7 0
x y z
x y z
x y z
− + + = − − + − = − − − =
Write the augmented matrix:
( )1 1
2 1 2
3 1 3
1 2 1
3 2 3
1 1 1 1431 2
23 7 0
1 1 1 1431 2
23 7 0
1 1 1 140 3 1
340 31
20 5140 3 1
0 0 0 0
R r
R r r
R r r
R r r
R r r
− − −−− − −
− − −−→ − = − − − − −
= + −→ − = − + − − −−
= + −→ − = − +
The matrix in the last step represents the system 5 2
4 3
0 0
x z
y z
− = − − = − =
or, equivalently,
5 2
4 3
0 0
x z
y z
= − = − =
The solution is 5 2x z= − , 4 3y z= − , z is any
real number or {( , , ) | 5 2, 4 3,x y z x z y z= − = − z
is any real number}.
Section 11.2: Systems of Linear Equations: Matrices
1101
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
54.
2 3 0
3 2 2 2
5 3 2
x y z
x y z
x y z
− − = + + = + + =
Write the augmented matrix:
1 3
2 1 2
3 1 3
3 2 37 413 13 1
2 213
3 02 1
3 2 2 2
5 31 2
5 31 2Interchange
3 2 2 2 and
3 02 1
5 31 23
0 13 7 42
0 13 7 4
5 31 2
0 1
0 0 0 0
r r
R r r
R r r
R r r
R r
− −
→ − −
= − +
→ − − − = − + − − − = − + → = −
( )
6413 13
7 41 2 113 13
01
0 51
0 0 0 0
R r r
→ = − +
The matrix in the last step represents the system 4 6
13 137 4
13 130 0
x z
y z
+ = + = =
or, equivalently, 4 6
13 137 4
13 130 0
x z
y z
= − + = − + =
The solution is 4 6
13 13x z= − + ,
7 4
13 13y z= − + ,
z is any real number or 4 6
( , , ) ,13 13
x y z x z = − +
7 4, is any real number
13 13y z z
= − +
.
55.
2 2 3 6
4 3 2 0
2 3 7 1
x y z
x y z
x y z
− + = − + =− + − =
Write the augmented matrix:
2 3 62
3 04 2
2 3 7 1
−
− − −
( )32
11 12
32
2 1 2
3 1 3
52
1 2 1
3 2 3
31 1
3 04 2
2 3 7 1
31 14
40 1 122
40 71
0 91
40 1 12
0 0 0 19
R r
R r r
R r r
R r r
R r r
− −→ = − − − = − + −→ − = + − − − = + −→ − = − +
There is no solution. The system is inconsistent.
56.
3 2 2 6
7 3 2 1
2 3 4 0
x y z
x y z
x y z
− + = − + = − − + =
Write the augmented matrix:
3 622
7 3 2 1
3 042
−
− − −
( )2 23 3
11 13
1 2
7 3 2 1
3 042
R r
− → − =− −
( )
2 23 3
2 1 25 83 3
3 1 35 83 3
2 23 3
5 83 2 33 3
1 27
0 15 2
0 4
1 2
0 15
0 0 0 19
R r r
R r r
R r r
− = − + −→ − = − + − − − −→ − = +
−
There is no solution. The system is inconsistent.
Chapter 11: Systems of Equations and Inequalities
1102
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
57.
6
3 2 5
3 2 14
x y z
x y z
x y z
+ − = − + = − + − =
Write the augmented matrix:
61 1 1
23 51
231 14
−
− − −
( )
2 1 2
3 1 3
234 15 5 2 25
715 5
234 1 2 15 5
3 2 36355
61 1 13
230 5 4
0 82 1
61 1 1
0 1
0 82 1
01
0 12
0 0
R r r
R r r
R r
R r r
R r r
−= − +
−→ − = − + − −
− → = − − − = − + −→ = − + −
( )71
5 5
234 55 5 3 33
11 3 15
42 3 25
01
0 1
20 0 1
0 01 1
0 0 3 1
20 0 1
R r
R r r
R r r
− −→ =
−
= + → = + −
The solution is 1, 3, 2x y z= = = − , or (1, 3, 2)− .
58.
4
2 3 4 15
5 2 12
x y z
x y z
x y z
− + = − − + = − + − =
Write the augmented matrix:
( )
2 1 2
3 1 3
2 2
1 1 1 4
3 152 4
25 1 12
1 1 1 42
0 7 1 25
0 6 7 32
1 1 1 4
0 7 1 2
0 6 7 32
R r r
R r r
R r
− −
− − −
− −= − +
→ −− = − + − − −
→ = −− −
( )
1 2 1
3 2 3
13 35
1 3 1
2 3 2
0 31 1
0 7 1 26
0 0 5 10
0 31 1
0 7 1 2
0 0 1 2
0 01 1
0 0 3 12
0 0 1 2
R r r
R r r
R r
R r r
R r r
−= +
→ − = − + − −
→ =− −
= + → = + −
The solution is 1, 3, 2x y z= = = − or (1, 3, 2)− .
59.
2 3
2 4 7
2 2 3 4
x y z
x y z
x y z
+ − = − − + = −− + − =
Write the augmented matrix:
31 2 1
4 72 1
2 32 4
−−
− − − −
( )
2 1 2
3 1 3
3 1 18 8 2 28
31 2 12
80 3 12
20 6 5
31 2 1
0 1
20 6 5
R r r
R r r
R r
−−= − +
−→ − = + −− −−
− → = − −−
( )
1314 4
3 1 1 2 18 8
3 2 311 114 4
1314 4
3 1 48 8 3 311
11 3 11 4
2 32 3 28
012
0 16
0 0
01
0 1
0 0 1 1
0 0 31
0 0 1
0 0 1 1
R r r
R r r
R r
R r r
R r r
− − = − + −→ = − + − − − − −→ = − − = +
→ = +
The solution is 1
3, , 12
x y z= − = = or 1
3, ,12
−
.
Section 11.2: Systems of Linear Equations: Matrices
1103
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
60. 4 3 8
3 3 12
6 1
x y z
x y z
x y z
+ − = − − + = + + =
Write the augmented matrix:
31 4 8
3 31 12
61 1 1
− −
−
( )
2 1 2
3 1 3
3612 12 213 13 13
9 4013 13
1 2 1361213 13
3 2 381 913 13
31 4 83
0 13 36 12
0 3 9 9
31 4 8
0 1
0 3 9 9
014
0 13
0 0
R r r
R r r
R r
R r r
R r r
− − = − + → − = − + −
− − − −→ = −
−
= − + − −→ = +
( )9 40
13 13
3612 133 313 13 81
19
91 3 1138
3 123 3 2131
9
01
0 1
0 0 1
0 0 31
0 0 1
0 0 1
R r
R r r
R r r
− − −→ = = − +
−→ = +
The solution is 8 1
3, ,3 9
x y z= = − = or 8 1
3, ,3 9
−
.
61.
23
32 1
84 2
3
x y z
x y z
x y
+ − =
− + = + =
Write the augmented matrix:
( )
23
83
1 1 23 3 9
11 13
83
3 1 1
2 1 1 1
04 2
1
2 1 1 1
04 2
R r
− −
− → − =
( )
1 1 23 3 9
5 5 5 2 1 23 3 9
3 1 3162 43 3 9
1 1 23 3 9
1 33 2 25
162 43 3 9
13 1
1 2 11 33 2
3 2 33
12
0 4
0
1
0 1 1
0
0 01
0 1 1
0 0 2 2
R r r
R r r
R r
R r r
R r r
− = − + −→ = − + − −→ = −− = − + − → − = − +
( )
( )
13
1 13 3 32
13
23 2 3 2
0 01
0 1 1
0 0 1 1
0 01
0 0 1
0 0 1 1
R r
R r r
−→ =− → = +
The solution is 1 2
, , 13 3
x y z= = = or 1 2
, , 13 3
.
62. 83
1
2 1
2
x y
x y z
x y z
+ =
− + = + + =
Write the augmented matrix:
83
1 1 0 1
2 1 1 1
1 2 1
−
( )
2 1 2
3 1 353
53
2 3
53 2 33
1 1 0 12
0 3 1 1
0 1 1
1 1 0 1Interchange
0 1 1 and
0 3 1 1
1 1 0 1
0 1 1 3
0 0 4 4
R r r
R r r
r r
R r r
= − +
→ − − = − +
→
− −
→ = +
Chapter 11: Systems of Equations and Inequalities
1104
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
( )
( )
( )
153 343
22 2 33
13
21 1 23
1 1 0 1
0 1 1
0 0 1 1
1 1 0 1
0 1 0
0 0 1 1
1 0 0
0 1 0
0 0 1 1
R r
R r r
R r r
→ =
→ = −
→ = −
The solution is 1 2
, , 13 3
x y z= = = or 1 2
, , 13 3
.
63.
4
2 0
3 2 6
2 2 2 1
x y z w
x y z
x y z w
x y z w
+ + + = − + = + + − = − − + = −
Write the augmented matrix:
1 1 1 1 4
0 02 1 1
3 62 1 1
2 21 2 1
− −
− − −
2 1 2
3 1 3
4 1 4
1 1 1 1 4 22 80 3 1 3
2 4 60 1
0 3 3 51
R r r
R r r
R r r
= − + − −− − → = − + − − −− = − +
− − −
( )
2 3
2 2
1 2 1
3 2 3
4 2 4
1 1 1 1 4Interchange2 4 60 1
and 2 80 3 1
0 3 3 51
1 1 1 1 4
0 61 2 4 2 80 3 1
0 3 3 51
20 31 1
0 61 2 4 30 0 5 10 10 30 0 3 13 13
r r
R r
R r r
R r r
R r r
− − −− → − −− −
− − − → = − − −− −
− − − −−− = − +
→ = + = +
( )13 35
1 3 1
2 3 2
4 3 4
20 31 1
0 61 2 4 0 0 1 2 2
0 0 3 13 13
0 0 01 1
0 0 01 2 20 0 1 2 2 30 0 0 7 7
R r
R r r
R r r
R r r
−−− → = − = +
→ = − + = − +
( )14 47
1 4 1
3 4 3
0 0 01 1
0 0 01 2 0 0 1 2 2
0 0 0 1 1
0 0 01 1
0 0 01 2 20 0 0 01
0 0 0 1 1
R r
R r r
R r r
− → = = + → = − +
The solution is 1, 2, 0, 1x y z w= = = = or
(1, 2, 0, 1).
64.
4
2 0
2 3 6
2 2 2 1
x y z w
x y z
x y z w
x y z w
+ + + = − + + = + + − =− + − + = −
Write the augmented matrix:
1 1 1 1 4
0 01 2 1
3 62 1 1
1 2 12 2
− − − − −
2 1 2
3 1 3
4 1 4
2 3
1 2 1
3
1 1 1 1 4
0 3 2 1 4 20 31 1 2 20 3 0 74
1 1 1 1 4Interchange0 31 1 2
and 0 3 2 1 4
0 3 0 74
0 61 2 4
0 31 1 2 0 0 5 10 10
0 0 3 13 13
R r r
R r r
R r r
r r
R r r
R
= + → = − + −− − = +
−− − → = − +
−− − → = −
2 3
4 2 4
3
3
r r
R r r
+ = − +
Section 11.2: Systems of Linear Equations: Matrices
1105
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
( )4 3 4
14 35
14 435
0 61 2 4
0 31 1 2 3 50 0 5 10 10
0 0 0 35 35
0 61 2 4
0 31 1 2 0 0 1 2 2
0 0 0 1 1
R r r
R r
R r
−− − → = −
− − =−− − →
= −
Write the matrix as the corresponding system: 2 4 6
3 2
2 2
1
x z w
y z w
z w
w
+ + = − − = − + = =
Substitute and solve: 2(1) 2
2 2
0
z
z
z
+ =+ =
=
0 3(1) 2
3 2
1
y
y
y
− − = −− = −
=
2(0) 4(1) 6
0 4 6
2
x
x
x
+ + =+ + =
=
The solution is 2x = , 1y = , 0z = , 1w = or
(2, 1, 0, 1).
65.
2 1
2 2 2
3 3 3
x y z
x y z
x y z
+ + = − + = + + =
Write the augmented matrix:
1 2 1 1
2 1 2 2
3 3 31
−
( )
2 1 2
3 1 3
3 2 3
1 2 1 12
0 5 0 0 3
0 5 0 0
1 2 1 1
0 5 0 0
0 0 0 0
R r r
R r r
R r r
= − +
→ − = − + −
→ − = − +
The matrix in the last step represents the system
2 1
5 0
0 0
x y z
y
+ + = − = =
Substitute and solve: 5 0
0
y
y
− ==
2(0) 1
1
x z
z x
+ + == −
The solution is 0, 1 ,y z x= = − x is any real
number or {( , , ) | 0, 1 ,x y z y z x= = − x is any
real number}.
66.
2 3
2 2 6
3 3 4
x y z
x y z
x y z
+ − = − + = − + =
Write the augmented matrix:
31 2 1
62 1 2
3 31 4
− − −
( )
2 1 2
3 1 3
3 2 3
31 2 12
0 5 0 4
0 5 4 1
31 2 1
0 5 0 4
0 0 0 1
R r r
R r r
R r r
−= − + → − = − + −
− → − = − +
There is no solution. The system is inconsistent.
67. 5
3 2 2 0
x y z
x y z
− + = + − =
Write the augmented matrix:
1 1 1 5
3 2 2 0
− −
( )
( )
( )
2 1 2
12 25
1 2 1
1 1 1 5 3
0 5 5 15
1 1 1 5
0 1 1 3
1 0 0 2
0 1 1 3
R r r
R r
R r r
− → = − + − −
− → = − −
→ = + − −
The matrix in the last step represents the system 2
3
x
y z
= − = −
or, equivalently, 2
3
x
y z
= = −
Thus, the solution is 2x = , 3y z= − , z is any
real number or {( , , ) | 2, 3,x y z x y z= = − z is
any real number}.
Chapter 11: Systems of Equations and Inequalities
1106
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
68. 2 4
3 1
x y z
x y z
+ − =− + + =
Write the augmented matrix:
2 1 1 4
1 1 3 1
− −
( )
( )
( )
1 2
2 1 2
12 25 3
3
43
1 2 153
interchange1 1 3 1
and 2 1 1 4
1 1 3 1 2
0 3 5 6
1 1 3 1
0 1 2
1 0 1
20 1
r r
R r r
R r
R r r
− − − → −− − − −
→ = − + − − −
→ = −
→ = +
The matrix in the last step represents the system 4
13 5
23
x z
y z
− = + =
or, equivalently,
41
3 5
23
x z
y z
= + = −
Thus, the solution is: 4
13
x z= + , 5
23
y z= − , z
is any real number or 4
( , , ) 1 ,3
x y z x z = +
52 , is any real number
3y z z
= −
.
69.
2 3 3
0
0
3 5
x y z
x y z
x y z
x y z
+ − = − − = − + + = + + =
Write the augmented matrix:
2 3 1 3
1 1 1 0
1 1 1 0
1 1 3 5
− − − −
1 2
2 1 2
3 1 3
4 1 4
1 1 1 0interchange2 3 1 3
and 1 1 1 0
1 1 3 5
1 1 1 02
0 5 1 3
0 0 0 0
0 2 4 5
r r
R r r
R r r
R r r
− − − → − − −
= − + → = + = − +
3 4
1 1 1 0
interchange0 5 1 3
and 0 2 4 5
0 0 0 0
r r
− − →
( )
( )
2 3 2
1 2 1
3 2 3
13 319 18
18
1 1 1 0
0 1 7 7 2
0 2 4 5
0 0 0 0
1 0 8 7
0 1 7 7
20 0 18 19
0 0 0 0
71 0 870 1 7
=0 0 1
0 0 0 0
R r r
R r r
R r r
R r
− − − − → = − + − − = +− − → = − +
− − −− →
The matrix in the last step represents the system 8 7
7 7
19
18
x z
y z
z
− = − − = − =
Substitute and solve: 19
7 718
7
18
y
y
− =
=
198 7
18
13
9
x
x
− = −
=
Thus, the solution is 13
9x = ,
7
18y = ,
19
18z = or
13 7 19, ,
9 18 18
.
70.
3 1
2 4 0
3 2 1
2 5
x y z
x y z
x y z
x y
− + = − − = − + = − =
Write the augmented matrix:
1 3 1 1
2 1 4 0
1 3 2 1
1 2 0 5
− − − −
−
Section 11.2: Systems of Linear Equations: Matrices
1107
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
2 1 2
3 1 2
4 1 2
2 4
1 2 1
4 2
1 3 1 12
0 5 6 2
0 0 1 0
0 1 1 4
1 3 1 1
interchange0 1 1 4
and 0 0 1 0
0 5 6 2
1 0 2 13
30 1 1 4
0 0 1 0 5
0 0 1 22
R r r
R r r
R r r
r r
R r r
R r
− = + − − → = − + = − + −
− − →
− − − = +− → = −
− −
4r
+
1 3 1
2 3 2
4 3 4
1 0 0 132
0 1 0 4
0 0 1 0
0 0 0 22
R r r
R r r
R r r
= +
→ = + = + −
There is no solution. The system is inconsistent.
71. 4 4
2 3 3
x y z w
x y z w
+ + − = − + + =
Write the augmented matrix:
1 2
4 1 1 1 4
1 1 2 3 3
interchange1 1 2 3 3
and 4 1 1 1 4 r r
− −
− → −
( )2 1 2
1 1 2 3 3 4
0 5 7 13 8R r r
− → = − + − − −
The matrix in the last step represents the system 2 3 3
5 7 13 8
x y z w
y z w
− + + = − − = −
The second equation yields 5 7 13 8
5 7 13 8
7 13 8
5 5 5
y z w
y z w
y z w
− − = −= + −
= + −
The first equation yields 2 3 3
3 2 3
x y z w
x y z w
− + + == + − −
Substituting for y: 8 7 13
3 2 35 5 5
3 2 7
5 5 5
x z w z w
x z w
= + − + + − −
= − − +
Thus, the solution is 3 2 7
5 5 5x z w= − − + ,
7 13 8
5 5 5y z w= + − , z and w are any real numbers or
3 2 7 7 13 8( , , , ) , ,
5 5 5 5 5 5x y z w x z w y z w
= − − + = + −
and are any real numbersz w
.
72. 4 5
2 5
4
x y
x y z w
z w
− + = − + − = + =
Write the augmented matrix:
4 1 0 0 52 1 1 1 50 0 1 1 4
− − −
( )
( )
( )
514 4
11 14
5144
1512 1 22 2
514 4
2 2
1 0 0
2 1 1 1 5 0 0 1 1 4
1 0 0
0 1 1 2
0 0 1 1 4
1 0 0
0 1 2 2 15 20 0 1 1 4
R r
R r r
R r
− −
→ − − = − −−
→ − − = − + − −
→ − − = −
( )1 12 2
11 2 14
1 0 50 1 2 2 15 0 0 1 1 4
R r r
− −
→ − − = +
11 3 12
2 3 2
1 0 0 1 30 1 0 4 7
20 0 1 1 4
R r r
R r r
− = + → − = +
The matrix in the last step represents the system 3
4 7 4
x wy w
z w
+ = − + = − + =
or, equivalently, 37 4
4
x wy wz w
= − − = − − = −
The solution is 3x w= − − , 7 4y w= − − , 4z w= − ,
w is any real number or {( , , , ) | 3 ,x y z w x w= − −
7 4 , 4 , is any real number}y w z w w= − − = −
Chapter 11: Systems of Equations and Inequalities
1108
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
73. Each of the points must satisfy the equation 2y ax bx c= + + .
(1,2) : 2
( 2, 7) : 7 4 2
(2, 3) : 3 4 2
a b c
a b c
a b c
= + +− − − = − +
− − = + +
Set up a matrix and solve:
( )
2 1 2
3 1 3
1512 262 2
1 12 2
1 2 1512 2
3 2 3
1 1 1 2
2 74 1
34 2 1
1 1 1 2 40 3 15 6 4
20 3 11
1 1 1 2
0 1
20 3 11
01
0 1 22 60 0
R r r
R r r
R r
R r r
R r r
− − −
= − +
→ − −− = − + − − −
= − → − − − − = − + → = + − −
( )1 12 2
11 53 322 2
11 3 12
12 3 22
1 0
0 1
0 0 1 3
1 0 0 2
0 1 0 1
0 0 1 3
R r
R r r
R r r
−
→ → = −
− = − + → = − +
The solution is 2, 1, 3a b c= − = = ; so the
equation is 22 3y x x= − + + .
74. Each of the points must satisfy the equation 2y ax bx c= + + .
(1, 1) : 1
(3, 1) : 1 9 3
(– 2,14) : 14 4 2
a b c
a b c
a b c
− − = + +− − = + +
= − +
Set up a matrix and solve:
2 1 2
3 1 3
1 1 1 1
9 3 1 1
4 1 142
1 1 1 1 90 8 6 8 40 3 186
R r r
R r r
−
− −
− = − + → − − = − + −−
12 264 4
3 33 2 3
1 13 3
1 2 14 43 3 1
3 35
11 3 13
41 3 23
1 1 1 1
0 1
0 0 5 10
01
0 1
0 0 1 2
0 01 1
0 0 1 4
0 0 1 2
R r
R r r
R r r
R r
R r r
R r r
− = −
− → = − + − = − + −→ =
= + → − = − +
The solution is 1, – 4, 2a b c= = = ; so the
equation is 2 4 2y x x= − + .
75. Each of the points must satisfy the equation 3 2( )f x ax bx cx d= + + + .
( 3) 112 : 27 9 3 112
( 1) 2 : 2
(1) 4 : 4
(2) 13 : 8 4 2 13
f a b c d
f a b c d
f a b c d
f a b c d
− = − − + − + = −− = − − + − + = −
= + + + == + + + =
Set up a matrix and solve:
27 9 3 1 112
21 1 1 1
1 1 1 1 4
8 134 2 1
− − −
−− −
3 1
1 1 1 1 4Interchange21 1 1 1
and 27 9 3 1 112
8 134 2 1
r r
−− − → − − −
2 1 2
3 1 3
4 1 4
1 1 1 1 4
0 02 2 2 2740 36 2824 8
4 60 7 19
R r r
R r r
R r r
= + → = + − = − +
− − − −
( )12 22
1 2 1
3 2 3
4 2 4
1 1 1 1 4
0 01 1 1 40 36 2824
4 60 7 19
0 0 31 1
0 01 1 1 368 400 0 24 4
60 0 3 15
R r
R r r
R r r
R r r
→ = −
− − − − = − +
→ = − + − − = + − − −
Section 11.2: Systems of Linear Equations: Matrices
1109
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
( )13 3245 5
3 3
0 0 31 1
0 01 1 1
0 0 1
60 0 3 15
R r
→ = − − − − −
( )
1 143 3
1 3 1
51 4 3 43 3
1 143 3
14 4551
3 3
11 4 13
2 4 2
13 4 33
0 01
0 01 1 1 60 0 1
250 0 0 5
0 01
0 01 1 1 0 0 1
0 0 0 51
0 0 0 31
40 0 01 0 0 0 01
0 0 0 51
R r r
R r r
R r
R r r
R r r
R r r
= − + → = + − − −− → = − − − = − + − → = − + = +
The solution is 3, 4, 0, 5a b c d= = − = = ; so the
equation is 3 2( ) 3 4 5f x x x= − + .
76. Each of the points must satisfy the equation 3 2( )f x ax bx cx d= + + + .
( 2) 10 : 8 4 2 10
( 1) 3 : 3
(1) 5 : 5
(3) 15 : 27 9 3 15
f a b c d
f a b c d
f a b c d
f a b c d
− = − − + − + = −− = − + − + =
= + + + == + + + =
Set up a matrix and solve:
3 1
2 1 2
3 1 3
4 1 4
104 18 231 1 1 1
51 1 1 1
27 9 3 151
51 1 1 1Interchange31 1 1 1
and 104 18 227 9 3 151
51 1 1 1
0 0 82 2 80 6 9 3012 270 18 12024 26
r r
R r r
R r r
R r r
−− −
− −
− − → − − − = +
→ = + = − + − − − −
( )12 22
51 1 1 1
0 01 1 4 0 6 9 3012
0 18 12024 26
R r
→ =
− − − −
( )
1 2 1
3 2 3
4 2 4
13 361
2
0 01 1 1
0 01 1 4 120 0 6 3 18
180 0 24 8 48
0 01 1 1
0 01 1 4
0 0 31
0 0 24 8 48
R r r
R r r
R r r
R r
= − + → = − + − − = +
− − − → = − − − − −
( )
12
1 3 1
1 4 3 42
12
14 420
12
11 4 12
2 4 2
13 4 32
0 01 4
0 01 1 4 24
0 0 31
0 0 0 12020
0 01 4
0 01 1 4
0 0 31
0 0 0 61
0 0 01 1
0 0 01 2 0 0 0 01
0 0 0 61
R r r
R r r
R r
R r r
R r r
R r r
= − + → = + − −
− −
→ = − − − = − + − → = − +
= +
The solution is 1, 2, 0, 6a b c d= = − = = ; so the
equation is 3 2( ) 2 6f x x x= − + .
77. Let x = the number of servings of salmon steak. Let y = the number of servings of baked eggs.
Let z = the number of servings of acorn squash. Protein equation: 30 15 3 78x y z+ + =
Carbohydrate equation: 20 2 25 59x y z+ + =
Vitamin A equation: 2 20 32 75x y z+ + =
Set up a matrix and solve:
( )
3 1
11 12
30 15 3 78
20 25 592
20 32 752
20 32 752Interchange
20 25 59 2 and r
30 15 3 78
10 16 37.51
20 25 59 2
30 15 3 78
r
R r
→
→ =
2 1 2
3 1 3
10 16 37.5120
0 198 295 691 30
0 285 477 1047
R r r
R r r
= − +
→ − − − = − + − − −
Chapter 11: Systems of Equations and Inequalities
1110
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
( )
( )
953 2 366
3457 345766 66
663 33457
10 16 37.51
0 198 295 691
0 0
10 16 37.51
0 198 295 691
0 0 1 1
R r r
R r
→ − − − = − + − −
→ − − − = −
Substitute 1z = and solve: 198 295(1) 691
198 396
2
y
y
y
− − = −− = −
=
10(2) 16(1) 37.5
36 37.5
1.5
x
x
x
+ + =+ =
=
The dietitian should serve 1.5 servings of salmon steak, 2 servings of baked eggs, and 1 serving of acorn squash.
78. Let x = the number of servings of pork chops. Let y = the number of servings of corn on the cob.
Let z = the number of servings of 2% milk. Protein equation: 23 3 9 47x y z+ + =
Carbohydrate equation: 16 13 58y z+ =
Calcium equation: 10 10 300 630x y z+ + =
Set up a matrix and solve:
13 110
3 1 313 2916 8 1
2 216
467 47516 8
13 2916
23 3 9 47
0 16 13 58
10 10 300 630
30 631 1 Interchange0 16 13 58
and 23 3 9 47
30 631 123
0 1
0 140220 681
1 0
0 1
r r
R r r
R r
→ = − + →
= − − −
→
( )
1 2 18
3 2 32659
4
467 47516 8
13 29 43 316 8 2659
4671 3 116
132 3 216
20
0 0 1402
1 0
0 1
0 0 1 2
1 0 0 1
0 1 0 2
0 0 1 2
R r r
R r r
R r
R r r
R r r
= − + = + − − → = − = − + → = − +
The dietitian should provide 1 serving of pork chops, 2 servings of corn on the cob, and 2 servings of 2% milk.
79. Let x = the amount invested in Treasury bills. Let y = the amount invested in Treasury bonds.
Let z = the amount invested in corporate bonds. Total investment equation: 10,000x y z+ + =
Annual income equation: 0.06 0.07 0.08 680x y z+ + =
Condition on investment equation: 0.5
2 0
z x
x z
=− =
Set up a matrix and solve: 10,0001 1 1
0.06 0.07 0.08 680
0 021
−
( )
2 1 2
3 1 3
2 2
10,0001 1 10.06
0 0.01 0.02 80 10,0000 31
10,0001 1 1
0 8000 1001 210,0000 31
R r r
R r r
R r
= − +
→ = − + −−−
→ = −−−
( )
1 2 1
3 2 3
3 3
1 3 1
2 3 2
0 20001 1
0 8000 1 2
0 0 20001
0 20001 1
0 8000 1 2
0 0 20001
0 0 40001
0 0 4000 12
0 0 20001
R r r
R r r
R r
R r r
R r r
−= − +
→ = + −− −
→ = −
= + → = − +
Carletta should invest $4000 in Treasury bills, $4000 in Treasury bonds, and $2000 in corporate bonds.
80. Let x = the fixed delivery charge; let y = the cost of each tree, and let z = the hourly labor charge. 1st subdivision: 250 166 7520x y z+ + =
2nd subdivision: 200 124 5945x y z+ + =
3rd subdivision: 300 200 8985x y z+ + =
Set up a matrix and solve:
250 166 75201
200 59451 124
300 200 89851
Section 11.2: Systems of Linear Equations: Matrices
1111
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
2 1 2
3 3 1
12 250
13 350
250 166 75201
0 50 1575 42
0 50 34 1465
250 166 75201
0 0.84 31.5 1
0 0.68 29.31
R r r
R r r
R r
R r
= −
→ = − = → =
( )
1 1 2
3 2 3
13 30.16
1 1 3
2 2 3
0 3551 44250
0 0.84 31.5 1
0 0 0.18 2.2
0 3551 44
0 0.84 31.5 1
0 0 13.751
0 0 250144
0 0 19.95 10.84
0 0 13.751
R r r
R r r
R r
R r r
R r r
−−= −
→ = − −−
→ =
= + → = −
The delivery charge is $250 per job, the cost for each tree is $19.95, and the hourly labor charge is $13.75.
81. Let x = the number of Deltas produced. Let y = the number of Betas produced.
Let z = the number of Sigmas produced. Painting equation: 10 16 8 240x y z+ + =
Drying equation: 3 5 2 69x y z+ + =
Polishing equation: 2 3 41x y z+ + =
Set up a matrix and solve:
( )
( )
1 2 1
2 1 2
3 1 3
12 22
10 16 8 240
3 5 2 69
2 3 1 41
1 1 2 33
3 5 2 69 3
2 3 1 41
1 1 2 333
0 2 4 30 2
0 1 3 25
1 1 2 33
0 1 2 15
0 1 3 25
1 0 4 48
0 1 2 15
0 0 1 10
R r r
R r r
R r r
R r
→ = − +
= − + → − − = − + − − → − − = − − → − −
− −
1 1 2
3 3 2
R r r
R r r
= − = −
( )3 3
1 3 1
2 3 2
1 0 4 48
0 1 2 15
0 0 1 10
1 0 0 84
0 1 0 5 2
0 0 1 10
R r
R r r
R r r
→ − − = −
= − + → = +
The company should produce 8 Deltas, 5 Betas, and 10 Sigmas.
82. Let x = the number of cases of orange juice produced; let y = the number of cases of
grapefruit juice produced; and let z = the number of cases of tomato juice produced. Sterilizing equation: 9 10 12 398x y z+ + =
Filling equation: 6 4 4 164x y z+ + =
Labeling equation: 2 58x y z+ + =
Set up a matrix and solve:
( )
1 3
2 1 2
3 1 3
1 12 24 8
9 10 39812
6 1644 4
581 2 1
581 2 1Interchange
6 164 4 4 and
9 10 39812
581 2 16
0 184 8 29
0 3 1248
581 2 1
0 23 1
0 3 1248
r r
R r r
R r r
R r
→
= − + → −− − = − + −−
→ = − − −
( )
12
1 2 114
3 2 3
12
1 13 34 5
01 122
0 23 18
0 0 5 60
01 12
0 23 1
0 0 1 12
R r r
R r r
R r
= − + → = + → =
11 3 12
12 3 24
0 0 61
0 0 20 1
0 0 1 12
R r r
R r r
= − + → = − +
The company should prepare 6 cases of orange juice, 20 cases of grapefruit juice, and 12 cases of tomato juice.
Chapter 11: Systems of Equations and Inequalities
1112
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
83. Rewrite the system to set up the matrix and solve:
2 2
4 1 1 4
3 1 1 3
3 4 1 1 3 4
4 8 2 0 2 4
8 5 5 8
4 3 3 4
0
I I
I I I I
I I I I
I I I I I I
− + − = = = + + = → = + + = + = − − =
2 1
12 22
3 1 3
4 1 4
0 0 02 4
0 0 5 81
0 3 01 4
0 01 1 1
0 0 5 81Interchange0 0 02 4
and 0 3 01 4
0 01 1 1
0 0 5 81
0 0 01 2 40 0 3 5
6 80 0 1
r r
R r
R r r
R r r
− −
→ − − = → = − + −− = − + − − −
( )
3 4
3 3
4 3 4
14 423
2823
0 0 5 81Interchange0 0 01 2
and 6 80 0 1
40 0 3 5
0 0 5 81
0 0 01 2 30 0 6 81
23 280 0 0
0 0 5 81
0 0 01 2
0 0 6 81
0 0 0 1
r r
R r
R r r
R r
→ − −−
−− = − → = − +
− − → = −
4423
1 4 116
3 4 323
2823
0 0 0150 0 01 2
60 0 01
0 0 0 1
R r r
R r r
= − + → = − +
The solution is 144
23I = , 2 2I = , 3
16
23I = ,
428
23I = .
84. Rewrite the system to set up the matrix and solve:
1 3 2 1 2 3
1 3 1 3
1 2 1 2
0
24 6 3 0 6 3 24
12 24 6 6 0 6 6 36
I I I I I I
I I I I
I I I I
= + − − = − − = → − − = − + − − = − − = −
( )
2 1 2
3 1 3
3 12 22 6
12
1 2 132
3 2 3
12
1 1 1 0
6 0 3 24
6 6 0 36
1 1 1 06
0 6 9 24 6
0 12 6 36
1 1 1 0
0 1 4
0 12 6 36
1 0 4
0 1 4 12
0 0 12 12
1 0 4
0 1
R r r
R r r
R r
R r r
R r r
− − − − − − − −
− − = + → − − − = + − − −
− − → = − − − −
= + → = +
→ ( )3 13 32 12
11 3 12
32 3 22
4
0 0 1 1
1 0 0 3.5
0 1 0 2.5
0 0 1 1
R r
R r r
R r r
= = − + → = − +
The solution is 1 2 33.5, 2.5, 1I I I= = = .
85. Let x = the amount invested in Treasury bills. Let y = the amount invested in corporate bonds.
Let z = the amount invested in junk bonds.
a. Total investment equation: 20,000x y z+ + =
Annual income equation: 0.07 0.09 0.11 2000x y z+ + =
Set up a matrix and solve:
( )
( )
( )
2 2
2 2 1
12 22
1 1 1 20,000
0.07 0.09 0.11 2000
1 1 1 20,000 100
7 9 11 200,000
1 1 1 20,000 7
0 2 4 60,000
1 1 1 20,000
0 1 2 30,000
R r
R r r
R r
→ =
→ = −
→ =
Section 11.2: Systems of Linear Equations: Matrices
1113
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
( )1 1 2
1 0 1 10,000
0 1 2 30,000R r r
− − → = −
The matrix in the last step represents the
system 10,000
2 30,000
x z
y z
− = − + =
Therefore the solution is 10,000x z= − + , 30,000 2y z= − , z is any real number.
Possible investment strategies:
Amount Invested At
7% 9% 11%
0 10,000 10,000
1000 8000 11,000
2000 6000 12,000
3000 4000 13,000
4000 2000 14,000
5000 0 15,000
b. Total investment equation:
25,000x y z+ + =
Annual income equation: 0.07 0.09 0.11 2000x y z+ + =
Set up a matrix and solve:
( )
( )
( )
( )
2 2
2 2 1
12 22
1 1 2
1 1 1 25,000
0.07 0.09 0.11 2000
1 1 1 25,000 100
7 9 11 200,000
1 1 1 25,000 7
0 2 4 25,000
1 1 1 25,000
0 1 2 12,500
1 0 112,500
0 1 2 12,500
R r
R r r
R r
R r r
→ =
→ = −
→ = −
→ = −
The matrix in the last step represents the
system 12,500
2 12,500
x z
y z
− = + =
Thus, the solution is 12,500x z= + , 2 12,500y z= − + , z is any real number.
Possible investment strategies:
Amount Invested At
7% 9% 11%
12,500 12,500 0
14,500 8500 2000
16,500 4500 4000
18,750 0 6250
c. Total investment equation:
30,000x y z+ + =
Annual income equation: 0.07 0.09 0.11 2000x y z+ + =
Set up a matrix and solve:
( )
( )
( )
( )
2 2
1 2 1
12 22
1 1 2
1 1 1 30,000
0.07 0.09 0.11 2000
1 1 1 30,000 100
7 9 11 200,000
1 1 1 30,000 7
0 2 4 10,000
1 1 1 30,000
0 1 2 5000
1 0 1 35,000
0 1 2 5000
R r
R r r
R r
R r r
→ =
→ = − −
→ = − −
→ = − −
The matrix in the last step represents the
system 35,000
2 5000
x z
y z
− = + = −
Thus, the solution is 35,000x z= + , 2 5000y z= − − , z is any real number.
However, y and z cannot be negative. From 2 5000y z= − − , we must have 0.y z= =
One possible investment strategy
Amount Invested At
7% 9% 11%
30,000 0 0
This will yield ($30,000)(0.07) = $2100,
which is more than the required income.
d. Answers will vary.
Chapter 11: Systems of Equations and Inequalities
1114
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
86. Let x = the amount invested in Treasury bills. Let y = the amount invested in corporate bonds.
Let z = the amount invested in junk bonds. Let I = income Total investment equation: 25,000x y z+ + =
Annual income equation: 0.07 0.09 0.11x y z I+ + =
Set up a matrix and solve:
( )
( )
( )
( )
2 2
1 2 1
12 22
1 1 2
1 1 1 25,000
0.07 0.09 0.11
1 1 1 25,000 100
7 9 11 100
1 1 1 25,000 7
0 2 4 100 175,000
1 1 1 25,000
0 1 2 50 87,500
1 0 1112,500 50
0 1 2 50 87,500
I
R rI
R r rI
R rI
IR r r
I
→ =
→ = − −
→ = − − −
→ = − −
The matrix in the last step represents the system 112,500 50
2 50 87,500
x z I
y z I
− = − + = −
Thus, the solution is 112,500 50x I z= − + , 50 87,500 2y I z= − − , z is any real number.
a. 1500I =
( )112,500 50
112,500 50 1500
37,500
x I z
z
z
= − += − += +
( )50 87,500 2
50 1500 87,500 2
12,500 2
y I z
z
z
= − −= − −= − −
z is any real number. Since y and z cannot be negative, we must have 0.y z= = Investing all of the money
at 7% yields $1750, which is more than the $1500 needed.
b. 2000I =
( )112,500 50
112,500 50 2000
12,500
x T z
z
z
= − += − += +
( )50 87,500 2
50 2000 87,500 2
12,500 2
y I z
z
z
= − −= − −= −
z is any real number.
Possible investment strategies:
Amount Invested At
7% 9% 11%
12,500 12,500 0
15,500 6500 3000
18,750 0 6250
c. 2500I =
( )112,500 50
112,500 50 2500
12,500
x T z
z
z
= − += − += − +
( )50 87,500 2
50 2500 87,500 2
37,500 2
y I z
z
z
= − −= − −= −
z is any real number.
Possible investment strategies:
Amount invested at
7% 9% 11%
0 12,500 12,500
1000 10,500 13,500
6250 0 18,750
d. Answers will vary. 87. Let x = the amount of supplement 1.
Let y = the amount of supplement 2.
Let z = the amount of supplement 3. 0.20 0.40 0.30 40 Vitamin C
0.30 0.20 0.50 30 Vitamin D
x y z
x y z
+ + = + + =
Multiplying each equation by 10 yields 2 4 3 400
3 2 5 300
x y z
x y z
+ + = + + =
Set up a matrix and solve:
( )
( )
( )
312
1 12
32
2 2 112
312
2 21 48
2 4 3 400
3 2 5 300
1 2 200
3 2 5 300
1 2 200 3
0 4 300
1 2 200
0 1 75
R r
R r r
R r
→ =
→ = −
− −
→ = − −
Section 11.3: Systems of Linear Equations: Determinants
1115
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
( )74
1 1 218
1 0 50 2
0 1 75R r r
→ = −
−
The matrix in the last step represents the system
74
18
50
75
x z
y z
+ =
− =
Therefore the solution is 7
504
x z= − ,
175
8y z= + , z is any real number.
Possible combinations:
Supplement 1 Supplement 2 Supplement 3
50mg 75mg 0mg
36mg 76mg 8mg
22mg 77mg 16mg
8mg 78mg 24mg
88. Let x = the amount of powder 1.
Let y = the amount of powder 2.
Let z = the amount of powder 3.
120.20 0.40 0.30 12 Vitamin B
0.30 0.20 0.40 12 Vitamin E
x y z
x y z
+ + = + + =
Multiplying each equation by 10 yields 2 4 3 120
3 2 4 120
x y z
x y z
+ + = + + =
Set up a matrix and solve:
( )
( )
32 2 12
1 1 2
2 4 3 120
3 2 4 120
2 4 3 120
0 4 0.5 60
2 0 2.5 60
0 4 0.5 60
R r r
R r r
→ = − − − −
→ = + − − −
The matrix in the last step represents the system 2 2.5 60
4 0.5 60
x z
y z
+ =− − = −
Thus, the solution is 30 1.25x z= − , 15 0.125y z= − , z is any real number.
Possible combinations:
Powder 1 Powder 2 Powder 3
30 units 15 units 0 units
20 units 14 units 8 units
10 units 13 units 16 units
0 units 12 units 24 units
89 – 91. Answers will vary. Section 11.3
1. ad bc−
2. 5 3
3 4− −
3. False; If ad=bc, the the det = 0.
4. False; The solution cannot be determined.
5. False; See Thm (15)
6. False; See Thm (14)
7. 6 4 6(3) ( 1)(4) 18 4 2231
= − − = + =−
8. 8 3 8(2) 4( 3) 16 12 284 2
− = − − = + =
9. 3 1 3(2) 4( 1) 6 4 24 2
− − = − − − = − + = −
10. 24 4(3) ( 5)(2) 12 10 25 3
− = − − − = − + = −−
11.
( ) [ ][ ]
3 4 25 51 1 1 151 1 3 4 22 22 1 1 221 2
3 1 ( 2) 2(5) 4 1( 2) 1(5)
2 1(2) 1( 1)
3( 8) 4( 7) 2(3)
24 28 6
10
− −− = − +− −−
= − − − − − − + − −
= − − − += − + +=
Chapter 11: Systems of Equations and Inequalities
1116
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
12.
[ ] [ ][ ]
31 25 6 5 61 16 5 1 3 ( 2)13 8 3 82 28 32
1 1(3) 2( 5) 3 6(3) 8( 5)
2 6(2) 8(1)
1(13) 3(58) 2(4)
13 174 8
169
−− −− = − + −
= − − − − −
− −= − −= − −= −
13.
[ ] [ ][ ]
4 1 20 6 0 61 16 0 4 ( 1) 21
3 34 1 4 131 4
4 1(4) 0( 3) 1 6(4) 1(0)
2 6( 3) 1( 1)
4( 4) 1(24) 2( 17)
16 24 34
26
−− −= − − +−− −−
= − − − + −
+ − − −= − + + −= − + −= −
14.
[ ] [ ][ ]
3 9 40 04 1 1 401 4 3 ( 9) 4
3 8 8 31 18 3 1
3 4(1) ( 3)(0) 9 1(1) 8(0)
4 1( 3) 8(4)
3(4) 9(1) 4( 35)
12 9 140
119
−= − − +
− −−
= − − + −
+ − −= + + −= + −= −
15. 8
4
x y
x y
+ = − =
1 1 1 1 21 1
8 1 8 4 124 1
81 4 8 41 4
x
y
D
D
D
= = − − = −−
= = − − = −−
= = − = −
Find the solutions by Cramer's Rule:
12 46 2
2 2yx
DDx y
D D
− −= = = = = =− −
The solution is (6, 2).
16. 2 5
3
x y
x y
+ = − =
1 2 1 2 31 1
5 2 5 6 113 1
51 3 5 231
x
y
D
D
D
= = − − = −−
= = − − = −−
= = − = −
Find the solutions by Cramer's Rule:
11 11 2 2
3 3 3 3yx
DDx y
D D
− −= = = = = =− −
The solution is 11 2
,3 3
.
17. 5 13
2 3 12
x y
x y
− = + =
5 1 15 2 1732
13 1 39 12 51312
5 13 60 26 342 12
x
y
D
D
D
−= = + =
−= = + =
= = − =
Find the solutions by Cramer's Rule:
51 343 2
17 17yx
DDx y
D D= = = = = =
The solution is (3, 2).
18. 3 5
2 3 8
x y
x y
+ = − = −
31 3 6 932
5 3 15 ( 24) 98 3
51 8 10 1882
x
y
D
D
D
= = − − = −−
= = − − − =− −
= = − − = −−
Find the solutions by Cramer's Rule:
9 181 2
9 9yx
DDx y
D D
−= = = − = = =− −
The solution is ( 1, 2)− .
Section 11.3: Systems of Linear Equations: Determinants
1117
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
19. 3 24
2 0
x
x y
= + =
3 0 6 0 61 2
024 48 0 480 2
3 24 0 24 2401
x
y
D
D
D
= = − =
= = − =
= = − = −
Find the solutions by Cramer's Rule:
48 248 4
6 6yx
DDx y
D D
−= = = = = = −
The solution is (8, 4)− .
20. 4 5 3
2 4
x y
y
+ = − − = −
54 8 0 80 2
3 5 6 ( 20) 264 2
34 16 0 160 4
x
y
D
D
D
= = − − = −−
−= = − − =− −
−= = − − = −−
Find the solutions by Cramer's Rule:
26 13 162
8 4 8yx
DDx y
D D
−= = = − = = =− −
The solution is 13
, 24
−
.
21. 3 6 24
5 4 12
x y
x y
− = + =
3 6 12 ( 30) 425 4
624 96 ( 72) 16812 4
3 24 36 120 845 12
x
y
D
D
D
−= = − − =
−= = − − =
= = − = −
Find the solutions by Cramer's Rule:
168 844 2
42 42yx
DDx y
D D
−= = = = = = −
The solution is (4, 2)− .
22. 2 4 16
3 5 9
x y
x y
+ = − = −
2 4 10 12 223 5
16 4 80 36 449 5
x
D
D
= = − − = −−
= = − + = −− −
162 18 48 663 9
yD = = − − = −−
Find the solutions by Cramer's Rule:
44 662 3
22 22yx
DDx y
D D
− −= = = = = =− −
The solution is (2, 3).
23. 3 2 4
6 4 0
x y
x y
− = − =
23 12 ( 12) 046
D −= = − − − =−
Since 0D = , Cramer's Rule does not apply.
24. 2 5
4 8 6
x y
x y
− + = − =
1 2 8 8 04 8
D −= = − =−
Since 0D = , Cramer's Rule does not apply.
25. 2 4 2
3 2 3
x y
x y
− = − + =
42 4 12 163 2
2 4 4 12 83 2
22 6 6 123 3
x
y
D
D
D
−= = + =
− −= = − + =
−= = + =
Find the solutions by Cramer's Rule:
8 1 12 3
16 2 16 4yx
DDx y
D D= = = = = =
The solution is 1 3
,2 4
.
Chapter 11: Systems of Equations and Inequalities
1118
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
26. 3 3 3
84 2
3
x y
x y
+ = + =
83
83
3 3 6 12 64 2
3 36 8 2
2
3 38 12 4
4
x
y
D
D
D
= = − = −
= = − = −
= = − = −
Find the solutions by Cramer's Rule:
2 1 4 2
6 3 6 3yx
DDx y
D D
− −= = = = = =− −
The solution is 1 2
,3 3
.
27. 2 3 1
10 10 5
x y
x y
− = − + =
32 20 ( 30) 5010 10
31 10 ( 15) 55 10
2 1 10 ( 10) 2010 5
x
y
D
D
D
−= = − − =
−−= = − − − =
−= = − − =
Find the solutions by Cramer's Rule:
5 1 20 2
50 10 50 5yx
DDx y
D D= = = = = =
The solution is 1 2
,10 5
.
28. 3 2 0
5 10 4
x y
x y
− = + =
3 2 30 ( 10) 405 10
0 2 0 ( 8) 8104
3 0 12 0 125 4
x
y
D
D
D
−= = − − =
−= = − − =
= = − =
Find the solutions by Cramer's Rule:
8 1 12 3
40 5 40 10yx
DDx y
D D= = = = = =
The solution is 1 3
,5 10
.
29. 2 3 6
1
2
x y
x y
+ = − =
12
12
32 2 3 51 1
6 3 3 156
2 21
621 6 5
1
x
y
D
D
D
= = − − = −−
= = − − = −−
= = − = −
Find the solutions by Cramer's Rule: 15
3 52 15 2 5
yxDD
x yD D
− −= = = = = =− −
The solution is 3
,2
1
.
30. 1
22
2 8
x y
x y
+ = − − =
12
12
1 1 1 21 2
2 1 4 8 48 2
2 4 ( 2) 681
x
y
D
D
D
= = − − = −−
−= = − = −−
−= = − − =
Find the solutions by Cramer's Rule:
4 62 3
2 2yx
DDx y
D D
−= = = = = = −− −
The solution is (2, 3)− .
31. 3 5 3
15 5 21
x y
x y
− = + =
3 5 15 ( 75) 9015 5
3 5 15 ( 105) 120521
3 3 63 45 1815 21
x
y
D
D
D
−= = − − =
−= = − − =
= = − =
Find the solutions by Cramer's Rule:
120 4 18 1
90 3 90 5yx
DDx y
D D= = = = = =
The solution is 4 1
,3 5
.
Section 11.3: Systems of Linear Equations: Determinants
1119
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
32. 2 1
1 3
2 2
x y
x y
− = − + =
12
3 12 2
32
2 11 1 2
1
1 1 1 31
2 2
2 13 1 4
1
x
y
D
D
D
−= = + =
− −= = − + =
−= = + =
Find the solutions by Cramer's Rule:
1 42
2 2yx
DDx y
D D= = = = =
The solution is 1
,2
2
.
33.
6
3 2 5
3 2 14
x y z
x y z
x y z
+ − = − + = − + − =
1 1 123 1
231
2 23 31 11 1 ( 1)2 23 31 1
1(4 3) 1( 6 1) 1(9 2)1 7 11
3
D−
−=−
− −= − + −− −
= − − − − − += + −= −
6 1 125 1
2314
2 25 51 16 1 ( 1)2 23 314 14
6(4 3) 1(10 14) 1( 15 28)6 4 13
3
xD−
−= −−
− −− −= − + −− −
= − − − − − += + −= −
61 1
3 5 1
21 14
5 3 3 51 11 6 ( 1)2 214 1 1 14
1(10 14) 6( 6 1) 1(42 5)
4 42 47
9
yD
−= −
−
− −= − + −− −
= − − − − − += − + −= −
61 1
23 5
31 14
2 25 3 5 31 1 63 314 1 14 1
1( 28 15) 1(42 5) 6(9 2)
13 47 66
6
zD −= −
− −− −= − +
= − + − + + += − − +=
Find the solutions by Cramer's Rule:
3 91 3
3 36
23
yx
z
DDx y
D DD
zD
− −= = = = = =− −
= = = −−
The solution is (1, 3, 2)− .
34.
4
2 3 4 15
5 2 12
x y z
x y z
x y z
− + = − − + = − + − =
1 1 132 4
5 1 2
3 34 2 4 21 ( 1) 12 25 51 1
1(6 4) 1( 4 20) 1(2 15)2 24 17
5
D−−=
−
− −= − − +− −
= − + − − + += − += −
1 1415 3 412 1 2
3 15 15 34 44 ( 1) 12 21 12 12 1
4(6 4) 1(30 48) 1( 15 36)8 18 215
xD−−
= − −−
− − − −= − − − +− −
= − − + − + − += − − += −
1 14
152 4
5 12 2
15 154 2 4 21 ( 4) 12 25 512 12
1(30 48) 4( 4 20) 1(24 75)
18 96 99
15
yD
−−=
−
− −= − − +− −
= − + − − + += − − += −
Chapter 11: Systems of Equations and Inequalities
1120
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
1 1 43 152
5 1 12
3 15 15 32 21 ( 1) ( 4)5 51 12 12 1
1( 36 15) 1(24 75) 4(2 15)21 99 68
10
zD− −− −=
− − − −= − − + −
= − + + + − += − + −=
Find the solutions by Cramer's Rule:
5 151 3
5 510
25
yx
z
DDx y
D DD
zD
− −= = = = = =− −
= = = −−
The solution is (1, 3, 2)− .
35.
2 3
2 4 7
2 2 3 4
x y z
x y z
x y z
+ − = − − + = −− + − =
1 2 142 1
2 32
4 41 2 1 21 2 ( 1)2 23 32 2
1(12 2) 2( 6 2) 1(4 8)10 8 422
3 2 147 1
34 2
4 47 71 13 2 ( 1)3 32 4 4 2
3(12 2) 2(21 4) 1( 14 16)30 34 266
x
D
D
−−=
− −
− −= − + −− −− −
= − − − + − −= + +=
− −−= −
−
− −− −= − − + −− −
= − − − − − − += − − −= −
31 1
72 1
2 34
7 71 2 1 21 ( 3) ( 1)2 23 34 4
1(21 4) 3( 6 2) 1(8 14)
17 12 6
11
yD
− −−=
− −
− −= − − + −− −− −
= − + − + − −= − +=
31 24 72
2 2 4
4 47 72 21 2 ( 3)2 22 4 4 2
1( 16 14) 2(8 14) 3(4 8)2 12 12
22
zD−
− −=−
− −− −= − + −− −
= − + − − − −= − + +=
Find the solutions by Cramer's Rule:
66 11 13
22 22 222
122
yx
z
DDx y
D DD
zD
−= = = − = = =
= = =
The solution is 1
3, , 12
−
.
36. 4 3 8
3 3 126 1
x y zx y z
x y z
+ − = − − + = + + =
31 43 31
61 1
3 3 3 31 11 4 ( 3)6 61 1 1 1
1( 6 3) 4(18 3) 3(3 1)9 60 1281
D−
= −
− −= − + −
= − − − − − += − − −= −
348312 161 1
3 31 12 12 18 4 ( 3)6 61 1 1 1
8( 6 3) 4(72 3) 3(12 1)72 276 39
243
xD−−
= −
− −= − − + −
= − − − − − − += − −= −
31 83 312
61 1
3 3 3 312 121 ( 8) ( 3)6 61 1 1 1
1(72 3) 8(18 3) 3(3 12)69 120 27216
yD−−
=
= − − + −
= − + − − −= + +=
Section 11.3: Systems of Linear Equations: Determinants
1121
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
1 4 8
3 1 12
1 1 1
3 31 12 12 11 4 ( 8)1 1 1 1 1 1
1( 1 12) 4(3 12) 8(3 1)
13 36 32
9
zD
−= −
− −= − + −
= − − − − − += − + −= −
Find the solutions by Cramer's Rule: 243 216 8
3 81 81 3
9 1
81 9
yx
z
DDx y
D D
Dz
D
−= = = = = = −− −−= = =−
The solution is 8 1
3, ,3 9
−
.
37.
2 3 1
3 2 0
2 4 6 2
x y z
x y z
x y z
− + = + − = − + =
2 31
23 1
4 62
2 23 31 11 ( 2) 36 64 2 2 4
1(6 8) 2(18 4) 3( 12 2)
2 44 42
0
D
−−=
−
− −= − − +− −
= − + + + − −= − + −=
Since 0D = , Cramer's Rule does not apply.
38.
2 5
3 2 4
2 2 4 10
x y z
x y
x y z
− + = + = − + − = −
1 1 2
3 02
22 4
0 3 0 32 21 ( 1) 22 24 2 4 2
1( 8 0) 1( 12 0) 2(6 4)
8 12 20
0
D
−=
− −
= − − +− − − −
= − − + − − + += − − +=
Since 0D = , Cramer's Rule does not apply.
39.
2 0
2 4 0
2 2 3 0
x y z
x y z
x y z
+ − = − + =− + − =
1 2 142 1
2 32
4 41 2 1 21 2 ( 1)2 23 32 2
1(12 2) 2( 6 2) 1(4 8)10 8 422
D−
−=− −
− −= − + −− −− −
= − − − + − −= + +=
0 2 1
40 0 [By Theorem (12)]1
0 32xD
−−= =
−
01 1
02 1 0 [By Theorem (12)]
2 0 3yD
−= =
− −
01 2
4 02 0 [By Theorem (12)]
2 02zD −= =
−
Find the solutions by Cramer's Rule:
0 00 0
22 220
022
yx
z
DDx y
D DD
zD
= = = = = =
= = =
The solution is (0, 0, 0).
40.
4 3 0
3 3 0
6 0
x y z
x y z
x y z
+ − = − + = + + =
31 4
3 31
61 1
3 3 3 31 11 4 ( 3)6 61 1 1 1
1( 6 3) 4(18 3) 3(3 1)
9 60 12
81
0 34
0 3 0 [By Theorem (12)]1
0 61x
D
D
−= −
− −= − + −
= − − − − − += − − −= −
−= =−
Chapter 11: Systems of Equations and Inequalities
1122
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
0 31
3 0 3 0 [By Theorem (12)]
0 61
01 4
3 0 0 [By Theorem (12)]1
01 1
y
z
D
D
−= =
= =−
Find the solutions by Cramer's Rule:
0 00 0
81 81
00
81
yx
z
DDx y
D D
Dz
D
= = = = = =− −
= = =−
The solution is (0, 0, 0).
41.
2 3 0
3 2 0
2 4 6 0
x y z
x y z
x y z
− + = + − = − + =
2 31
23 1
4 62
2 23 31 11 ( 2) 34 46 62 2
1(6 8) 2(18 4) 3( 12 2)
2 44 42
0
D
−−=
−
− −= − − +− −
= − + + + − −= − + −=
Since 0D = , Cramer's Rule does not apply.
42.
2 0
3 2 0
2 2 4 0
x y z
x y
x y z
− + = + = − + − =
1 1 2
3 02
22 4
0 3 0 32 21 ( 1) 22 24 2 4 2
1( 8 0) 1( 12 0) 2(6 4)
8 12 20
0
D
−=
− −
= − − +− − − −
= − − + − − + += − − +=
Since 0D = , Cramer's Rule does not apply.
43. 4
1 2 3
x y z
u v w =
By Theorem (11), the value of a determinant changes sign if any two rows are interchanged.
Thus,
1 2 3
4u v w
x y z
= − .
44. 4
1 2 3
x y z
u v w =
By Theorem (14), if any row of a determinant is multiplied by a nonzero number k, the value of the determinant is also changed by a factor of k.
Thus, 2 2(4) 8
2 4 6 1 2 3
x y z x y z
u v w u v w= = = .
45. Let 4
1 2 3
x y z
u v w = .
3 6 9 3 1 2 3 [Theorem (14)]
3( 1) [Theorem (11)]
1 2 3
3(4) 12
x y z x y z
u v w u v w
x y z
u v w
− − − = −
= − −
= =
46. Let 4
1 2 3
x y z
u v w =
( )2 2 3
1 2 3 1 2 3[Theorem (15)]
( 1) 1 2 3 [Theorem (11)]
( 1)( 1) [Theorem (11)]
1 2 3
1 2 3
x u y v z w x y zR r r
u v w u v w
x y z
u v w
x y z
u v w
x y z
u v w
− − − == +
= −
= − −
= 4=
Section 11.3: Systems of Linear Equations: Determinants
1123
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
47. Let 4
1 2 3
x y z
u v w =
1 2 3
3 6 9
2 2 2
1 2 3
2 3 6 9 [Theorem (14)]
3 6 9
2( 1) 1 2 3 [Theorem (11)]
x y z
u v w
x y z
u v w
x y z
u v w
− − −
= − − −
− − −= −
1 3 1
3 6 9
2( 1)( 1) [Theorem (11)]
1 2 3
[Theorem (15)]2( 1)( 1)
( 3 )1 2 3
x y z
u v w
x y z
u v wR r r
− − −= − −
= − −= − +
2( 1)( 1)(4) 8= − − =
48. Let 4
1 2 3
x y z
u v w =
3 1 3
[Theorem (15)] ( )
1 2 2 1 2 3
4
x y z x x y z
u v w u u v wC c c
−− =
= +
=
49. Let 4
1 2 3
x y z
u v w =
1 2 3
2 2 2
1 2 3
1 2 3
2 [Theorem (14)]
1 2 3
2( 1) 1 2 3 [Theorem (11)]
1 2 3
x y z
u v w
x y z
u v w
x y z
u v w
− − −
=− − −
= −− − −
2 3 2
2( 1)( 1) 1 2 3 [Theorem (11)]
1 2 3
[Theorem (15)]2( 1)( 1)
( )1 2 3
x y z
u v w
x y z
u v wR r r
= − − − − −
= − −= − +
2( 1)( 1)(4)
8
= − −=
50. Let 4
1 2 3
x y z
u v w =
1 3 1
2 3 2
3 6 9
3 1 3 2 3 3
1 2 3
[Theorem (15)]3 1 3 2 3 3
( =3 )1 2 3
[Theorem (15)]3 3 3
( )1 2 3
x y z
u v w
x y z
u v wR r r
x y z
u v wR r r
+ + +− − −
= − − −+
== − +
3 [Theorem (14)]
1 2 3
3(4)
12
x y z
u v w=
==
51. Solve for x:
534
3 4 5
5
5
x x
x x
x
x
=
− =− =
= −
52. Solve for x:
2
2
1 23
3 2
1 0
( 1)( 1) 0
x
x
x
x
x x
= −
− = −
− =− + =
1 0 or 1 0
1 or 1
x x
x x
− = + == = −
Chapter 11: Systems of Equations and Inequalities
1124
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
53. Solve for x:
( ) ( ) ( )
1 1
34 2 2
51 2
3 32 4 2 41 1 25 52 1 1 2
15 4 20 2 8 3 2
11 22 11 2
11 13
13
11
x
x
x
x
x
x
=−
− + =− −
− − + + + =− + =
=
=
54. Solve for x:
( ) ( ) ( )
3 2 4
51 0
20 1
5 51 13 2 4 02 20 01 1
3 2 5 2 2 4 1 0
6 15 4 4 0
6 7 0
6 7
7
6
x
x x
x
x
x
x
x
=−
− + =− −
− − − − + =− − + + =
− − =− =
= −
55. Solve for x:
( ) ( ) ( )
( )
2
2
32
01 7
26 1
0 01 12 3 72 26 61 1
2 2 2 3 1 6 7
2 4 3 18 7
2 18 0
2 9 0
x
x
x xx
x x x
x x
x x
x x
=−
− + =− −
− − − + − =
− + + − =
− − =− + =
0 or 9x x= = −
56. Solve for x:
( ) ( ) ( )
( )
2
2
1 2
31 4
0 1 2
3 31 11 2 40 01 2 2 1
2 3 1 2 2 1 4
2 3 2 2 4
2 0
2 1 0
x
x x
x xx x
x x x
x x x
x x
x x
= −
− + = −
− − + = −
− − + = −
+ =+ =
1
0 or 2
x x= = −
57. Expanding the determinant:
( )
1 1
2 2
1 1 1 1
2 2 2 2
1 2 1 2 1 2 2 1
1
1 0
1
1 11 0
1 1
( ) ( ) 0
x y
x y
x y
y x x yx y
y x x y
x y y y x x x y x y
=
− + =
− − − + − =
( )( )
1 2 2 1 2 1 1 2
2 1 2 1 1 2 2 1
( )
( )
x y y y x x x y x y
y x x x y x y x y y
− + − = −
− = − + −
( )
( )
2 1 1 2 1
2 1 1 2 2 1 1 2 1
2 1 1
2 1 2 1 1 2 1 2 1 1
( )
( ) ( )
( )
( )
y x x y x x
x y x y x y y y x x
x x y y
x y y x y x y y x y x
− − −= − + − − −
− −= − + − − +
( )( )
( )
2 1 1 2 1 2 1 1
2 1 1 2 1 1
2 11 1
2 1
( ) ( ) ( )
( ) ( )( )
( )( ) ( )
x x y y y y x y y x
x x y y y y x x
y yy y x x
x x
− − = − − −
− − = − −−
− = −−
This is the 2-point form of the equation for a line.
Section 11.3: Systems of Linear Equations: Determinants
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
58. Any point ( , )x y on the line containing 2 2( , )x y
and 3 3( , )x y satisfies:
2 2
3 3
1
1 0
1
x y
x y
x y
=
If the point 1 1( , )x y is on the line containing
2 2( , )x y and 3 3( , )x y [the points are collinear],
then 1 1
2 2
3 3
1
1 0
1
x y
x y
x y
= .
Conversely, if 1 1
2 2
3 3
1
1 0
1
x y
x y
x y
= , then 1 1( , )x y is
on the line containing 2 2( , )x y and 3 3( , )x y , and
the points are collinear.
59. If the vertices of a triangle are (2, 3), (5, 2), and (6, 5), then:
[ ]
[ ]
[ ]
2 5 61
3 2 52
1 1 1
2 5 3 5 3 212 5 6
1 1 1 1 1 12
12(2 5) 5(3 5) 6(3 2)
21
2( 3) 5( 2) 6(1)21
6 10 625
D =
= − +
= − − − + −
= − − − +
= − + +
=
The area of the triangle is 5 5= square units.
60. Expanding the determinant:
[ ]
2
2
2
2 22
2 2
2 2 2 2 2
2
2
1
1
1
1 11
1 1
( ) ( ) 1( )
( ) ( )( ) ( )
( )
( ) ( ) ( )
( )( )( )
x x
y y
z z
y y y yx x
z z z z
x y z x y z y z z y
x y z x y z y z yz y z
y z x xy xz yz
y z x x y z x y
y z x y x z
= − +
= − − − + −
= − − − + + −
= − − − + = − − − −= − − −
61. If 0,a = then 0b ≠ and 0c ≠ since
0ad bc− ≠ , and the system is by s
cx dy t
= + =
.
The solution of the system is ,s
yb
=
( )sbt dt dy tb sd
xc c bc
−− −= = = . Using Cramer’s
Rule, we get 0 b
D bcc d
= = − ,
x
s bD sd tb
t d= = − ,
00y
sD sc sc
c t= = − = − , so
xD ds tb td sdx
D bc bc
− −= = =−
and
yD sc sy
D bc b
−= = =−
, which is the solution. Note
that these solutions agree if 0.d =
If 0,b = then 0a ≠ and 0d ≠ since
0ad bc− ≠ , and the system is ax s
cx dy t
= + =
.
The solution of the system is ,s
xa
=
t cx at csy
d ad
− −= = . Using Cramer’s Rule, we
get 0a
D adc d
= = , 0
x
sD sd
t d= = , and
Chapter 11: Systems of Equations and Inequalities
1126
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
y
a sD at cs
c t= = − , so xD sd s
xD ad a
= = =
and yD at csy
D ad
−= = , which is the solution.
Note that these solutions agree if 0.c =
If 0,c = then 0a ≠ and 0d ≠ since
0ad bc− ≠ , and the system is
ax by s
dy t
+ = =
.
The solution of the system is ,t
yd
=
s by sd tbx
a ad
− −= = . Using Cramer’s Rule, we
get 0
a bD ad
d= = , x
s bD sd tb
t d= = − ,
and 0y
a sD at
t= = , so xD sd tb
xD ad
−= = and
yD at ty
D ad d= = = , which is the solution. Note
that these solutions agree if 0.b =
If 0,d = then 0b ≠ and 0c ≠ since
0ad bc− ≠ , and the system is
ax by s
cx t
+ = =
.
The solution of the system is ,t
xc
=
s ax cs aty
b bc
− −= = . Using Cramer’s Rule, we
get 00
a bD bc bc
c= = − = − ,
00x
s bD tb tb
t= = − = − , and
y
a sD at cs
c t= = − , so xD tb t
xD bc c
−= = =−
and
yD at cs cs aty
D bc bc
− −= = =−
, which is the
solution. Note that these solutions agree if 0.a =
62. Evaluating the determinant to show the relationship:
13 12 11
23 22 21
33 32 31
22 21 23 21 23 2213 12 11
32 31 33 31 33 32
13 22 31 21 32 12 23 31 21 33
11 23 32 22 33
( ) ( )
( )
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
a a a a a a a a a a
a a a a a
= − +
= − − −+ −
13 22 31 13 21 32 12 23 31 12 21 33
11 23 32 11 22 33
11 22 33 11 23 32 12 21 33 12 23 31
13 21 32 13 22 31
11 22 33 23 32 12 21 33 23 31
( ) ( )
a a a a a a a a a a a a
a a a a a a
a a a a a a a a a a a a
a a a a a a
a a a a a a a a a a
= − − ++ −
= − + + −− +
= − − + −
13 21 32 22 31
22 23 21 23 21 2211 12 13
32 33 31 33 31 32
22 23 21 23 21 2211 12 13
32 33 31 33 31 32
11 12 13
21 22 23
31 32 33
( )a a a a a
a a a a a aa a a
a a a a a a
a a a a a aa a a
a a a a a a
a a a
a a a
a a a
− −
= − + −
= − − +
= −
63. Evaluating the determinant to show the relationship:
11 12 13
21 22 23
31 32 33
22 23 21 23 21 2211 12 13
32 33 31 33 31 32
11 22 33 23 32 12 21 33 23 31
13 21 32 22 31
11 22 33 23 32 12 21 33
( ) ( )
( )
( ) (
a a a
ka ka ka
a a a
ka ka ka ka ka kaa a a
a a a a a a
a ka a ka a a ka a ka a
a ka a ka a
ka a a a a ka a a a
= − +
= − − −+ −
= − − −
()
23 31
13 21 32 22 31
11 22 33 23 32 12 21 33 23 31
13 21 32 22 31
22 23 21 23 21 2211 12 13
32 33 31 33 31 32
11 12 13
21 22 23
)
( )
( ) ( )
( )
a
ka a a a a
k a a a a a a a a a a
a a a a a
a a a a a ak a a a
a a a a a a
a a a
k a a a
a
+ −= − − −
+ −
= − +
=
31 32 33a a
Section 11.4: Matrix Algebra
1127
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
64. Set up a 3 by 3 determinant in which the first column and third column are the same and evaluate:
11 12 11
21 22 21
31 32 31
22 21 21 21 21 2211 12 11
32 31 31 31 31 32
11 22 31 32 21 12 21 31 31 21
11 21 32 31 22
11 22 31 11 32 21 12 11 21 32 11 31 22
( ) ( )
( )
(0)
0
a a a
a a a
a a a
a a a a a aa a a
a a a a a a
a a a a a a a a a a
a a a a a
a a a a a a a a a a a a a
= − +
= − − −+ −
= − − + −=
65. Evaluating the determinant to show the relationship:
11 21 12 22 13 23
21 22 23
31 32 33
22 23 21 2311 21 12 22
32 33 31 33
21 2213 23
31 32
11 21 22 33 23 32
12 22 21 33 23 31
13 23 21 32 22 31
11 22 33
( ) ( )
( )
( )( )
( )( )
( )( )
(
a ka a ka a ka
a a a
a a a
a a a aa ka a ka
a a a a
a aa ka
a a
a ka a a a a
a ka a a a a
a ka a a a a
a a a
+ + +
= + − +
+ +
= + −− + −+ + −
= 23 32 21 22 33 23 32
12 21 33 23 31 22 21 33 23 31
13 21 32 22 31 23 21 32 22 31
11 22 33 23 32 21 22 33
21 23 32 12 21 33 23 31
22 21 33 22 23 31
13 21 32
) ( )
( ) ( )
( ) ( )
( )
( )
(
a a ka a a a a
a a a a a ka a a a a
a a a a a ka a a a a
a a a a a ka a a
ka a a a a a a a
ka a a ka a a
a a a
− + −− − − −+ − + −
= − +− − −− ++ − 22 31 23 21 32
23 22 31
11 22 33 23 32 12 21 33 23 31
13 21 32 22 31
22 23 21 23 21 2211 12 13
32 33 31 33 31 32
11 12 13
21 22 23
31 32 33
)
( ) ( )
( )
a a ka a a
ka a a
a a a a a a a a a a
a a a a a
a a a a a aa a a
a a a a a a
a a a
a a a
a a a
+−
= − − −+ −
= − +
=
Section 11.4
1. square
2. True
3. columns; rows
4. False
5. inverse
6. singular
7. True
8. 1A B−
9. 0 3 5 4 1 0
1 2 6 2 3 2
0 4 3 1 5 0
1 ( 2) 2 3 6 ( 2)
4 4 5
1 5 4
A B−
+ = + − − + + − +
= + − + + − −
= −
10. 0 3 5 4 1 0
1 2 6 2 3 2
0 4 3 1 5 0
1 ( 2) 2 3 6 ( 2)
4 2 5
3 1 8
A B−
− = − − − − − − −
= − − − − − − −
= −
11. 0 3 5
4 41 2 6
4 0 4 3 4( 5)
4 1 4 2 4 6
0 12 20
4 8 24
A−
= ⋅ ⋅ −
= ⋅ ⋅ ⋅ −
=
12. 4 1 0
3 32 3 2
3 4 3 1 3 0
3 2 3 3 3 2
12 3 0
6 9 6
B − = − − − − ⋅ − ⋅ − ⋅ = − ⋅ − − ⋅ − ⋅ −
− − = −
Chapter 11: Systems of Equations and Inequalities
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13. 0 3 5 4 1 0
3 2 3 21 2 6 2 3 2
0 9 15 8 2 0
3 6 18 4 6 4
8 7 15
7 0 22
A B− − = − − −
− = − − − − − =
14. 0 3 5 4 1 0
2 4 2 41 2 6 2 3 2
0 6 10 16 4 0
2 4 12 8 12 8
16 10 10
6 16 4
A B−
+ = + − − −
= + − − −
= −
15. 4 1
0 3 56 2
1 2 62 3
0(4) 3(6) ( 5)( 2) 0(1) 3(2) ( 5)(3)
1(4) 2(6) 6( 2) 1(1) 2(2) 6(3)
28 9
4 23
AC
− = ⋅ −
+ + − − + + − = + + − + +
− =
16. 4 1
4 1 06 2
2 3 22 3
4(4) 1(6) 0( 2) 4(1) 1(2) 0(3)
2(4) 3(6) ( 2)( 2) 2(1) 3(2) ( 2)(3)
22 6
14 2
BC
= ⋅ − − −
+ + − + + = − + + − − − + + −
= −
17. 4 1
0 3 56 2
1 2 62 3
4(0) 1(1) 4(3) 1(2) 4( 5) 1(6)
6(0) 2(1) 6(3) 2(2) 6( 5) 2(6)
2(0) 3(1) 2(3) 3(2) 2( 5) 3(6)
1 14 14
2 22 18
3 0 28
CA
− = ⋅ −
+ + − + = + + − + − + − + − − +
− = −
18. 4 1
4 1 06 2
2 3 22 3
4(4) 1( 2) 4(1) 1(3) 4(0) 1( 2)
6(4) 2( 2) 6(1) 2(3) 6(0) 2( 2)
2(4) 3( 2) 2(1) 3(3) 2(0) 3( 2)
14 7 2
20 12 4
14 7 6
CB
= ⋅ − − −
+ − + + − = + − + + − − + − − + − + −
− = − − −
19. 4 1
0 3 5 4 1 0( ) 6 2
1 2 6 2 3 22 3
4 14 4 5
6 21 5 4
2 3
15 21 16
22 34 22
11 7 22
C A B
− + = + − − −
− = ⋅ − −
− = − −
20. 4 1
0 3 5 4 1 0( ) 6 2
1 2 6 2 3 22 3
4 14 4 5
6 21 5 4
2 3
50 3
18 21
A B C
− + = + − − −
− = ⋅ − −
− =
21. 2
4 10 3 5 1 0
3 6 2 31 2 6 0 1
2 3
28 9 3 0
4 23 0 3
25 9
4 20
AC I
− − = ⋅ − −
− = −
− =
Section 11.4: Matrix Algebra
1129
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
22. 3
4 1 1 0 00 3 5
5 6 2 5 0 1 01 2 6
2 3 0 0 1
1 14 14 5 0 0
2 22 18 0 5 0
3 0 28 0 0 5
6 14 14
2 27 18
3 0 33
CA I
− + = ⋅ + −
− = − +
− = −
23. CA CB−
4 1 4 1
0 3 5 4 1 06 2 6 2
1 2 6 2 3 22 3 2 3
1 14 14 14 7 2
2 22 18 20 12 4
3 0 28 14 7 6
13 7 12
18 10 14
17 7 34
− = ⋅ − ⋅ − − − −
− − = − − − − − − − = − − −
24. 4 1 4 1
0 3 5 4 1 06 2 6 2
1 2 6 2 3 22 3 2 3
28 9 22 6
4 23 14 2
50 3
18 21
AC BC+
− = ⋅ + ⋅ − − − −
− = + −
− =
25. 11
12
13
14
21
22
23
24
2(2) ( 2)(3) 2
2(1) ( 2)( 1) 4
2(4) ( 2)(3) 2
2(6) ( 2)(2) 8
1(2) 0(3) 2
1(1) 0( 1) 1
1(4) 0(3) 4
1(6) 0(2) 6
a
a
a
a
a
a
a
a
= + − = −= + − − == + − == + − == + == + − == + == + =
2 2 2 1 4 6 2 4 2 8
1 0 3 1 3 2 2 1 4 6
− − = −
26. 11
12
13
14
21
22
23
24
4( 6) 1(2) 22
4(6) 1(5) 29
4(1) 1(4) 8
4(0) 1( 1) 1
2( 6) 1(2) 10
2(6) 1(5) 17
2(1) 1(4) 6
2(0) 1( 1) 1
a
a
a
a
a
a
a
a
= − + = −= + == + == + − = −= − + = −= + == + == + − = −
4 1 6 6 1 0 22 29 8 1
2 1 2 5 4 1 10 17 6 1
− − − = − − −
27. 1 2
1 2 31 0
0 1 42 4
1(1) 2( 1) 3(2) 1(2) 2(0) 3(4)
0(1) ( 1)( 1) 4(2) 0(2) ( 1)(0) 4(4)
5 14
9 16
− −
+ − + + + = + − − + + − +
=
28. 1 1
2 8 13 2
3 6 00 5
1(2) ( 1)(3) 1(8) ( 1)(6) 1( 1) ( 1)(0)
( 3)(2) 2(3) ( 3)(8) 2(6) ( 3)( 1) 2(0)
0(2) 5(3) 0(8) 5(6) 0( 1) 5(0)
− − −
+ − + − − + − = − + − + − − + + + − +
1 2 1
0 12 3
15 30 0
− − = −
29. 1 0 1 1 3
2 4 1 6 2
3 6 1 8 1
1(1) 0(6) 1(8) 1(3) 0(2) 1( 1)
2(1) 4(6) 1(8) 2(3) 4(2) 1( 1)
3(1) 6(6) 1(8) 3(3) 6(2) 1( 1)
9 2
34 13
47 20
−
+ + + + − = + + + + − + + + + − =
Chapter 11: Systems of Equations and Inequalities
1130
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30. 4 2 3 2 6
0 1 2 1 1
1 0 1 0 2
4(2) ( 2)(1) 3(0) 4(6) ( 2)( 1) 3(2)
0(2) 1(1) 2(0) 0(6) 1( 1) 2(2)
1(2) 0(1) 1(0) 1(6) 0( 1) 1(2)
− − −
+ − + + − − + = + + + − + − + + − + − +
6 32
1 3
2 4
= − −
31. 2 1
1 1A
=
Augment the matrix with the identity and use row operations to find the inverse:
( )
1 2
2 1 2
2 1 1 0
1 1 0 1
Interchange1 1 0 1
and 2 1 1 0
1 1 0 1 2
0 1 1 2
r r
R r r
→
→ = − + − −
( )
( )
2 2
1 2 1
1 1 0 1
0 1 1 2
1 0 1 1
0 1 1 2
R r
R r r
→ = − −
− → = − + −
Thus, 1 1 1
1 2A− −
= − .
32. 3 1
2 1A
− = −
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
1 2 1
2 1 2
3 1 1 0
2 1 0 1
1 0 1 1
2 1 0 1
1 0 1 1 2
0 1 2 3
R r r
R r r
− −
→ = + −
→ = +
Thus, 1 1 1
2 3A−
=
.
33. 6 5
2 2A
=
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
1 2
2 1 2
111 122
2 2
52
1 2 1
6 5 1 0
2 2 0 1
Interchange2 2 0 1
and 6 5 1 0
2 2 0 1 3
0 1 1 3
1 1 0
0 1 1 3
1 0 1
0 1 1 3
r r
R r r
R r
R r
R r r
→ → = − + − −
= → − = −
−→ = − + −
Thus, 5
1 21
1 3A− −
= − .
34. 4 1
6 2A
− = −
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
( )
( )
1 114 4
1 14
1 14 4
2 1 2312 2
1 14 4
2 2
112
1 24
4 1 1 0
6 2 0 1
1 0
6 2 0 1
1 0 6
0 1
1 0 2
0 1 3 2
1 0 1
0 1 3 2
R r
R r r
R r
R r r
− −
− −→ = − −
− −→ = − + −
− −→ = − − −
− −→ = + − −
Thus, 1
1 21
3 2A− − −
= − − .
Section 11.4: Matrix Algebra
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
35. 2 1
where 0.A aa a
= ≠
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
( )
( )
1 112 2
1 12
1 12 2
2 1 21 12 2
1 122 2
2 22
11
1 2 122
2 1 1 0
0 1
1 0
0 1
1 0
0 1
1 0
0 1 1
1 0 1
0 1 1
aa
a
a
a a
R ra a
R a r ra a
R r
R r r
→ =
→ = − + −
→ = − −
→ = − + −
Thus, 1
12
1
1a
a
A− −=
− .
36. 3
where 0.2
bA b
b
= ≠
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
( )
( )
2 1 2
3 11
1 1
3 1
2 2
323
1 2 1
3 1 0
2 0 1
3 1 0
0 1 1 1
1 0
0 1 1 1
1 0
0 1 1 1
1 0
0 1 1 1
b bb
b b
b bb
b
b
bR r r
R r
R r
R r r
→ = − + − −
→ =
− −
→ = − −
−→ = − +
−
Thus, 32
1
1 1b bA− −
= −
.
37.
1 1 1
0 2 1
2 3 0
A
− = − − −
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
3 1 3
1 1 12 22 2 2
1 12 2
1 2 11 12 2
3 2 3512 2
1 12 2
1 1 1 1 0 0
0 2 1 0 1 0
2 3 0 0 0 1
1 1 1 1 0 0
0 2 1 0 1 0 2
0 5 2 2 0 1
1 1 1 1 0 0
0 1 0 0
0 5 2 2 0 1
1 0 1 0
0 1 0 0 5
0 0 2 1
1 0 1
R r r
R r
R r r
R r r
− − − −
− → − = + −
− → − − = − − −
= + → − − = + − −
−→ ( )1 1
3 32 2
11 3 12
12 3 22
0
0 1 0 0 2
0 0 1 4 5 2
1 0 0 3 3 1
0 1 0 2 2 1
0 0 1 4 5 2
R r
R r r
R r r
− − = − − −
− = − + → − − = + − −
Thus, 1
3 3 1
2 2 1
4 5 2
A−−
= − − − −
.
38.
1 0 2
1 2 3
1 1 0
A
= − −
Augment the matrix with the identity and use row operations to find the inverse:
( )
2 1 2
3 1 3
5 1 1 12 22 2 2 2
1 0 2 1 0 0
1 2 3 0 1 0
1 1 0 0 0 1
1 0 2 1 0 0
0 2 5 1 1 0
0 1 2 1 0 1
1 0 2 1 0 0
0 1 0
0 1 2 1 0 1
R r r
R r r
R r
− −
= + → = − + − − −
→ = − − −
Chapter 11: Systems of Equations and Inequalities
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
( )5 1 13 2 32 2 2
1 1 12 2 2
1 0 2 1 0 0
0 1 0
0 0 1
R r r
→ = + −
( )5 1 13 32 2 2
1 3 1
52 3 22
1 0 2 1 0 0
0 1 0 2
0 0 1 1 1 2
1 0 0 3 2 42
0 1 0 3 2 5
0 0 1 1 1 2
R r
R r r
R r r
→ = −
− − = − + → − − = − + −
Thus, 1
3 2 4
3 2 5
1 1 2
A−− −
= − − −
.
39.
1 1 1
3 2 1
3 1 2
A
= −
Augment the matrix with the identity and use row operations to find the inverse:
1 1 1 1 0 0
3 2 1 0 1 0
3 1 2 0 0 1
−
2 1 2
3 1 3
1 1 1 1 0 03
0 1 4 3 1 0 3
0 2 1 3 0 1
R r r
R r r
= − + → − − − = − + − − −
( )2 2
1 1 1 1 0 0
0 1 4 3 1 0
0 2 1 3 0 1
R r
→ − = − − − −
( )13 37
3 2 17 7 7
5 317 7 7
1 3 19 1 47 7 7
2 3 23 2 17 7 7
1 0 3 2 1 0
0 1 4 3 1 0
0 0 1
1 0 03
0 1 0 4
0 0 1
R r
R r r
R r r
− − → − = − − = +
→ − = − + −
Thus,
5 317 7 7
1 9 1 47 7 73 2 17 7 7
A−
−
= − −
.
40.
3 3 1
1 2 1
2 1 1
A
= −
Augment the matrix with the identity and use row operations to find the inverse:
3 3 1 1 0 0
1 2 1 0 1 0
2 1 1 0 0 1
−
( )
1 2
2 1 2
3 1 3
2 1 12 23 3 3
1 23 3
2 13 3
7 53 3
1 2 1 0 1 0Interchange
3 3 1 1 0 0 and
2 1 1 0 0 1
1 2 1 0 1 03
0 3 2 1 3 0 2
0 5 1 0 2 1
1 2 1 0 1 0
0 1 1 0
0 5 1 0 2 1
1 0 1 0
0 1 1 0
0 0 3 1
r r
R r r
R r r
R r
→ −
= − + → − − − = − + − − −
→ − = − − − − − −
→ −
−
( )
1 2 1
3 2 3
1 23 3
32 13 33 3 7
5 9 37 7 7
3 4 17 7 7 1
1 3 131 1 27 7 7 2
2 3 235 9 37 7 7
2
5
1 0 1 0
0 1 1 0
0 0 1
1 0 0
0 1 0
0 0 1
R r r
R r r
R r
R r r
R r r
= − + = +
− − → − = − − = +
→ − = − + −
Thus,
3 4 17 7 7
1 1 1 27 7 7
5 9 37 7 7
A−
− = − −
.
Section 11.4: Matrix Algebra
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41. 2 8
5
x y
x y
+ = + =
Rewrite the system of equations in matrix form: 2 1 8
, ,1 1 5
xA X B
y
= = =
Find the inverse of 1 and solve A X A B−= :
From Problem 29, 1 1 1
1 2A− −
= − , so
1 1 1 8 3
1 2 5 2X A B− −
= = = − .
The solution is 3, 2x y= = or (3, 2) .
42. 3 8
2 4
x y
x y
− =− + =
Rewrite the system of equations in matrix form: 3 1 8
, ,2 1 4
xA X B
y
− = = = −
Find the inverse of 1 and solve A X A B−= :
From Problem 30, 1 1 1
2 3A−
=
, so
1 1 1 8 12
2 3 4 28X A B−
= = =
.
The solution is 12, 28x y= = or (12, 28) .
43. 2 0
5
x y
x y
+ = + =
Rewrite the system of equations in matrix form: 2 1 0
, ,1 1 5
xA X B
y
= = =
Find the inverse of 1 and solve A X A B−= :
From Problem 29, 1 1 1
1 2A− −
= − , so
1 1 1 0 5
1 2 5 10X A B− − −
= = = − .
The solution is 5, 10x y= − = or ( 5,10)− .
44. 3 4
2 5
x y
x y
− =− + =
Rewrite the system of equations in matrix form: 3 1 4
, ,2 1 5
xA X B
y
− = = = −
Find the inverse of 1 and solve A X A B−= :
From Problem 30, 1 1 1
2 3A−
=
, so
1 1 1 4 9
2 3 5 23X A B−
= = =
.
The solution is 9, 23x y= = or (9, 23) .
45. 6 5 7
2 2 2
x y
x y
+ = + =
Rewrite the system of equations in matrix form: 6 5 7
, ,2 2 2
xA X B
y
= = =
Find the inverse of 1 and solve A X A B−= :
From Problem 31, 5
1 21
1 3A− −
= − , so
51 2
7 21
2 11 3X A B− − = = = −−
.
The solution is 2, 1x y= = − or (2, 1)− .
46. 4 0
6 2 14
x y
x y
− + = − =
Rewrite the system of equations in matrix form: 4 1 0
, ,6 2 14
xA X B
y
− = = = −
Find the inverse of 1 and solve A X A B−= :
From Problem 32, 1
1 21
3 2A− − −
= − − , so
11 2
0 71
14 283 2X A B− − − − = = = −− −
.
The solution is 7, 28x y= − = − or ( 7, 28)− − .
Chapter 11: Systems of Equations and Inequalities
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47. 6 5 13
2 2 5
x y
x y
+ = + =
Rewrite the system of equations in matrix form: 6 5 13
, ,2 2 5
xA X B
y
= = =
Find the inverse of 1 and solve A X A B−= :
From Problem 31, 5
1 21
1 3A− −
= − , so
5 11 22
131
5 21 3X A B− − = = = −
.
The solution is 1
, 22
x y= = or 1
, 22
.
48. 4 5
6 2 9
x y
x y
− + = − = −
Rewrite the system of equations in matrix form: 4 1 5
, ,6 2 9
xA X B
y
− = = = − −
Find the inverse of 1 and solve A X A B−= :
From Problem 32, 1
1 21
3 2A− − −
= − − , so
1 11 2 2
51
93 2 3X A B− − − − = = = −− −
.
The solution is 1
, 32
x y= − = or 1
, 32
−
.
49. 2 3
0x y
aax ay a
+ = −≠ + = −
Rewrite the system of equations in matrix form: 2 1 3
, ,x
A X Ba a y a
− = = = −
Find the inverse of 1 and solve A X A B−= :
From Problem 33, 1
12
1
1a
a
A− −=
− , so
11
2
1 3 2
11a
a
X A Ba
− − − − = = = −−
.
The solution is 2, 1x y= − = or ( 2, 1)− .
50. 3 2 3
02 2 2
bx y bb
bx y b
+ = +≠ + = +
Rewrite the system of equations in matrix form: 3 2 3
, ,2 2 2
b x bA X B
b y b
+ = = = +
Find the inverse of 1 and solve A X A B−= :
From Problem 34, 32
1
1 1b bA− −
= −
, so
321 2 3 2
2 2 11 1b b b
X A Bb
− − + = = = +−
.
The solution is 2, 1x y= = or (2, 1).
51. 7
20
5
x yaa
ax ay
+ = ≠ + =
Rewrite the system of equations in matrix form: 72 1
, ,5ax
A X Ba a y
= = =
Find the inverse of 1 and solve A X A B−= :
From Problem 33, 1
12
1
1a
a
A− −=
− , so
1 271
32
1
1 5a aa
a a
X A B− − = = =
− .
The solution is 2 3
,x ya a
= = or 2 3
,a a
.
52. 3 14
02 10
bx yb
bx y
+ =≠ + =
Rewrite the system of equations in matrix form: 3 14
, ,2 10
b xA X B
b y
= = =
Find the inverse of 1 and solve A X A B−= :
From Problem 34, 32
1
1 1b bA− −
= −
, so
3 221 14
10 41 1bb bX A B− −
= = = − .
The solution is 2
, 4x yb
= = or 2
, 4b
.
Section 11.4: Matrix Algebra
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53.
0
2 1
2 3 5
x y z
y z
x y
− + = − + = −− − = −
Rewrite the system of equations in matrix form: 1 1 1 0
0 2 1 , , 1
2 3 0 5
x
A X y B
z
− = − = = − − − −
Find the inverse of 1 and solve A X A B−= :
From Problem 35, 1
3 3 1
2 2 1
4 5 2
A−−
= − − − −
, so
1
3 3 1 0 2
2 2 1 1 3
4 5 2 5 5
X A B−− −
= = − − − = − − −
.
The solution is 2, 3, 5x y z= − = = or ( 2, 3, 5)− .
54.
2 6
2 3 5
6
x z
x y z
x y
+ =− + + = − − =
Rewrite the system of equations in matrix form: 1 0 2 6
1 2 3 , , 5
1 1 0 6
x
A X y B
z
= − = = − −
Find the inverse of 1 and solve A X A B−= :
From Problem 36, 1
3 2 4
3 2 5
1 1 2
A−− −
= − − −
, so
1
3 2 4 6 4
3 2 5 5 2
1 1 2 6 1
X A B−− −
= = − − − = − −
.
The solution is 4, 2, 1x y z= = − = or (4, 2,1)− .
55.
2
2 2
12 3
2
x y z
y z
x y
− + = − + =− − =
Rewrite the system of equations in matrix form:
12
1 1 1 2
0 2 1 , , 2
2 3 0
x
A X y B
z
− = − = = − −
Find the inverse of 1 and solve A X A B−= :
From Problem 35, 1
3 3 1
2 2 1
4 5 2
A−−
= − − − −
, so
12
1 12
12
3 3 1 2
2 2 1 2
4 5 2 1
X A B− − = = − − = − − −
The solution is 1 1
, , 12 2
x y z= = − = or
1 1, , 1
2 2 −
.
56.
2 2
32 3
22
x z
x y z
x y
+ =− + + = −
− =
Rewrite the system of equations in matrix form:
32
1 0 2 2
1 2 3 , ,
1 1 0 2
x
A X y B
z
= − = = − −
Find the inverse of 1 and solve A X A B−= :
From Problem 36, 1
3 2 4
3 2 5
1 1 2
A−− −
= − − −
, so
1 32
12
3 2 4 2 1
3 2 5 1
1 1 2 2
X A B− − − = = − − − = − −
.
The solution is 1
1, 1,2
x y z= = − = or 1
1, 1,2
−
.
Chapter 11: Systems of Equations and Inequalities
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57.
9
3 2 8
3 2 1
x y z
x y z
x y z
+ + = + − = + + =
Rewrite the system of equations in matrix form: 1 1 1 9
3 2 1 , , 8
3 1 2 1
x
A X y B
z
= − = =
Find the inverse of 1 and solve A X A B−= :
From Problem 37,
5 317 7 7
1 9 1 47 7 73 2 17 7 7
A−
−
= − −
, so
5 3 3417 7 7 7
1 9 851 47 7 7 73 2 1 127 7 7 7
9
8
1
X A B−
− − = = − = −
.
The solution is 34 85 12
, ,7 7 7
x y z= − = = or
34 85 12, ,
7 7 7 −
.
58.
3 3 8
2 5
2 4
x y z
x y z
x y z
+ + = + + = − + =
Rewrite the system of equations in matrix form: 3 3 1 8
1 2 1 , , 5
2 1 1 4
x
A X y B
z
= = = −
Find the inverse of 1 and solve A X A B−= :
From Problem 38,
3 4 17 7 7
1 1 1 27 7 75 9 37 7 7
A−
−
= − −
, so
3 84 17 7 7 7
1 51 1 27 7 7 75 9 3 177 7 7 7
8
5
4
X A B−
− = = − = −
.
The solution is 8 5 17
, ,7 7 7
x y z= = = or
8 5 17, ,
7 7 7
.
59.
2
73 2
310
3 23
x y z
x y z
x y z
+ + = + − = + + =
Rewrite the system of equations in matrix form:
73
103
1 1 1 2
3 2 1 , ,
3 1 2
x
A X y B
z
= − = =
Find the inverse of 1 and solve A X A B−= :
From Problem 37,
5 317 7 7
1 9 1 47 7 73 2 17 7 7
A−
−
= − −
, so
5 31 17 7 7 3
1 9 71 47 7 7 3
23 102 137 7 7 3
2
1X A B−
−
= = − = −
.
The solution is 1 2
, 1,3 3
x y z= = = or 1 2
, 1,3 3
.
60.
3 3 1
2 0
2 4
x y z
x y z
x y z
+ + = + + = − + =
Rewrite the system of equations in matrix form: 3 3 1 1
1 2 1 , , 0
2 1 1 4
x
A X y B
z
= = = −
Find the inverse of 1 and solve A X A B−= :
From Problem 38,
3 4 17 7 7
1 1 1 27 7 75 9 37 7 7
A−
−
= − −
, so
3 4 17 7 7
1 1 1 27 7 75 9 37 7 7
1 1
0 1
4 1
X A B−
− = = − = − −
.
The solution is 1, 1, 1x y z= = − = or (1, 1,1)− .
Section 11.4: Matrix Algebra
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61. 4 2
2 1A
=
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
12 1 21 2
2
1 112 4
1 11 42
4 2 1 0
2 1 0 1
4 2 1 0
0 0 1
1 0
0 0 1
R r r
R r
→ = − + −
→ = −
There is no way to obtain the identity matrix on the left. Thus, this matrix has no inverse.
62. 12
3
6 1A
−= −
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
12
12
2 1 2
1 116 3
1 13
3 1 0
6 1 0 1
3 1 0 2
0 0 2 1
1 0
0 0 2 1
R r r
R r
− −
−→ = +
− −
→ = −
There is no way to obtain the identity matrix on the left. Thus, this matrix has no inverse.
63. 15 3
10 2A
=
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
22 1 22 3
3
1 15 15 1
1 11523
15 3 1 0
10 2 0 1
15 3 1 0
0 0 1
1 0
0 0 1
R r r
R r
→ = − + −
→ =
−
There is no way to obtain the identity matrix on the left; thus, there is no inverse.
64. 3 0
4 0A
− =
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
42 1 24 3
3
13 1
1 1343
3 0 1 0
4 0 0 1
3 0 1 0
0 0 1
1 0 0
0 0 1
R r r
R r
−
− → = +
−
→ = −
There is no way to obtain the identity matrix on the left; thus, there is no inverse.
65.
3 1 1
1 4 7
1 2 5
A
− − = − −
Augment the matrix with the identity and use row operations to find the inverse:
3 1 1 1 0 0
1 4 7 0 1 0
1 2 5 0 0 1
− − − −
( )
1 3
2 1 2
3 1 3
1 1 12 26 6 6
1 2 5 0 0 1Interchange
1 4 7 0 1 0 and
3 1 1 1 0 0
1 2 5 0 0 1
0 6 12 0 1 1 3
0 7 14 1 0 3
1 2 5 0 0 1
0 1 2 0
0 7 14 1 0 3
r r
R r r
R r r
R r
→ − − − −
= − + → − − − = +
→ − = −
1 23 3
1 2 11 16 6
3 2 37 116 6
1 0 1 02
0 1 2 0 7
0 0 0 1
R r r
R r r
= − +
→ − = − +
There is no way to obtain the identity matrix on the left; thus, there is no inverse.
Chapter 11: Systems of Equations and Inequalities
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66. 1 1 3
2 4 1
5 7 1
A
− = − −
Augment the matrix with the identity and use row operations to find the inverse:
1 1 3 1 0 0
2 4 1 0 1 0
5 7 1 0 0 1
− − −
( )
2 1 2
3 1 3
7 1 1 12 26 3 6 6
11 2 16 3 6
1 2 17 1 16 3 6
3 2 3
1 1 3 1 0 02
0 6 7 2 1 0 5
0 12 14 5 0 1
1 1 3 1 0 0
0 1 0
0 12 14 5 0 1
1 0 0
0 1 0 12
0 0 0 1 2 1
R r r
R r r
R r
R r r
R r r
− = − + → − − = + −
−
→ − − = − − − = − + → − − = − +
There is no way to obtain the identity matrix on the left; thus, there is no inverse.
67. 25 61 12
18 2 4
8 35 21
A
− = −
Thus, 1
0.01 0.05 0.01
0.01 0.02 0.01
0.02 0.01 0.03
A−−
≈ − −
68. 18 3 4
6 20 14
10 25 15
A
− = − −
Thus, 1
0.26 0.29 0.20
1.21 1.63 1.20
1.84 2.53 1.80
A−− −
≈ − −
69.
44 21 18 6
2 10 15 5
21 12 12 4
8 16 4 9
A
− = − − −
Thus, 1
0.02 0.04 0.01 0.01
0.02 0.05 0.03 0.03
0.02 0.01 0.04 0.00
0.02 0.06 0.07 0.06
A−
− − − − ≈ − −
.
70.
16 22 3 5
21 17 4 8
2 8 27 20
5 15 3 10
A
− − = − −
Thus, 1
0.01 0.04 0.00 0.03
0.02 0.02 0.01 0.01
0.04 0.02 0.04 0.06
0.05 0.02 0.00 0.09
A−
− = − − −
.
71. 25 61 12 10
18 12 7 ; 9
3 4 1 12
A B
− = − = − −
Enter the matrices into a graphing utility and use 1A B− to solve the system. The result is shown
below:
Thus, the solution to the system is 4.57x ≈ , 6.44y ≈ − , 24.07z ≈ − or (4.57, 6.44, 24.07)− − .
Section 11.4: Matrix Algebra
1139
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72. 25 61 12 15
18 12 7 ; 3
3 4 1 12
A B
− = − = − −
Enter the matrices into a graphing utility and use 1A B− to solve the system. The result is shown
below:
Thus, the solution to the system is 4.56x ≈ , 6.06y ≈ − , 22.55z ≈ − or (4.56, 6.06, 22.55)− − .
73. 25 61 12 21
18 12 7 ; 7
3 4 1 2
A B
− = − = − −
Thus, the solution to the system is 1.19x ≈ − , 2.46y ≈ , 8.27z ≈ or ( 1.19, 2.46, 8.27)− .
74. 25 61 12 25
18 12 7 ; 10
3 4 1 4
A B
− = − = − −
Thus, the solution to the system is 2.05x ≈ − ,
3.88y ≈ , 13.36z ≈ or ( 2.05, 3.88, 13.36)− .
75. 2 3 11
5 7 24
x y
x y
+ = + =
Multiply each side of the first equation by 5, and each side of the second equation by 2− . Then add the equations to eliminate x:
10 15 55
10 14 48
x y
x y
+ =− − = −
7y =
Substitute and solve for x:
( )2 3 7 11
2 21 11
2 10
5
x
x
x
x
+ =+ =
= −= −
The solution of the system is 5, 7x y= − = or
using ordered pairs ( )5,7− .
76. 2 8 8
7 13
x y
x y
+ = − + = −
Multiply each side of the second equation by 2− . Then add the equations to eliminate x: 2 8 8
2 14 26
x y
x y
+ = −− − =
6 18
3
y
y
− == −
Substitute and solve for x:
( )7 3 13
21 13
8
x
x
x
+ − = −− = −
=
The solution of the system is 8, 3x y= = − or
using ordered pairs ( )8, 3− .
77. 2 4 2
3 5 2 17
4 3 22
x y z
x y z
x y
− + =− + − = − = −
Write the augmented matrix:
1 2 4 2
3 5 172
3 04 22
− − − − −
2 1 2
3 1 3
1 2 4 20 10 231 30 5 16 30 4
R r rR r r
− − = + − − = − +
2 2
1 2 4 20 10 2310 5 16 30
R r − − − = − − −
3 2 3
1 2 4 20 10 2310 0 34 85 5R r r
− − − = − +
52 3 3
1 2 4 20 10 231
/ 340 0 1 R r
− − −
=
1 3 1
2 3 252
0 81 2 40 01 2 10
0 0 1
R r rR r r
−− = − + = +
Chapter 11: Systems of Equations and Inequalities
1140
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
1 2 1
52
0 01 4 20 01 2
0 0 1
R r r − = +
The solution is 5
4, 2,2
x y z= − = = or 5
4,2,2
−
.
78. 2 3 2
4 3 6
6 2 2
x y z
x z
y z
+ − = − + = − =
Write the augmented matrix:
32 1 2
0 3 64
0 6 2 2
− − −
3 12 2 1 11 1 / 2
0 3 64
0 6 2 2
R r − − =
−
3 12 2
2 1 2
1 1
0 6 5 10 4
0 6 2 2
R r r
− − − = − +
−
( )
3 12 2
5 52 26 3
1 1
/ 60 1
0 6 2 2
R r
− − − − = − −
3 12 2
5 56 3
3 2 3
1 1
0 160 0 3 12 R r r
− − − −
= − +
3 12 2
5 56 3
3 3
1 1
0 1/30 0 1 4 R r
− − − −
=
312 1 3 12
553 2 3 26
01 1
0 01
0 0 1 4
R r r
R r r
= + = +
332 1 2 12
53
0 01
0 01
0 0 1 4
R r r − = − +
The solution is 3 5
, , 42 3
x y z= − = = or
3 5, , 4
2 3 −
.
79. 5 4 2
5 4 3
7 13 4 17
x y z
x y z
x y z
− + = − + − = + − =
Write the augmented matrix:
5 1 4 2
5 31 4
7 13 174
− − − −
1 2
2 1
5 31 4
5 1 4 2
7 13 174
R r
R r
− − = −
− = −
2 1 2
3 1 3
5 31 4
0 16 1724 5
0 48 32 38 7
R r r
R r r
− −
− = − + − = − +
2 172 23 24
5 31 4
/ 240 1
0 48 32 38
R r
− − − =
−
2 173 24
3 2 3
5 31 4
0 1480 0 0 4 R r r
− − − = − +
The last row of our matrix is a contradiction. Therefore, the system is inconsistent. The solution set is { } , or ∅ .
80. 3 2 2
2 6 7
2 2 14 17
x y z
x y z
x y z
+ − = + + = − + − =
Write the augmented matrix:
3 2 1 2
6 72 1
172 2 14
−
− −
1 1 27 91 1
6 72 1
172 2 14
R r r − = −
− −
2 1 2
3 1 3
7 91 1
0 20 251 2
0 0 0 1 2
R r r
R r r
−
−− = − + − = − +
The last row of our matrix is a contradiction. Therefore, the system is inconsistent. The solution set is { } , or ∅ .
Section 11.4: Matrix Algebra
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81. 2 3 4
3 2 3
5 6
x y z
x y z
y z
− + =− + − = − − + =
Write the augmented matrix:
32 1 4
3 32 1
0 5 61
− − −− −
3 12 2 1 11 2 / 2
3 32 1
0 5 61
R r − = − −−
−
3 12 2
5 12 1 22 2
1 2
30 3
0 5 61
R r r
− − = + −
3 12 2
61 25 5 2 25
1 2
0 1
0 5 61
R r
− − − = − −
3 12 2
615 5
3 2 3
1 2
0 150 0 0 0 R r r
− − −
= +
1 135 5 1 2 12
615 5
01
0 1
0 0 0 0
R r r = + − −
Since the last row yields an identity, and no contradictions exist in the other rows, there are an infinite number of solutions. The solution is
1 1
5 5x z= − + ,
1 6
5 5y z= − , and z is any real
number. That is,
( )
}
1 1 1 6, , | , ,
5 5 5 5
is any real number
x y z x z y z
z
= − + = −
82. 4 3 2 6
3 2
9 6
x y z
x y z
x y z
− + + = + − = − + + =
Write the augmented matrix:
3 64 2
3 1 1 2
9 61 1
−
− −
1 3
3 1
9 61 1
3 1 1 2
3 64 2
R r
R r
=
− − − =
2 1 2
3 1 3
9 61 1
0 26 204 3
0 39 6 30 4
R r r
R r r
− −− = − + = +
( )1022 213 13
9 61 1
/ 260 1
0 39 6 30
R r
= −
10213 13
3 2 3
9 61 1
0 1390 0 0 0 R r r
= − +
5 1213 13 1 2 1
10213 13
01 9
0 1
0 0 0 0
R r r − − = − +
Since the last row yields an identity, and no contradictions exist in the other rows, there are an infinite number of solutions. The solution is
5 12
13 13x z= − ,
2 10
13 13y z= − + , and z is any real
number. That is,
( )
}
5 12 2 10, , | , ,
13 13 13 13
is any real number
x y z x z y z
z
= − = − +
83. a. 6 9 80.00
; 3 12 277.80
A B
= =
b. 6 9 80.00
3 12 277.80
6(80.00) 9(277.80) 2980.20
3(80.00) 12(277.80) 3573.60
AB
=
+ = = +
Nikki’s total tuition is $2980.20, and Joe’s total tuition is $3573.60.
84. a. 4000 3000 0.011
; 2500 3800 0.006
A B
= =
b. 4000 3000 0.011
2500 3800 0.006
4000(0.011) 3000(0.006) 62.00
2500(0.011) 3800(0.006) 50.30
AB
=
+ = = +
After one month, Jamal’s loans accrued
Chapter 11: Systems of Equations and Inequalities
1142
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
$62.00 in interest, and Stephanie’s loans accrued $50.30 in interest.
c. 4000 3000 1 0.011
( )2500 3800 1 0.006
4000 3000 1.011
2500 3800 1.006
4000(1.011) 3000(1.006)
2500(1.011) 3800(1.006)
7062.00
6350.30
A C B + = +
=
+ = + =
Jamal’s loan balance after one month was $7062.00, and Stephanie’s loan balance was $6350.30.
85. a. The rows of the 2 by 3 matrix represent stainless steel and aluminum. The columns represent 10-gallon, 5-gallon, and 1-gallon.
The 2 by 3 matrix is: 500 350 400
700 500 850
.
The 3 by 2 matrix is:
500 700
350 500
400 850
.
b. The 3 by 1 matrix representing the amount of
material is:
15
8
3
.
c. The days usage of materials is: 15
500 350 400 11,5008
700 500 850 17,0503
⋅ =
Thus, 11,500 pounds of stainless steel and 17,050 pounds of aluminum were used that day.
d. The 1 by 2 matrix representing cost is:
[ ]0.10 0.05 .
e. The total cost of the day’s production was:
[ ] [ ]11,5000.10 0.05 2002.50
17,050
⋅ =
.
The total cost of the day’s production was $2002.50.
86. a. The rows of the 2 by 3 matrix represent the location. The columns represent the type of car sold. The 2 by 3 matrix for January is:
400 250 50
450 200 140
. The 2 by 3 matrix for
February is: 350 100 30
350 300 100
.
b. Adding the matrices: 400 250 50 350 100 30
450 200 140 350 300 100
750 350 80
800 500 240
+
=
c. The 3 by 1 matrix representing profit:
100
150
200
.
d. Multiplying to find the profit at each location: 100
750 350 80 143,500150
800 500 240 203,000200
⋅ =
.
The city location has a two-month profit of $143,500. The suburban location has a two-month profit of $203,000.
87. a.
2 1 1
1 1 0
1 1 1
K =
Augment the matrix with the identity and use row operations to find the inverse:
2 1 1 1 0 0
1 1 0 0 1 0
1 1 1 0 0 1
1 2
2 1 2
3 1 3
1 1 0 0 1 0Interchange
2 1 1 1 0 0 and
1 1 1 0 0 1
1 1 0 0 1 02
0 1 1 1 2 0
0 0 1 0 1 1
r r
R r r
R r r
→
= − + → − − = − + −
Section 11.4: Matrix Algebra
1143
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
( )
( )
( )
2 2
2 2 3
1 1 2
1 1 0 0 1 0
0 1 1 1 2 0
0 0 1 0 1 1
1 1 0 0 1 0
0 1 0 1 1 1
0 0 1 0 1 1
1 0 0 1 0 1
0 1 0 1 1 1
0 0 1 0 1 1
R r
R r r
R r r
→ − − = − − → − = + −
− → − = − −
Thus, 1
1 0 1
1 1 1
0 1 1
K −−
= − −
.
b. 1
47 34 33 1 0 1
44 36 27 1 1 1
47 41 20 0 1 1
M E K −−
= ⋅ = − −
13 1 20
8 9 19
6 21 14
=
because
11
12
13
21
22
23
31
32
33
47(1) 34( 1) 33(0) 13
47(0) 34(1) 33( 1) 1
47( 1) 34(1) 33(1) 20
44(1) 36( 1) 27(0) 8
44(0) 36(1) 27( 1) 9
44( 1) 36(1) 27(1) 19
47(1) 41( 1) 20(0) 6
47(0) 41(1) 20( 1) 21
a
a
a
a
a
a
a
a
a
= + − + == + + − == − + + == + − + == + + − == − + + == + − + == + + − == 47( 1) 41(1) 20(1) 14− + + =
c. 13 ;1 ; 20 ; 8 ; 9 ;
19 ;6 ; 21 ;14
M A T H I
S F U N
→ → → → →→ → → →
The message: Math is fun.
88. 2
0.4 0.2 0.1 0.4 0.2 0.1
0.5 0.6 0.5 0.5 0.6 0.5
0.1 0.2 0.4 0.1 0.2 0.4
0.27 0.22 0.18
0.55 0.56 0.55
0.18 0.22 0.27
P
= =
because
11
12
13
21
22
23
0.4(0.4) 0.2(0.5) 0.1(0.1) 0.27
0.4(0.2) 0.2(0.6) 0.1(0.2) 0.22
0.4(0.1) 0.2(0.5) 0.1(0.4) 0.18
0.5(0.4) 0.6(0.5) 0.5(0.1) 0.55
0.5(0.2) 0.6(0.6) 0.5(0.2) 0.56
0.5(0.1) 0.6(0.5) 0.5
a
a
a
a
a
a
= + + == + + == + + == + + == + + == + +
31
32
33
(0.4) 0.55
0.1(0.4) 0.2(0.5) 0.4(0.1) 0.18
0.1(0.2) 0.2(0.6) 0.4(0.2) 0.22
0.1(0.1) 0.2(0.5) 0.4(0.4) 0.27
a
a
a
== + + == + + == + + =
Each entry represents the probability that a grandchild has a certain income level given his or her grandparents’ income level.
89. a b
Ac d
=
If 0D ad bc= − ≠ , then 0a ≠ and 0d ≠ , or 0b ≠ and 0c ≠ . Assuming the former, then
( )
( )
11
1 1
1
2 1 2
1
1 0
0 1
1 0
0 1
1 0
0 1
1 0
0 1
ba a
a
ba a
bc ca a
ba a
ad bc ca a
a b
c d
R rc d
R c r rd
−
→ = ⋅
→ = − ⋅ + − −
→
−
( )
( )
1
2 2
1( )
1 2 1
1 0
0 1
1 0
0 1
1 0
0 1
1 0
0 1
ba a a
ad bcc aad bc ad bc
bc ba a ad bc ad bc b
ac aad bc ad bc
d bad bc ad bc
c aad bc ad bc
d bD D
c aD D
R r
R r r
−−− −
−− −
−− −
−− −
−− −
→ = ⋅ + → = − ⋅ + → − → −
Thus, 1 1d bD D
c aD D
d bA
c aD− − −
= = − −
where D ad bc= − .
90. Answers will vary.
91. Since the product is found by multiplying the components from the columns of the first matrix
Chapter 11: Systems of Equations and Inequalities
1144
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
by the components in the rows of the second matrix and then adding those products, then the number of columns in the first must equal the number of rows in the second.
92. For real numbers this you multiply both sides by
the multiplicative inverse of c, 1c , c not equal 0.
Since c times 1c results in 1, the multiplicative
identity, then a = b. For matrices, this would hold true as long as C has an inverse. Then you
would multiply both sides by 1C− to get:
1 1
AC BC
ACC BCC
AI BI
A B
− −
==
==
93. If the inverse of A exists:
1 1
0
0
0
0
AX
A AX A
IX
X
− −
==
==
If the inverse of A does not exist, then A is singular and you would not be able to multiply A by its inverse to give the identity inverse and X would have no solution.
Section 11.5
1. True
2. True
3. ( )( )
4 3 2 2 2
22
3 6 3 3 2 1
3 1
x x x x x x
x x
+ + = + +
= +
4. True
5. The rational expression 2 1
x
x − is proper, since
the degree of the numerator is less than the degree of the denominator.
6. The rational expression 3
5 2
1
x
x
+−
is proper, since
the degree of the numerator is less than the degree of the denominator.
7. The rational expression 2
2
5
4
x
x
+−
is improper, so
perform the division:
2 2
2
1
4 5
4
9
x x
x
− +
−
The proper rational expression is: 2
2 2
5 91
4 4
x
x x
+ = +− −
8. The rational expression 2
2
3 2
1
x
x
−−
is improper, so
perform the division:
2 2
2
3
1 3 2
3 3
1
x x
x
− −
−
The proper rational expression is: 2
2 2
3 2 13
1 1
x
x x
− = +− −
9. The rational expression 3
2
5 2 1
4
x x
x
+ −−
is improper,
so perform the division:
2 3
3
5
4 5 2 1
5 20
22 1
x
x x x
x x
x
− + −−
−
The proper rational expression is: 3
2 2
5 2 1 22 1
4 45
x x x
x xx
+ − −− −
= +
10. The rational expression 4 2
3
3 2
8
x x
x
+ −+
is improper,
so perform the division:
3 4 2
4
2
3
8 3 2
3 24
24 2
x
x x x
x x
x x
+ + −+
− −
The proper rational expression is: 4 2 2
3 3
3 2 24 2
8 83
x x x x
x xx
+ − − −+ +
= +
Section 11.5: Partial Fraction Decomposition
1145
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
11. The rational expression 2
2
( 1)
( 4)( 3) 12
x x x x
x x x x
− −=+ − + −
is improper, so
perform the division:
2 2
2
1
12 0
122 12
x x x x
x xx
+ − − ++ −
− +
The proper rational expression is:
2
( 1) 2 12 2( 6)
( 4)( 3) ( 4)( 3)121 1
x x x x
x x x xx x
− − + − −+ − + −+ −
= + = +
12. The rational expression 2 3
2 2
2 ( 4) 2 8
1 1
x x x x
x x
+ +=+ +
is
improper, so perform the division:
2 3
3
2
1 2 8
2 26
x
x x x
x xx
+ ++
The proper rational expression is:
2
2 2
2 ( 4) 6
1 12
x x x
x xx
++ +
= +
13. Find the partial fraction decomposition: 4
( 1) 1
4
( 1) 1( 1) ( 1)
4 ( 1)
A B
x x x x
A B
x x x xx x x x
A x Bx
= +− −
+ − − − = −
= − +
Let 1x = , then 4 (0)
4
A B
B
= +=
Let 0x = , then 4 ( 1) (0)
4
A B
A
= − += −
4 4 4
( 1) 1x x x x
−= +− −
14. Find the partial fraction decomposition 3
( 2)( 1) 2 1
x A B
x x x x= +
+ − + −
Multiplying both sides by ( 2)( 1)x x+ − , we
obtain: 3 ( 1) ( 2)x A x B x= − + +
Let 1x = , then 3(1) (0) (3)
3 31
A B
B
B
= +==
Let 2x = − , then 3(– 2) ( 3) (0)
3 62
A B
A
A
= − +− = −
=
3 2 1
( 2)( 1) 2 1
x
x x x x= +
+ − + −
15. Find the partial fraction decomposition:
2 2
2 22 2
2
1
( 1) 1
1
( 1) 1( 1) ( 1)
1 ( 1) ( )
A Bx C
xx x x
A Bx C
xx x xx x x x
A x Bx C x
+= ++ +
+ + + + + = +
= + + +
Let 0x = , then 21 (0 1) ( (0) )(0)
1
A B C
A
= + + +=
Let 1x = , then 21 (1 1) ( (1) )(1)
1 2
1 2(1)
1
A B C
A B C
B C
B C
= + + += + += + +
+ = −
Let 1x = − , then 21 (( 1) 1) ( ( 1) )( 1)
1 (1 1) ( )( 1)
1 2
1 2(1)
1
A B C
A B C
A B C
B C
B C
= − + + − + −= + + − + −= + −= + −
− = −
Solve the system of equations: 11
2 21
B CB C
BB
+ = −− = −
= −= −
1 1
0CC
− + = −=
2 2
1 1
( 1) 1
x
xx x x
−= ++ +
16. Find the partial fraction decomposition:
2 2
1
1( 1)( 4) 4
A Bx C
xx x x
+= +++ + +
Multiplying both sides by 2( 1)( 4)x x+ + , we
obtain: 21 ( 4) ( )( 1)A x Bx C x= + + + +
Chapter 11: Systems of Equations and Inequalities
1146
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Let 1x = − , then 1 (5) ( ( 1) )(0)
5 1
1
5
A B C
A
A
= + − +=
=
Let 1x = , then
( )
21 (1 4) ( (1) )(1 1)
1 5 ( )(2)
1 5 1/ 5 2 2
1 1 2 2
0 2 2
0
A B C
A B C
B C
B C
B C
B C
= + + + += + += + += + += += +
Let 0x = , then
( )
21 (0 4) ( (0) )(0 1)
1 4
1 4 1/ 5
41
51
5
A B C
A C
C
C
C
= + + + += += +
= +
=
Since 0B C+ = , we have that 1
05
1
5
B
B
+ =
= −
1 1 15 5 5
2 2
1
1( 1)( 4) 4
x
xx x x
− += +
++ + +
17. Find the partial fraction decomposition:
( 1)( 2) 1 2
x A B
x x x x= +
− − − −
Multiplying both sides by ( 1)( 2)x x− − , we
obtain: ( 2) ( 1)x A x B x= − + −
Let 1x = , then 1 (1 2) (1 1)
1
1
A B
A
A
= − + −= −= −
Let 2x = , then 2 (2 2) (2 1)
2
A B
B
= − + −=
1 2
( 1)( 2) 1 2
x
x x x x
−= +− − − −
18. Find the partial fraction decomposition: 3
( 2)( 4) 2 4
x A B
x x x x= +
+ − + −
Multiplying both sides by ( 2)( 4)x x+ − , we
obtain: 3 ( 4) ( 2)x A x B x= − + +
Let 2x = − , then 3( 2) ( 2 4) ( 2 2)
6 6
1
A B
A
A
− = − − + − +− = −
=
Let 4x = , then 3(4) (4 4) (4 2)
12 6
2
A B
B
B
= − + +==
3 1 2
( 2)( 4) 2 4
x
x x x x= +
+ − + −
19. Find the partial fraction decomposition: 2
2 21 1( 1) ( 1) ( 1)
x A B C
x xx x x= + +
− +− + −
Multiplying both sides by 2( 1) ( 1)x x− + , we
obtain: 2 2( 1)( 1) ( 1) ( 1)x A x x B x C x= − + + + + −
Let 1x = , then 2 2
2
1 (1 1)(1 1) (1 1) (1 1)
1 (0)(2) (2) (0)
1 2
1
2
A B C
A B C
B
B
= − + + + + −
= + +=
=
Let 1x = − , then 2 2
2
( 1) ( 1 1)( 1 1) ( 1 1) ( 1 1)
1 ( 2)(0) (0) ( 2)
1 4
1
4
A B C
A B C
C
C
− = − − − + + − + + − −
= − + + −=
=
Let 0x = , then 2 20 (0 1)(0 1) (0 1) (0 1)
0
1 1 3
2 4 4
A B C
A B C
A B C
A
= − + + + + −= − + += +
= + =
3 1 124 2 4
2 21 1( 1) ( 1) ( 1)
x
x xx x x= + +
− +− + −
20. Find the partial fraction decomposition:
2 2
1
2( 2)
x A B C
x xx x x
+ = + +−−
Multiplying both sides by 2( 2)x x − , we obtain: 21 ( 2) ( 2)x Ax x B x Cx+ = − + − +
Section 11.5: Partial Fraction Decomposition
1147
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Let 0x = , then 20 1 (0)(0 2) (0 2) (0)
1 2
1
2
A B C
B
B
+ = − + − += −
= −
Let 2x = , then 22 1 (2)(2 2) (2 2) (2)
3 4
3
4
A B C
C
C
+ = − + − +=
=
Let 1x = , then 2
2
1 3 32
2 4 4
A B C
A B C
A
= − − += − + −
= − − + − = −
3 314 2 4
2 2
1
2( 2)
x
x xx x x
− −+ = + +−−
21. Find the partial fraction decomposition:
3 2
2 2
1 1
8 ( 2)( 2 4)
1
2( 2)( 2 4) 2 4
x x x x
A Bx C
xx x x x x
=− − + +
+= +−− + + + +
Multiplying both sides by 2( 2)( 2 4)x x x− + + ,
we obtain: 21 ( 2 4) ( )( 2)A x x Bx C x= + + + + −
Let 2x = , then
( )21 2 2(2) 4 ( (2) )(2 2)
1 12
1
12
A B C
A
A
= + + + + −
=
=
Let 0x = , then
( )
( )
2
23
1 0 2(0) 4 ( (0) )(0 2)
1 4 2
1 4 1/12 2
2
13
A B C
A C
C
C
C
= + + + + −
= −= −
− =
= −
Let 1x = , then
( )
( )
21 1 2(1) 4 ( (1) )(1 2)
1 7
11 7 1/12
31
12
A B C
A B C
B
B
= + + + + −
= − −
= − +
= −
( )
1 1112 312
3 2
1 112 12
2
1
28 2 4
4
2 2 4
x
xx x x
x
x x x
− −= +
−− + +− +
= +− + +
22. Find the partial fraction decomposition:
3 2 2
2 4 2 4
11 ( 1)( 1) 1
x x A Bx C
xx x x x x x
+ + += = +−− − + + + +
Multiplying both sides by 2( 1)( 1)x x x− + + , we
obtain: 22 4 ( 1) ( )( 1)x A x x Bx C x+ = + + + + −
Let 1x = , then
( ) ( )22(1) 4 1 1 1 (1) (1 1)
6 3
2
A B C
A
A
+ = + + + + −
==
Let 0x = , then
( )22(0) 4 0 0 1 ( (0) )(0 1)
4
4 2
2
A B C
A C
C
C
+ = + + + + −
= −= −= −
Let 1x = − , then
( )22( 1) 4 ( 1) ( 1) 1 ( ( 1) )( 1 1)
2 2 2
2 2 2 2( 2)
2 4
2
A B C
A B C
B
B
B
− + = − + − + + − + − −
= + −= + − −= −= −
3 2
2 4 2 2 2
11 1
x x
xx x x
+ − −= +−− + +
23. Find the partial fraction decomposition: 2
2 2 2 21 1( 1) ( 1) ( 1) ( 1)
x A B C D
x xx x x x= + + +
− +− + − +
Multiplying both sides by 2 2( 1) ( 1)x x− + , we obtain:
2 2 2
2 2
( 1)( 1) ( 1)
( 1) ( 1) ( 1)
x A x x B x
C x x D x
= − + + +
+ − + + −
Chapter 11: Systems of Equations and Inequalities
1148
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Let 1x = , then 2 2 2
2 2
1 (1 1)(1 1) (1 1)
(1 1) (1 1) (1 1)
1 4
1
4
A B
C D
B
B
= − + + +
+ − + + −=
=
Let 1x = − , then 2 2 2
2 2
( 1) ( 1 1)( 1 1) ( 1 1)
( 1 1) ( 1 1) ( 1 1)
1 4
1
4
A B
C D
D
D
− = − − − + + − +
+ − − − + + − −=
=
Let 0x = , then 2 2 2
2 2
0 (0 1)(0 1) (0 1)
(0 1) (0 1) (0 1)
0
1 1 1
4 4 2
A B
C D
A B C D
A C B D
A C
= − + + +
+ − + + −= − + + +
− = +
− = + =
Let 2x = , then 2 2 2
2 2
2 (2 1)(2 1) (2 1)
(2 1) (2 1) (2 1)
4 9 9 3
9 3 4 9
1 1 39 3 4 9
4 4 2
13
2
A B
C D
A B C D
A C B D
A C
A C
= − + + +
+ − + + −= + + +
+ = − −
+ = − − =
+ =
Solve the system of equations: 1213 2
A C
A C
− =
+ =
4 1
14
3 14 2
14
A
A
C
C
=
=
+ =
= −
1 1 1 124 4 4 4
2 2 2 21 1( 1) ( 1) ( 1) ( 1)
x
x xx x x x
−= + + +
− +− + − +
24. Find the partial fraction decomposition:
2 2 2 2
1
2( 2) ( 2)
x A B C D
x xx x x x
+ = + + +−− −
Multiplying both sides by 2 2( 2)x x − , we obtain: 2 2 2 21 ( 2) ( 2) ( 2)x Ax x B x Cx x Dx+ = − + − + − +
Let 0x = , then 2 2
2 2
0+1 (0)(0 2) (0 2)
(0) (0 2) (0)
1 4
1
4
A B
C D
B
B
= − + −
+ − +=
=
Let 2x = , then 2 2
2 2
2 1 (2)(2 2) (2 2)
(2) (2 2) (2)
3 4
3
4
A B
C D
D
D
+ = − + −
+ − +=
=
Let 1x = , then 2 2
2 2
1 1 (1)(1 2) (1 2)
(1) (1 2) (1)
22
1 32 1
4 4
A B
C D
A B C D
A C B D
A C
+ = − + −+ − +
= + − +− = − −
− = − − =
Let 3x = , then 2 2
2 2
3 1 (3)(3 2) (3 2)
(3) (3 2) (3)
4 3 9 9
3 9 4 9
1 33 9 4 9 3
4 43 1
A B
C D
A B C D
A C B D
A C
A C
+ = − + −+ − +
= + + ++ = − −
+ = − − = −
+ = −
Solve the system of equations: 1
3 1
A C
A C
− = + = −
( )
1
1
3 1
1 3 1
A C
A C
A C
C C
− == +
+ = −+ + = −
4 2
1
2
C
C
= −
= −
Section 11.5: Partial Fraction Decomposition
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1 11 1
2 2A C= + = − + =
31 1 12 4 2 4
2 2 2 2
1
2( 2) ( 2)
x
x xx x x x
−+ = + + +−− −
25. Find the partial fraction decomposition:
2 2
3
2 1( 2)( 1) ( 1)
x A B C
x xx x x
− = + ++ ++ + +
Multiplying both sides by 2( 2)( 1)x x+ + , we
obtain: 23 ( 1) ( 2)( 1) ( 2)x A x B x x C x− = + + + + + +
Let 2x = − , then 22 3 ( 2 1) ( 2 2)( 2 1) ( 2 2)
5
5
A B C
A
A
− − = − + + − + − + + − +− =
= −
Let 1x = − , then 21 3 ( 1 1) ( 1 2)( 1 1) ( 1 2)
44
A B C
C
C
− − = − + + − + − + + − +− =
= −
Let 0x = , then 20 3 (0 1) (0 2)(0 1) (0 2)
3 2 2
3 5 2 2( 4)
2 10
5
A B C
A B C
B
B
B
− = + + + + + +− = + +− = − + + −
==
2 2
3 5 5 4
2 1( 2)( 1) ( 1)
x
x xx x x
− − −= + ++ ++ + +
26. Find the partial fraction decomposition: 2
2 22 1( 2)( 1) ( 1)
x x A B C
x xx x x
+ = + ++ −+ − −
Multiplying both sides by 2( 2)( 1)x x+ − , we
obtain: 2 2( 1) ( 2)( 1) ( 2)x x A x B x x C x+ = − + + − + +
Let 2x = − , then 2 2( 2) ( 2) ( 2 1) ( 2 2)( 2 1)
( 2 2)
2 9
2
9
A B
C
A
A
− + − = − − + − + − −+ − +
=
=
Let 1x = , then 2 21 1 (1 1) (1 2)(1 1) (1 2)
2 3
2
3
A B C
C
C
+ = − + + − + +=
=
Let 0x = , then 2 20 0 (0 1) (0 2)(0 1) (0 2)
0 2 2
2 2
2 2 142 2
9 3 9
7
9
A B C
A B C
B A C
B
B
+ = − + + − + += − += +
= + =
=
72 229 9 3
2 22 1( 2)( 1) ( 1)
x x
x xx x x
+ = + ++ −+ − −
27. Find the partial fraction decomposition:
2 2 2 2
4
( 4) 4
x A B Cx D
xx x x x
+ += + ++ +
Multiplying both sides by 2 2( 4)x x + , we obtain: 2 2 24 ( 4) ( 4) ( )x Ax x B x Cx D x+ = + + + + +
Let 0x = , then
( )2 2 20 4 (0)(0 4) (0 4) (0) (0)
4 41
A B C D
BB
+ = + + + + +==
Let 1x = , then 2 2 21 4 (1)(1 4) (1 4) ( (1) )(1)
5 5 55 5 5
A B C DA B C DA C D
+ = + + + + += + + += + + +
5 0A C D+ + =
Let 1x = − , then 2 2
2
1 4 ( 1)(( 1) 4) (( 1) 4)
( ( 1) )( 1)3 5 53 5 5
A B
C DA B C DA C D
− + = − − + + − ++ − + −
= − + − += − + − +
5 2A C D− − + = −
Let 2x = , then 2 2 22 4 (2)(2 4) (2 4) ( (2) )(2)
6 16 8 8 46 16 8 8 4
A B C DA B C DA C D
+ = + + + + += + + += + + +
16 8 4 2A C D+ + = −
Chapter 11: Systems of Equations and Inequalities
1150
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Solve the system of equations: 5 05 2
2 21
A C DA C D
DD
+ + =− − + = −
= −= −
5 1 0
1 5A C
C A+ − =
= −
16 8(1 5 ) 4( 1) 216 8 40 4 2
24 61
4
A AA A
A
A
+ − + − = −+ − − = −
− = −
=
1 5 1
1 5 14 4 4
C = − = − = −
( )
1 14 4
2 2 2 2
1 14 4
2 2
14 1
( 4) 4
41
4
xx
xx x x x
x
x x x
− −+ = + ++ +
− += + +
+
28. Find the partial fraction decomposition: 2
2 2 2 2
10 2
1( 1) ( 2) ( 1) 2
x x A B Cx D
xx x x x
+ += + +−− + − +
Multiply both sides by 2 2( 1) ( 2)x x− + : 2 2 2
2
10 2 ( 1)( 2) ( 2)
( )( 1)
x x A x x B x
Cx D x
+ = − + + +
+ + −
Let 1x = , then
( )
2 2 2
2
10(1) 2(1) (1 1)(1 2) (1 2)
(1) (1 1)
12 3
4
A B
C D
B
B
+ = − + + +
+ + −==
Let 0x = , then
( )
2 2 2
2
10(0) 2(0) (0 1)(0 2) (0 2)
(0) (0 1)
0 2 2
0 2 8
2 8
2 8
A B
C D
A B D
A D
A D
D A
+ = − + + +
+ + −= − + += − + +
− == −
Let 1x = − , then 2 2
2
2
10( 1) 2( 1) ( 1 1)(( 1) 2)
(( 1) 2)
( ( 1) )( 1 1)
8 6 3 4 4
8 6 12 4 4
6 4 4 4
A
B
C D
A B C D
A C D
A C D
− + − = − − − +
+ − +
+ − + − −= − + − += − + − +
− − + = −
Let 2x = , then 2 2 2
2
10(2) 2(2) (2 1)(2 2) (2 2)
( (2) )(2 1)
44 6 6 2
44 6 24 2
6 2 20
A B
C D
A B C D
A C D
A C D
+ = − + + +
+ + −= + + += + + +
+ + =
Solve the system of equations (Substitute for D): 2 8
6 4 4 4
6 4 4(2 8) 4
2 4 28
2 14
D A
A C D
A C A
A C
A C
= −− − + = −
− − + − = −− =− =
( )6 2 20
6 2 2 8 20
8 2 28
A C D
A C A
A C
+ + =+ + − =
+ =
Add the equations and solve:
2 14
8 2 28
A C
A C
− =+ =
9 42
14
3
A
A
=
=
14 282 14 14
3 314
3
C A
C
= − = − = −
= −
14 42 8 2 8
3 3D A
= − = − =
14 14 423 3 3
2 2 2 2
10 2 4
1( 1) ( 2) ( 1) 2
xx x
xx x x x
− ++ = + +−− + − +
Section 11.5: Partial Fraction Decomposition
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29. Find the partial fraction decomposition: 2
2 2
2 3
1( 1)( 2 4) 2 4
x x A Bx C
xx x x x x
+ + += +++ + + + +
Multiplying both sides by 2( 1)( 2 4)x x x+ + + ,
we obtain: 2 22 3 ( 2 4) ( )( 1)x x A x x Bx C x+ + = + + + + +
Let 1x = − , then 2 2( 1) 2( 1) 3 (( 1) 2( 1) 4)
( ( 1) )( 1 1)
2 3
2
3
A
B C
A
A
− + − + = − + − ++ − + − +
=
=
Let 0x = , then
( )
2 20 2(0) 3 (0 2(0) 4) ( (0) )(0 1)
3 4
3 4 2/3
1
3
A B C
A C
C
C
+ + = + + + + += += +
=
Let 1x = , then
( ) ( )
2 2
14 2 23 3 3
13
1 2(1) 3 (1 2(1) 4) ( (1) )(1 1)
6 7 2 2
6 7 2 / 3 2 2 1/ 3
2 6
A B C
A B C
B
B
B
+ + = + + + + += + += + +
= − − =
=
2 1 123 3 3
2 2
2 13 3
2
2 3
1( 1)( 2 4) 2 4
( 1)
1 2 4
xx x
xx x x x x
x
x x x
++ + = +++ + + + +
+= +
+ + +
30. Find the partial fraction decomposition: 2
2 2
11 18
( 3 3) 3 3
x x A Bx C
xx x x x x
− − += ++ + + +
Multiplying both sides by 2( 3 3)x x x+ + 2( 1)( 2 4)x x x+ + + , we obtain:
2 211 18 ( 3 3) ( )x x A x x Bx C x− − = + + + +
Let 0x = , then
( ) ( )2 20 11(0) 18 0 3(0) 3 (0) (0)
18 3
6
A B C
A
A
− − = + + + +
− == −
Let 1x = , then
( ) ( )2 21 11(1) 18 1 3(1) 3 (1) (1)
28 7
28 7( 6)
14
A B C
A B C
B C
B C
− − = + + + +
− = + +− = − + ++ =
Let 1x = − , then
( )( )
2 2( 1) 11( 1) 18 ( 1) 3( 1) 3
( 1) ( 1)
6
6 6
0
A
B C
A B C
B C
B C
− − − − = − + − +
+ − + −− = + −− = − + −
− =
Add the last two equations and solve: 14
0
B C
B C
+ =− =
2 14
7
B
B
==
14
7 14
7
B C
C
C
+ =+ =
=
2
2 2
11 18 6 7 7
( 3 3) 3 3
x x x
xx x x x x
− − − += ++ + + +
31. Find the partial fraction decomposition:
(3 2)(2 1) 3 2 2 1
x A B
x x x x= +
− + − +
Multiplying both sides by (3 2)(2 1)x x− + , we
obtain: (2 1) (3 2)x A x B x= + + −
Let 12x = − , then
( )( ) ( )( )12 1/ 2 1 3 1/ 2 2
21 7
2 21
7
A B
B
B
− = − + + − −
− = −
=
Let 2
3x = , then
( )( ) ( )( )22 2 / 3 1 3 2 / 3 2
32 7
3 32
7
A B
A
A
= + + −
=
=
2 17 7
(3 2)(2 1) 3 2 2 1
x
x x x x= +
− + − +
Chapter 11: Systems of Equations and Inequalities
1152
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32. Find the partial fraction decomposition: 1
(2 3)(4 1) 2 3 4 1
A B
x x x x= +
+ − + −
Multiplying both sides by (2 3)(4 1)x x+ − , we
obtain: 1 (4 1) (2 3)A x B x= − + +
Let 3
2x = − , then
3 31 4 1 2 3
2 2
1 7
1
7
A B
A
A
= − − + − +
= −
= −
Let 1
4x = , then
1 11 4 1 2 3
4 4
71
22
7
A B
B
B
= − + +
=
=
1 27 71
(2 3)(4 1) 2 3 4 1x x x x
−= +
+ − + −
33. Find the partial fraction decomposition:
2 ( 3)( 1) 3 12 3
x x A B
x x x xx x= = +
+ − + −+ −
Multiplying both sides by ( 3)( 1)x x+ − , we
obtain: ( 1) ( 3)x A x B x= − + +
Let 1x = , then 1 (1 1) (1 3)
1 41
4
A B
B
B
= − + +=
=
Let 3x = − , then 3 ( 3 1) ( 3 3)
3 43
4
A B
A
A
− = − − + − +− = −
=
3 14 4
2 3 12 3
x
x xx x= +
+ −+ −
34. Find the partial fraction decomposition: 2 2
2
8 8
( 1)( 2)( 3)( 1)( 5 6)
1 2 3
x x x x
x x xx x xA B C
x x x
− − − −=+ + ++ + +
= + ++ + +
Multiplying both sides by ( 1)( 2)( 3)x x x+ + + ,
we obtain: 2 8 ( 2)( 3) ( 1)( 3)
( 1)( 2)
x x A x x B x x
C x x
− − = + + + + ++ + +
Let 1x = − , then 2( 1) ( 1) 8 ( 1 2)( 1 3)
( 1 1)( 1 3)
( 1 1)( 1 2)
6 2
3
A
B
C
A
A
− − − − = − + − ++ − + − +
+ − + − +− =
= −
Let 2x = − , then 2( 2) ( 2) 8 ( 2 2)( 2 3)
( 2 1)( 2 3) ( 2 1)( 2 2)
22
AB
CB
B
− − − − = − + − ++ − + − +
+ − + − +− = −
=
Let 3x = − , then 2( 3) ( 3) 8 ( 3 2)( 3 3)
( 3 1)( 3 3) ( 3 1)( 3 2)
4 22
AB
CC
C
− − − − = − + − ++ − + − +
+ − + − +==
2
2
8 3 2 2
1 2 3( 1)( 5 6)
x x
x x xx x x
− − −= + ++ + ++ + +
35. Find the partial fraction decomposition: 2
2 2 2 2 2
2 3
( 4) 4 ( 4)
x x Ax B Cx D
x x x
+ + + += ++ + +
Multiplying both sides by 2 2( 4)x + , we obtain: 2 2
2 3 2
2 3 2
2 3 ( )( 4)
2 3 4 4
2 3 (4 ) 4
x x Ax B x Cx D
x x Ax Bx Ax B Cx D
x x Ax Bx A C x B D
+ + = + + + +
+ + = + + + + +
+ + = + + + + +
0A = ; 1B = ;
4 2
4(0) 2
2
A C
C
C
+ =+ =
=
4 3
4(1) 3
1
B D
D
D
+ =+ =
= −
2
2 2 2 2 2
2 3 1 2 1
( 4) 4 ( 4)
x x x
x x x
+ + −= ++ + +
Section 11.5: Partial Fraction Decomposition
1153
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36. Find the partial fraction decomposition: 3
2 2 2 2 2
1
( 16) 16 ( 16)
x Ax B Cx D
x x x
+ + += ++ + +
Multiplying both sides by 2 2( 16)x + , we obtain: 3 2
3 3 2
3 3 2
1 ( )( 16)
1 16 16
1 (16 ) 16
x Ax B x Cx D
x Ax Bx Ax B Cx D
x Ax Bx A C x B D
+ = + + + +
+ = + + + + +
+ = + + + + +
1A = ; 0B = ;
16 0
16(1) 0
16
A C
C
C
+ =+ =
= −
16 1
16(0) 1
1
B D
D
D
+ =+ =
=
3
2 2 2 2 2
1 16 1
( 16) 16 ( 16)
x x x
x x x
+ − += ++ + +
37. Find the partial fraction decomposition:
3 2
7 3 7 3
( 3)( 1)2 3
3 1
x x
x x xx x x
A B C
x x x
+ +=− +− −
= + +− +
Multiplying both sides by ( 3)( 1)x x x− + , we
obtain: 7 3 ( 3)( 1) ( 1) ( 3)x A x x Bx x Cx x+ = − + + + + −
Let 0x = , then 7(0) 3 (0 3)(0 1) (0)(0 1) (0)(0 3)
3 3
1
A B C
A
A
+ = − + + + + −= −= −
Let 3x = , then 7(3) 3 (3 3)(3 1) (3)(3 1) (3)(3 3)
24 12
2
A B C
B
B
+ = − + + + + −==
Let 1x = − , then 7( 1) 3 ( 1 3)( 1 1) ( 1)( 1 1)
( 1)( 1 3)
4 4
1
A B
C
C
C
− + = − − − + + − − ++ − − −
− == −
3 2
7 3 1 2 1
3 12 3
x
x x xx x x
+ − −= + +− +− −
38. Find the partial fraction decomposition: 3 3
5 4 4
2 3 4
1 1
( 1)
1
x x
x x x x
A B C D E
x xx x x
+ +=− −
= + + + +−
Multiplying both sides by 4 ( 1)x x − , we obtain: 3 3 2
4
1 ( 1) ( 1) ( 1)
( 1)
x Ax x Bx x Cx x
D x Ex
+ = − + − + −
+ − +
Let 0x = , then 4 3 2
4
0 1 0 (0 1) 0 (0 1) 0(0 1)
(0 1) 0
A B C
D E
+ = ⋅ − + ⋅ − + ⋅ −
+ − + ⋅
1
1
D
D
= −= −
Let 1x = , then 4 3 2
4
1 1 1 (1 1) 1 (1 1) 1(1 1)
(1 1) 1
A B C
D E
+ = ⋅ − + ⋅ − + ⋅ −
+ − + ⋅
2 E=
Let 1x = − , then 4 3 2
4
( 1) 1 ( 1) ( 1 1) ( 1) ( 1 1)
( 1)( 1 1) ( 1 1) ( 1)
A B
C D E
− + = − − − + − − −
+ − − − + − − + −
0 2 2 2 2
0 2 2 2 2( 1) 2
2 2 2 4
2
A B C D E
A B C
A B C
A B C
= − + − += − + − − +
− + = −− + = −
Let 2x = , then 4 3 2
4
2 1 2 (2 1) 2 (2 1) 2(2 1)
(2 1) 2
A B C
D E
+ = ⋅ − + ⋅ − + ⋅ −
+ − + ⋅
9 8 4 2 16
9 8 4 2 ( 1) 16(2)
8 4 2 22
4 2 11
A B C D E
A B C
A B C
A B C
= + + + += + + + − +
+ + = −+ + = −
Let 2x = − , then 4 3 2
4
( 22) 1 ( 2) ( 2 1) ( 2) ( 2 1)
( 2)( 2 1) ( 2 1) ( 2)
A B
C D E
− + = − − − + − − −
+ − − − + − − + −
7 24 12 6 3 16
7 24 12 6 3( 1) 81(2)
42 24 12 6
14 8 4 2
A B C D E
A B C
A B C
A B C
− = − + − +− = − + − − +
− = − +− = − +
Chapter 11: Systems of Equations and Inequalities
1154
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Solve the system of equations by using a matrix equation:
2
4 2 11
8 4 2 14
A B C
A B C
A B C
− + = − + + = − − + = −
1
1 1 1 2
4 2 1 11
8 4 2 14
1 1 1 2 2
4 2 1 11 1
8 4 2 14 1
A
B
C
A
B
C
−
− − = − − −
− − − = − = − − − −
So, 2A = − , 1B = − , and 1C = − . Thus,
3 3
5 4 4
2 3 4
1 1
( 1)
2 1 1 1 2
1
x x
x x x x
x xx x x
+ +=− −
− − − −= + + + +−
39. Perform synthetic division to find a factor:
2 1 4 5 2
2 4 2
1 2 1 0
− −−
−
3 2 2
2
4 5 2 ( 2)( 2 1)
( 2)( 1)
x x x x x x
x x
− + − = − − +
= − −
Find the partial fraction decomposition:
2 2
3 2 2
2
4 5 2 ( 2)( 1)
2 1 ( 1)
x x
x x x x x
A B C
x x x
=− + − − −
= + +− − −
Multiplying both sides by 2( 2)( 1)x x− − , we
obtain: 2 2( 1) ( 2)( 1) ( 2)x A x B x x C x= − + − − + −
Let 2x = , then 2 22 (2 1) (2 2)(2 1) (2 2)
4
A B C
A
= − + − − + −=
Let 1x = , then
2 21 (1 1) (1 2)(1 1) (1 2)
1
1
A B C
C
C
= − + − − + −= −= −
Let 0x = , then 2 20 (0 1) (0 2)(0 1) (0 2)
0 2 2
0 4 2 2( 1)
2 6
3
A B C
A B C
B
B
B
= − + − − + −= + −= + − −
− == −
2
3 2 2
4 3 1
2 14 5 2 ( 1)
x
x xx x x x
− −= + +− −− + − −
40. Perform synthetic division to find a factor:
1 1 1 5 3
1 2 3
1 2 3 0
−−
−
3 2 2
2
5 3 ( 1)( 2 3)
( 3)( 1)
x x x x x x
x x
+ − + = − + −
= + −
Find the partial fraction decomposition:
2 2
3 2 2
2
1 1
5 3 ( 3)( 1)
3 1 ( 1)
x x
x x x x x
A B C
x x x
+ +=+ − + + −
= + ++ − −
Multiplying both sides by 2( 3)( 1)x x+ − , we
obtain: 2 21 ( 1) ( 3)( 1) ( 3)x A x B x x C x+ = − + + − + +
Let 3x = − , then 2 2( 3) 1 ( 3 1) ( 3 3)( 3 1)
( 3 3)
10 16
5
8
A B
C
A
A
− + = − − + − + − −+ − +
=
=
Let 1x = , then 2 21 1 (1 1) (1 3)(1 1) (1 3)
2 4
1
2
A B C
C
C
+ = − + + − + +=
=
Section 11.5: Partial Fraction Decomposition
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Let 0x = , then 2 20 1 (0 1) (0 3)(0 1) (0 3)
1 3 3
5 11 3 3
8 2
93
83
8
A B C
A B C
B
B
B
+ = − + + − + += − +
= − +
=
=
5 3 128 8 2
3 2 2
1
3 15 3 ( 1)
x
x xx x x x
+ = + ++ −+ − + −
41. Find the partial fraction decomposition: 3
2 3 2 2 2 2 3( 16) 16 ( 16) ( 16)
x Ax B Cx D Ex F
x x x x
+ + += + ++ + + +
Multiplying both sides by 2 3( 16)x + , we obtain: 3 2 2 2
3 4 2 3 2
3 5 4 3 2
( )( 16) ( )( 16)
( )( 32 256) 16 16
32 32 256
x Ax B x Cx D xEx F
x Ax B x x Cx DxCx D Ex F
x Ax Bx Ax Bx Ax
= + + + + ++ +
= + + + + ++ + + +
= + + + +3 2 256
16 16B Cx DxCx D Ex F
+ + ++ + + +
3 5 4 3 2(32 ) (32 )
(256 16 )
(256 16 )
x Ax Bx A C x B D x
A C E x
B D F
= + + + + ++ + +
+ + +
0; 0A B= = ; 32 1
32(0) 1
1
A C
C
C
+ =+ =
=
32 0
32(0) 0
0
B D
D
D
+ =+ =
=
256 16 0
256(0) 16(1) 0
16
A C E
E
E
+ + =+ + =
= −
256 16 0
256(0) 16(0) 0
0
B D F
F
F
+ + =+ + =
=
3
2 3 2 2 2 3
16
( 16) ( 16) ( 16)
x x x
x x x
−= ++ + +
42. Find the partial fraction decomposition: 2
2 3 2 2 2 2 3( 4) 4 ( 4) ( 4)
x Ax B Cx D Ex F
x x x x
+ + += + ++ + + +
Multiplying both sides by 2 3( 4)x + , we obtain: 2 2 2 2
2 4 2 3 2
2 5 4 3 2
3 2
( )( 4) ( )( 4)
( )( 8 16)
4 4
8 8 16 16
4 4
x Ax B x Cx D x
Ex F
x Ax B x x Cx Dx
Cx D Ex F
x Ax Bx Ax Bx Ax B
Cx Dx Cx D Ex
= + + + + ++ +
= + + + + ++ + + +
= + + + + +
+ + + + +2 5 4 3 2(8 ) (8 )
(16 4 ) (16 4 )
F
x Ax Bx A C x B D x
A C E x B D F
+
= + + + + ++ + + + + +
0; 0A B= = ; 8 0
8(0) 0
0
A C
C
C
+ =+ =
=
8 1
8(0) 1
1
B D
D
D
+ =+ =
=
16 4 0
16(0) 4(0) 0
0
A C E
E
E
+ + =+ + =
=
16 4 0
16(0) 4(1) 0
4
B D F
F
F
+ + =+ + =
= −
2
2 3 2 2 2 3
1 4
( 4) ( 4) ( 4)
x
x x x
−= ++ + +
43. Find the partial fraction decomposition:
2
4 4
( 3)(2 1) 3 2 12 5 3
A B
x x x xx x= = +
− + − +− −
Multiplying both sides by ( 3)(2 1)x x− + , we
obtain: 4 (2 1) ( 3)A x B x= + + −
Let 1
2x = − , then
1 14 2 1 3
2 2
74 28
7
A B
B
B
= − + + − −
= −
= −
Chapter 11: Systems of Equations and Inequalities
1156
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Let 3x = , then 4 (2(3) 1) (3 3)
4 7
4
7
A B
A
A
= + + −=
=
847 7
2
4
3 2 12 5 3 x xx x
−= +
− +− −
44. Find the partial fraction decomposition:
2
4 4
( 2)(2 1) 2 2 12 3 2
x x A B
x x x xx x= = +
+ − + −+ −
Multiplying both sides by ( 2)(2 1)x x+ − , we
obtain: 4 (2 1) ( 2)x A x B x= − + +
Let 1
2x = , then
1 1 14 2 1 2
2 2 2
52
34
5
A B
B
B
= − + +
=
=
Let 2x = − , then 4( 2) (2( 2) 1) ( 2 2)
8 5
8
5
A B
A
A
− = − − + − +− = −
=
8 45 5
2
4
2 2 12 3 2
x
x xx x= +
+ −+ −
45. Find the partial fraction decomposition:
4 2 2
2
2 3 2 3
9 ( 3)( 3)
3 3
x x
x x x x x
A B C D
x x xx
+ +=− − +
= + + +− +
Multiplying both sides by 2 ( 3)( 3)x x x− + , we
obtain:
2 2
2 3 ( 3)( 3) ( 3)( 3)
( 3) ( 3)
x Ax x x B x x
Cx x Dx x
+ = − + + − +
+ + + −
Let 0x = , then
2 2
2 0 3 0(0 3)(0 3) (0 3)(0 3)
0 (0 3) 0 (0 3)
3 9
1
3
A B
C D
B
B
⋅ + = ⋅ − + + − +
+ ⋅ + + ⋅ −= −
= −
Let 3x = , then
2 2
2 3 3 3(3 3)(3 3) (3 3)(3 3)
3 (3 3) 3 (3 3)
9 54
1
6
A B
C D
C
C
⋅ + = ⋅ − + + − +
+ ⋅ + + ⋅ −=
=
Let 3x = − , then
2
2
2( 3) 3 ( 3)( 3 3)( 3 3)
( 3 3)( 3 3)
( 3) ( 3 3)
( 3) ( 3 3)
3 54
1
18
A
B
C
D
D
D
− + = − − − − ++ − − − +
+ − − +
+ − − −− = −
=
Let 1x = , then
( ) ( ) ( )
2 2
2 1 3 1(1 3)(1 3) (1 3)(1 3)
1 (1 3) 1 (1 3)
5 8 8 4 2
5 8 8 1/ 3 4 1/ 6 2 1/18
8 2 15 8
3 3 916
89
2
9
A B
C D
A B C D
A
A
A
A
⋅ + = ⋅ − + + − +
+ ⋅ + + ⋅ −= − − + −= − − − + −
= − + + −
− =
= −
2 1 1 19 3 6 18
4 2 2
2 3
3 39
x
x x xx x x
− −+ = + + +− +−
46. Find the partial fraction decomposition: 2 2
4 2 2
2
9 9
2 8 ( 2)( 2)( 2)
2 2 2
x x
x x x x x
A B Cx D
x x x
+ +=− − + − +
+= + +− + +
Multiplying both sides by 2( 2)( 2)( 2)x x x+ − + ,
we obtain: 2 2 29 ( 2)( 2) ( 2)( 2)
( )( 2)( 2)
x A x x B x x
Cx D x x
+ = + + + − ++ + − +
Section 11.6: Systems of Nonlinear Equations
1157
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Let 2x = , then 2 2 22 9 (2 2)(2 2) (2 2)(2 2)
( (2) )(2 2)(2 2)
13 24
13
24
A B
C D
A
A
+ = + + + − ++ + − +
=
=
Let 2x = − , then 2 2
2
( 2) 9 (( 2) 2)( 2 2)
( 2 2)(( 2) 2)
( ( 2) )( 2 2)( 2 2)
13 24
13
24
A
B
C D
B
B
− + = − + − +
+ − − − ++ − + − − − +
= −
= −
Let 0x = , then 2 2 20 9 (0 2)(0 2) (0 2)(0 2)
( (0) )(0 2)(0 2)
9 4 4 4
13 139 4 4 4
24 24
144
37
6
A B
C D
A B D
D
D
D
+ = + + + − ++ + − +
= − −
= − − −
= −
= −
Let 1x = , then 2 2 21 9 (1 2)(1 2) (1 2)(1 2)
( (1) )(1 2)(1 2)
10 9 3 3 3
39 13 710 3
8 8 23 0
0
A B
C D
A B C D
C
C
C
+ = + + + − ++ + − +
= − − −
= + − +
==
713 132624 24
4 2 2
9
2 22 8 2
x
x xx x x
−−+ = + +− +− − +
Section 11.6
1. 3 2y x= +
The graph is a line. x-intercept:
0 3 2
3 2
2
3
x
x
x
= += −
= −
y-intercept: ( )3 0 2 2y = + =
2
−2
y
x
(0, 2)
2,0
3 −
2
−2
2. 2 4y x= −
The graph is a parabola. x-intercepts:
2
2
0 4
4
2, 2
x
x
x x
= −
== − =
y-intercept: 20 4 4y = − = −
The vertex has x-coordinate:
( )0
02 2 1
bx
a= − = − = .
The y-coordinate of the vertex is 20 4 4y = − = − .
Chapter 11: Systems of Equations and Inequalities
1158
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
3. 2 2
2 2
22
2 2
1
1
11 1
y x
x y
yx
= −
− =
− =
The graph is a hyperbola with center (0, 0), transverse axis along the x-axis, and vertices at ( 1,0)− and (1,0) . The asymptotes are y x= −
and y x= . y
x−5(−1, 0)
5
−5
5(1, 0)
4. 2 2
2 2
22
22
2 2
4 4
4 44 4
14
12 1
x y
x y
xy
yx
+ =
+ =
+ =
+ =
The graph is an ellipse with center (0,0) , major
axis along the x-axis, vertices at ( 2,0)− and
(2,0) . The graph also has y-intercepts at (0, 1)−
and (0,1) . y
x−5(−2, 0)
5
−5
5
(2, 0)
(0, −1)
(0, 1)
5. 2 1
1
y x
y x
= +
= +
(0, 1) and (1, 2) are the intersection points.
Solve by substitution: 2
2
1 1
0
( 1) 0
0 or 1
1 2
x x
x x
x x
x x
y y
+ = +
− =− =
= == =
Solutions: (0, 1) and (1, 2)
6. 2 1
4 1
y x
y x
= +
= +
(0, 1) and (4, 17) are the intersection points.
Solve by substitution: 2
2
1 4 1
4 0
( 4) 0
0 or 4
1 17
x x
x x
x x
x x
y y
+ = +
− =− =
= == =
Solutions: (0, 1) and (4, 17)
Section 11.6: Systems of Nonlinear Equations
1159
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
7. 236
8
y x
y x
= −
= −
(2.59, 5.41) and (5.41, 2.59) are the intersection points.
Solve by substitution: 2
2 2
2
2
36 8
36 64 16
2 16 28 0
8 14 0
8 64 56
2
8 2 2
2
4 2
x x
x x x
x x
x x
x
− = −
− = − +
− + =
− + =
± −=
±=
= ±
( )( )
If 4 2, 8 4 2 4 2
If 4 2, 8 4 2 4 2
x y
x y
= + = − + = −
= − = − − = +
Solutions: ( ) ( )4 2, 4 2 and 4 2, 4 2+ − − +
8. 24
2 4
y x
y x
= −
= +
(–2, 0) and (–1.2, 1.6) are the intersection points.
Solve by substitution: 2
2 2
2
4 2 4
4 4 16 16
5 16 12 0
( 2)(5 6) 0
62 or
58
0 or 5
x x
x x x
x x
x x
x x
y y
− = +
− = + +
+ + =+ + =
= − = −
= =
Solutions: ( ) 6 82, 0 and ,
5 5 − −
9. 2
y x
y x
=
= −
(1, 1) is the intersection point.
Solve by substitution:
2
2
2
4 4
5 4 0
( 4)( 1) 0
4 or 1
2 or =1
x x
x x x
x x
x x
x x
y y
= −
= − +
− + =− − =
= == −
Eliminate (4, –2); we must have 0y ≥ .
Solution: (1, 1)
Chapter 11: Systems of Equations and Inequalities
1160
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
10. 6
y x
y x
=
= −
(4, 2) is the intersection point.
Solve by substitution:
2
2
6
36 12
13 36 0
( 4)( 9) 0
4 or 9
2 or 3
x x
x x x
x x
x x
x x
y y
= −
= − +
− + =− − =
= == = −
Eliminate (9, –3); we must have 0y ≥ .
Solution: (4, 2)
11. 2
2
2
x y
x y y
=
= −
(0, 0) and (8, 4) are the intersection points.
Solve by substitution: 2
2
2 2
4 0
( 4) 0
0 or =4
0 or =8
y y y
y y
y y
y y
x x
= −
− =− =
==
Solutions: (0, 0) and (8, 4)
12. 2
1
6 9
y x
y x x
= −
= − +
(2, 1) and (5, 4) are the intersection points.
Solve by substitution: 2
2
6 9 1
7 10 0
( 2)( 5) 0
2 or =5
1 or =4
x x x
x x
x x
x x
y y
− + = −
− + =− − =
==
Solutions: (2, 1) and (5, 4)
13. 2 2
2 2
4
2 0
x y
x x y
+ =
+ + =
(–2, 0) is the intersection point.
Substitute 4 for 2 2x y+ in the second equation.
2
2 4 0
2 4
2
4 ( 2) 0
x
x
x
y
+ == −= −
= − − =
Solution: (–2, 0)
Section 11.6: Systems of Nonlinear Equations
1161
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14. 2 2
2 2
8
4 0
x y
x y y
+ =
+ + =
(–2, –2) and (2, –2) are the intersection points.
Substitute 8 for 2 2x y+ in the second equation.
2
8 4 0
4 8
2
8 ( 2) 2
y
y
y
x
+ == −= −
= ± − − = ±
Solution: (–2, –2) and (2, –2)
15. 2 2
3 5
5
y x
x y
= −
+ =
(1, –2) and (2, 1) are the intersection points.
Solve by substitution: 2 2
2 2
2
2
(3 5) 5
9 30 25 5
10 30 20 0
3 2 0
( 1)( 2) 0
x x
x x x
x x
x x
x x
+ − =
+ − + =
− + =
− + =− − =
1 or 2
2 1
x x
y y
= == − =
Solutions: (1, –2) and (2, 1)
16. 2 2 10
2
x y
y x
+ =
= +
(1, 3) and (–3, –1) are the intersection points.
Solve by substitution: 2 2
2 2
2
( 2) 10
4 4 10
2 4 6 0
2( 3)( 1) 0
x x
x x x
x x
x x
+ + =
+ + + =
+ − =+ − =
3 or 1
1 3
x x
y y
= − == − =
Solutions: (–3, –1) and (1, 3)
17. 2 2
2
4
4
x y
y x
+ =
− =
(–1, 1.73), (–1, –1.73), (0, 2), and (0, –2) are the intersection points.
2Substitute 4 for in the first equation:x y+ 2
2
2 2
4 4
0
( 1) 00 or 1
4 3
2 3
x x
x x
x x
x x
y y
y y
+ + =
+ =+ =
= = −= == ± = ±
Solutions: ( ) ( )(0, – 2), (0, 2), 1, 3 , 1, 3− − −
Chapter 11: Systems of Equations and Inequalities
1162
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
18. 2 2
2
16
2 8
x y
x y
+ =
− =
(–3.46, 2), (0, –4), and (3.46, 2) are the intersection points.
Substitute 2 8y + for 2x in the first equation. 2
2
2 8 16
2 8 0
( 4)( 2) 0
y y
y y
y y
+ + =
+ − =+ − =
2 2
4 or 2
0 or 12
0 2 3
y y
x x
x x
= − =
= =
= = ±
Solutions: ( ) ( )(0, – 4), 2 3, 2 , 2 3, 2−
19. 2 2
4
8
xy
x y
=
+ =
(–2, –2) and (2, 2) are the intersection points.
Solve by substitution: 2
2
22
4 2
4 2
2 2
2
2
48
168
16 8
8 16 0
( 4) 0
4 0
4
2 or 22 or 2
xx
xx
x x
x x
x
x
x
x xy y
+ =
+ =
+ =− + =
− =− =
== = −= = −
Solutions: (–2, –2) and (2, 2)
20. 2
1
x y
xy
=
=
(1, 1) is the intersection point.
Solve by substitution:
2
3
2
1
1
1
(1) 1
xx
x
x
y
=
==
= =
Solution: (1, 1)
21. 2 2
2
4
9
x y
y x
+ =
= −
No solution; Inconsistent.
Solve by substitution: 2 2 2
2 4 2
4 2
( 9) 4
18 81 4
17 77 0
x x
x x x
x x
+ − =
+ − + =
− + =
2 17 289 4(77)
2
17 19
2
x± −
=
± −=
There are no real solutions to this expression. Inconsistent.
Section 11.6: Systems of Nonlinear Equations
1163
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22. 1
2 1
xy
y x
= = +
(–1, –1) and (0.5, 2) are the intersection points.
Solve by substitution:
2
(2 1) 1
2 1 0
( 1)(2 1) 0
11 or
21 2
x x
x x
x x
x x
y y
+ =
+ − =+ − =
= − =
= − =
Solutions: (–1, –1) and 1
, 22
23. 2 4
6 13
y x
y x
= −
= −
(3, 5) is the intersection point.
Solve by substitution: 2
2
2
2
4 6 13
6 9 0
( 3) 0
3 0
3
(3) 4 5
x x
x x
x
x
x
y
− = −
− + =
− =− =
=
= − =
Solution: (3,5)
24. 2 2 10
3
x y
xy
+ =
=
(1, 3), (3, 1), (–3, –1), and (–1, –3) are the intersection points.
Solve by substitution:
( )22
2
4 2
4 2
2 2
2
3
9
10
10
9 10
10 9 0
( 9)( 1) 0
( 3)( 3)( 1)( 1) 0
x
x
x
x
x x
x x
x x
x x x x
+ =
+ =
+ =
− + =
− − =− + − + =
3 or –3 or 1 or 1
1 1 3 –3
x x x x
y y y y
= = = = −= = − = =
Solutions: (3, 1), (–3, –1), (1, 3), (–1, –3)
25. Solve the second equation for y, substitute into the first equation and solve:
2 22 18
44
x y
xy yx
+ =
= =
( )( )
22
22
4 2
4 2
4 2
2 2
42 18
162 18
2 16 18
2 18 16 0
9 8 0
8 1 0
xx
xx
x x
x x
x x
x x
+ =
+ =
+ =
− + =
− + =
− − =
2 28 or 1
or 18= 2 2
x x
xx
= == ±= ± ±
Chapter 11: Systems of Equations and Inequalities
1164
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
4If 2 2 : 2
2 24
If 2 2 : 22 2
4If 1: 4
14
If 1: 41
x y
x y
x y
x y
= = =
= − = = −−
= = =
= − = = −−
Solutions:
( ) ( )2 2, 2 , 2 2, 2 , (1, 4), ( 1, 4)− − − −
26. Solve the second equation for y, substitute into the first equation and solve:
2 2 21
7 7
x y
x y y x
− =
+ = = −
( )( )
22
2 2
7 21
49 14 21
14 70
5
7 5 2
x x
x x x
x
x
y
− − =
− − + =
=== − =
Solution: (5, 2)
27. Substitute the first equation into the second equation and solve:
2 2
2 1
2 1
y x
x y
= +
+ =
( )
( )
22
2 2
2
2 2 1 1
2 4 4 1 1
6 4 0
2 3 2 0
x x
x x x
x x
x x
+ + =
+ + + =
+ =+ =
2 0 or 3 2 0
20 or
3
x x
x x
= + =
= = −
If 0 : 2(0) 1 1
2 2 4 1If : 2 1 1
3 3 3 3
x y
x y
= = + =
= − = − + = − + = −
Solutions: 2 1
(0, 1), ,3 3
− −
28. Solve the second equation for x and substitute into the first equation and solve:
2 24 162 2 2 2
x yy x x y
− = − = = −
( )
2 2
2 2
32
(2 2) 4 16
4 8 4 4 168 12
3
22 2 5
y y
y y yy
y
x
− − =− + − =
− =
= −
= − − = −
Solutions: ( )32
5,− −
29. Solve the first equation for y, substitute into the second equation and solve:
2 2
1 0 1
6 5
x y y x
x y y x
+ + = = − − + + − = −
2 2
2 2
2
( 1) 6( 1) 5
2 1 6 6 5
2 5 0(2 5) 0
x x x x
x x x x x
x xx x
+ − − + − − − = −+ + + − − − = −
− =− =
520 or x x= =
If 0 : (0) 1 15 5 7
If : 12 2 2
x y
x y
= = − − = −
= = − − = −
Solutions: ( )5 72 2
(0, 1), ,− −
30. Solve the second equation for y, substitute into the first equation and solve:
2 22 84
4
x xy y
xy yx
− + = = =
( ) ( )( )( )
22
22
4 2
4 2
2 2
4 42 8
162 4 8
2 16 12
6 8 0
4 2 0
( 2)( 2) 2 2 0
x xx x
xx
x x
x x
x x
x x x x
− + =
− + =
+ =− + =
− − =
− + − + =
2 or 2 or 2 or 2
2 2 2 2 2 2
x x x x
y y y y
= = − = = −= = − = = −
Solutions:
( ) ( )( 2, 2), 2, 2 2 , 2, 2 2 , (2, 2)− − − −
Section 11.6: Systems of Nonlinear Equations
1165
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31. Solve the second equation for y, substitute into the first equation and solve:
2 24 3 9 15
2 52 3 5
3 3
x xy y
x y y x
− + =
+ = = − +
2
2
2 2 2
2
2 5 2 54 3 9 15
3 3 3 3
4 2 5 4 20 25 15
10 25 10 0
x x x x
x x x x x
x x
− − + + − + =
+ − + − + =
− + =
22 5 2 0
(2 1)( 2) 0
x x
x x
− + =− − =
1
or 22
x x= =
1 2 1 5 4If :
2 3 2 3 3
2 5 1If 2 : (2)
3 3 3
x y
x y
= = − + =
= = − + =
Solutions: 1 4 1
, , 2,2 3 3
32. 22 3 6 2 4 0
2 3 4 0
y xy y x
x y
− + + + =
− + =
Solve the second equation for x, substitute into the first equation and solve: 2 3 4 0
2 3 4
3 4
2
x y
x y
yx
− + == −
−=
2
2 2
2
2
3 4 3 42 3 6 2 4
2 2
92 6 6 3 4 4
25
15 02
5 30 0
5 ( 6) 0
y yy y y
y y y y y
y y
y y
y y
− − − + + = −
− + + + − = −
− + =
− + =− − =
0 or 6y y= =
( )3 0 4If 0 : 2
2y x
−= = −
( )3 6 4If 6 : 7
2y x
−= = =
Solutions: (–2, 0), (7, 6)
33. Multiply each side of the second equation by 4 and add the equations to eliminate y:
2 2 2 2
42 2 2 2
2
2
4 7 4 7
3 31 12 4 124
13 117
9 3
x y x y
x y x y
x
xx
− = − ⎯⎯→ − = −
+ = ⎯⎯→ + ==
== ±
If 3x = : 2 2 23(3) 31 4 2y y y+ = = = ±
If 3x = − : 2 2 23( 3) 31 4 2y y y− + = = = ±
Solutions: (3, 2), (3, –2), (–3, 2), (–3, –2)
34. 2 2
2 2
3 2 5 0
2 2 0
x y
x y
− + =
− + =
2 2
2 2
3 2 5
2 2
x y
x y
− = −
− = −
Multiply each side of the second equation by –2 and add the equations to eliminate y:
2 2
2 2
2
2
3 2 5
4 2 4
1
1 1
x y
x y
x
xx
− = −− + =
− = −== ±
2 2 2
2 2 2
If 1:
2(1) 2 4 2
If 1:
2( 1) 2 4 2
x
y y y
x
y y y
=
− = − = = ±= −
− − = − = = ±
Solutions: (1, 2), (1, –2), (–1, 2), (–1, –2)
35. 2 2
2 2
7 3 5 0
3 5 12
x y
x y
− + =
+ =
2 2
2 2
7 3 5
3 5 12
x y
x y
− = −
+ =
Multiply each side of the first equation by 5 and each side of the second equation by 3 and add the equations to eliminate y:
2 2
2 2
2
2
35 15 25
9 15 36
44 11
1
41
2
x y
x y
x
x
x
− = −
+ =
=
=
= ±
Chapter 11: Systems of Equations and Inequalities
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22 2
22 2
12
12
If :
1 9 33 5 12
2 4 2
If :
1 9 33 5 12
2 4 2
x
y y y
x
y y y
=
+ = = = ±
= −
− + = = = ±
Solutions: 1 3 1 3 1 3 1 3
, , , , , , ,2 2 2 2 2 2 2 2
− − − −
36. 2 2
2 2
3 1 0
2 7 5 0
x y
x y
− + =
− + =
2 2
2 2
3 1
2 7 5
x y
x y
− = −
− = −
Multiply each side of the first equation by –2 and add the equations to eliminate x:
2 2
2 2
2
2
2 6 2
2 7 5
3
3
3
x y
x y
y
y
y
− + =
− = −
− = −
=
= ±
( )
( )
22 2
22 2
If 3 :
3 3 1 8 2 2
If 3 :
3 3 1 8 2 2
y
x x x
y
x x x
=
− = − = = ±
= −
− − = − = = ±
Solutions:
( ) ( ) ( )( )2 2, 3 , 2 2, 3 , 2 2, 3 ,
2 2, 3
− −
− −
37. Multiply each side of the second equation by 2 and add the equations to eliminate xy:
2 2
22 2
2
2
2 10 2 10
3 2 6 2 4
7 14
2
2
x xy x xy
x xy x xy
x
x
x
+ = ⎯⎯→ + =
− = ⎯⎯→ − =
=
=
= ±
( )
( ) ( )
2
2
If 2 :
3 2 2 2
42 4 2 2
2
If 2 :
3 2 2 2
42 4 2 2
2
x
y
y y y
x
y
y y y
=
− ⋅ =
− ⋅ = − = =
= −
− − − =
− ⋅ = − = = −
Solutions: ( ) ( )2, 2 2 , 2, 2 2− −
38. 2
2
5 13 36 0
7 6
xy y
xy y
+ + =
+ =
2
2
5 13 36
7 6
xy y
xy y
+ = −
+ =
Multiply each side of the second equation by –5 and add the equations to eliminate xy:
2
2
2
2
5 13 36
5 35 30
22 66
3
3
xy y
xy y
y
y
y
+ = −
− − = −
− = −
=
= ±
( ) ( )2
If 3 :
3 7 3 6
153 15 5 3
3
y
x
x x x
=
+ =
− ⋅ = − = = −
( ) ( )2
If 3 :
3 7 3 6
153 15 5 3
3
y
x
x x x
= −
− + − =
− ⋅ = − = =
Solutions: ( ) ( )5 3, 3 , 5 3, 3− −
Section 11.6: Systems of Nonlinear Equations
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39. 2 2
2 2
2 2
2 8 0
x y
x y
+ =
− + =
2 2
2 2
2 2
2 8
x y
x y
+ =
− = −
Multiply each side of the first equation by 2 and add the equations to eliminate y:
2 2
2 2
2
2
4 2 4
2 8
5 44
5
x y
x y
x
x
+ =− = −
= −
= −
No real solution. The system is inconsistent.
40. 2 2
2 2
4 0
2 3 6
y x
x y
− + =
+ =
2 2
2 2
4
2 3 6
x y
x y
− =
+ =
Multiply each side of the first equation by 2− and add the equations to eliminate x:
2 2
2 2
2
2
2 2 8
2 3 6
5 22
5
x y
x y
y
y
− + = −+ =
= −
= −
No real solution. The system is inconsistent.
41. 2 2
2 2
2 16
4 24
x y
x y
+ =
− =
Multiply each side of the second equation by 2 and add the equations to eliminate y:
2 2
2 2
2
2
2 16
8 2 48
9 6464
98
3
x y
x y
x
x
x
+ =− =
=
=
= ±
22 2
2
8If :
3
8 802 16 2
3 9
40 2 10
9 3
x
y y
y y
=
+ = =
= = ±
22 2
2
8If :
3
8 802 16 2
3 9
40 2 10
9 3
x
y y
y y
= −
− + = =
= = ±
Solutions:
8 2 10 8 2 10 8 2 10, , , , , ,
3 3 3 3 3 3
8 2 10,
3 3
− −
− −
42. 2 2
2 2
4 3 4
2 6 3
x y
x y
+ =
− = −
Multiply each side of the first equation by 2 and add the equations to eliminate y:
2 2
2 2
2
2
8 6 8
2 6 3
10 5
1
2
2
2
x y
x y
x
x
x
+ =
− = −
=
=
= ±
2
2 2
2
2
2 2
2
2If :
2
24 3 4 3 2
2
2 6
3 3
2If :
2
24 3 4 3 2
2
2 6
3 3
x
y y
y y
x
y y
y y
=
+ = =
= = ±
= −
− + = =
= = ±
Solutions:
2 6 2 6 2 6, , , , , ,
2 3 2 3 2 3
2 6,
2 3
− −
− −
Chapter 11: Systems of Equations and Inequalities
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43. 2 2
2 2
5 23 0
3 17
x y
x y
− + = + =
2 2
2 2
5 23
3 17
x y
x y
− = − + =
Multiply each side of the second equation by 2 and add the equations to eliminate y:
2 2
2 2
2
2
5 23
6 214
11 11
1
1
x y
x y
x
x
x
− = −
+ =
=
== ±
22 2 2
22 2 2
If 1:
3 1 1 17 4
4(1)
1
2If 1:
3 1 1 17 4
4( 1)
1
2
x
yy y
y
x
yy y
y
=
+ = = =
= ±
= −
+ = = =−
= ±
Solutions: 1 1 1 1
1, , 1, , 1, , 1,2 2 2 2
− − − −
44. 2 2
2 2
2 31 0
6 72 0
x y
x y
− + = − + =
2 2
2 2
2 31
6 72
x y
x y
− = − − = −
Multiply each side of the first equation by –3 and add the equations to eliminate x:
2 2
2 2
2
2
6 93
6 72
2 1
2
2
y y
x y
y
y
y
− + =
− = −
=
=
= ±
( )
( )
2 2 2
2
2 2 2
2
2 3 2 1If 2 : 1
22
4 2
2 3 2 1If 2 : 1
22
4 2
yx x
x x
yx x
x x
= − = − =
= = ±
= − − = − =−
= = ±
Solutions:
( ) ( ) ( ) ( )2, 2 , 2, 2 , 2, 2 , 2, 2− − − −
45. 4 4
4 4
1 66
2 219
x y
x y
+ = − =
Multiply each side of the first equation by –2 and add the equations to eliminate x:
4 4
4 4
4
4
2 1212
2 219
147
2
x y
x y
y
y
− − = −
− =
− =
= −
There are no real solutions. The system is inconsistent.
Section 11.6: Systems of Nonlinear Equations
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46. Add the equations to eliminate y:
4 4
4 4
4
4
4
1 11
1 14
2 5
2
52
5
x y
x y
x
x
x
− =
+ =
=
=
= ±
44 4 4
4
4 4
44 4 4
4
4 4
2 1 1 1 3If : 4
5 225
2 2
3 3
2 1 1 1 3If : 4
5 225
2 2
3 3
xy y
y y
xy y
y y
= + = =
= = ±
= − + = =
−
= = ±
Solutions:
4 4 4 4 4 4
4 4
2 2 2 2 2 2, , , , , ,
5 3 5 3 5 3
2 2,
5 3
− −
− −
47. 2 2
2
3 2 0
6
x xy y
x xy
− + =
+ =
Subtract the second equation from the first to
eliminate the 2x term. 2
2
4 2 6
2 3
xy y
xy y
− + = −
− =
Since 0y ≠ , we can solve for x in this equation
to get 2 32
yx
y+= , 0y ≠
Now substitute for x in the second equation and solve for y.
2
22 2
4 2 2
2
6
3 36
2 2
6 9 36
24
x xy
y yy
y y
y y y
y
+ =
+ ++ =
+ + ++ =
( )( )
4 2 4 2 2
4 2
4 2
2 2
6 9 2 6 24
3 12 9 0
4 3 0
3 1 0
y y y y y
y y
y y
y y
+ + + + =
− + =
− + =
− − =
Thus, 3y = ± or 1y = ± .
If 1: 2 1 2
If 1: 2( 1) 2
y x
y x
= = ⋅ == − = − = −
If 3 : 3
If 3 : 3
y x
y x
= =
= − = −
Solutions: ( ) ( )(2, 1), (–2, 1), 3, 3 , 3, 3− − −
48. 2 22 0
6 0
x xy y
xy x
− − =
+ + =
Factor the first equation, solve for x, substitute into the second equation and solve:
2 22 0
( 2 )( ) 0
2 or
x xy y
x y x y
x y x y
− − =− + =
= = −
Substitute 2x y= and solve:
2
2
6 0
(2 ) 2 6
2 2 6 0
2( 3) 0
xy x
y y y
y y
y y
+ + =+ = −
+ + =
+ + =
21 1 4(1)(3) (No real solution)
2(1)y
− ± −=
Substitute x y= − and solve:
2
6 0
( ) 6
6 0
( 3)( 2) 0
3 or =2
If 3 : 3
If 2 : 2
xy x
y y y
y y
y y
y y
y x
y x
+ + =− ⋅ + − = −
− − + =− − − =
= −= − == = −
Solutions: (3, –3), (– 2, 2)
Chapter 11: Systems of Equations and Inequalities
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49. 2 2 2 0
21 0
y y x x
xy
y
+ + − − =
− + + =
Multiply each side of the second equation by –y and add the equations to eliminate y:
( )
2 2
2
2
2 0
2 0
2 0
2 0
y y x x
y y x
x x
x x
+ + − − =
− − − + =
− =− =
0 or 2x x= =
2 2 2
2 2 2
If 0 :
0 0 2 0 2 0
( 2)( 1) 0 2 or 1
If 2 :
2 2 2 0 0
( 1) 0 0 or 1
Note: 0 because of division by zero.
x
y y y y
y y y y
x
y y y y
y y y y
y
=
+ + − − = + − = + − = = − =
=
+ + − − = + = + = = = −
≠
Solutions: (0, –2), (0, 1), (2, –1)
50. 3 2 2
2
2
2 3 4 0
2 0
x x y y
y yx
x
− + + − = −− + =
Multiply each side of the second equation by 2x− and add the equations to eliminate x: 3 2 2
3 2 2
2 3 4 0
2 0
4 4 0
4 4
1
x x y y
x x y y
y
y
y
− + + − =
− + − + =
− ===
3 2 2 3 2
2
If 1:
2 1 3 1 4 0 2 0
( 2) 0 0 or 2
Note: 0 because of division by zero.
y
x x x x
x x x x
x
=
− + + ⋅ − = − =
− = = =≠
Solution: (2, 1)
51. Rewrite each equation in exponential form: 3
5
log 3
log (4 ) 5 4
x
x
y y x
y y x
= → =
= → =
Substitute the first equation into the second and solve:
3 5
5 3
3 2
4
4 0
( 4) 0
x x
x x
x x
=
− =
− =
3 20 or 4 0 or 2x x x x= = = = ± The base of a logarithm must be positive, thus
0 and 2x x≠ ≠ − .
3If 2 : 2 8x y= = =
Solution: (2, 8)
52. Rewrite each equation in exponential form: 3
2
log (2 ) 3 2
log (4 ) 2 4
x
x
y y x
y y x
= =
= =
Multiply the first equation by 2 then substitute the first equation into the second and solve:
3 2
3 2
2
2
2 0
(2 1) 0
x x
x x
x x
=
− =
− =
2 1 10 or or 0
2 2x x x x= = = =
The base of a logarithm must be positive, thus 0x ≠ .
21 1 1 1
If : 42 2 4 16
x y y = = = =
Solution: 1 1
,2 16
53. Rewrite each equation in exponential form: 44ln ln 4ln 4ln y yx y x e e y= = = =
23 3 3
3 3
2 2log 2log log2 2 2
log 2 2log
3 3 3 3 3 9y y y
x y
x y+
= +
= = ⋅ = ⋅ =
So we have the system 4
29
x y
x y
=
=
Therefore we have :
2 4 2 4 2 29 9 0 (9 ) 0y y y y y y= − = − = 2 (3 )(3 ) 0
0 or 3 or 3
y y y
y y y
+ − == = − =
Since ln y is undefined when 0y ≤ , the only
Section 11.6: Systems of Nonlinear Equations
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
solution is 3y = . 4 4If 3 : 3 81y x y x= = = =
Solution: ( )81, 3
54. Rewrite each equation in exponential form: 55ln ln 5ln 5ln y yx y x e e y= = = =
22 2 2
2 2
3 2log 2log log3 3 2
log 3 2log
2 2 2 2 2 8y y y
x y
x y+
= +
= = ⋅ = ⋅ =
So we have the system 5
28
x y
x y
=
=
Therefore we have
( )
2 5
2 5
2 3
8
8 0
8 0
y y
y y
y y
=
− =
− =
30 or 8 0 2y y y= − = =
Since ln y is undefined when 0y ≤ , the only
solution is 2y = . 5 5If 2 : 2 32y x y x= = = =
Solution: ( )32, 2
55.
( ) ( )( ) ( )
2 2
2
22 31 12 2 2
2 21 1 12 2 2
3 2 0
1 0
x x y y
y yx
x
x y
x y
+ + − + = −+ + = + + − = + + − =
56.
( ) ( )( )
2 2
2 2 51 12 2 2
2 912 4
2 0
21 0
y y x x
xy
y
x y
x y
+ + − − =
− + + = − + + = = − + +
57. Graph: 1 2(2 / 3); ( )y x y e x= ∧ = ∧ −
Use INTERSECT to solve:
4.7–4.7
3.1
-3.1 Solution: 0.48, 0.62x y= = or (0.48, 0.62)
58. Graph: 1 2(3 / 2); ( )y x y e x= ∧ = ∧ −
Use INTERSECT to solve:
4.7–4.7
3.1
–3.1 Solution: 0.65, 0.52x y= = or (0.65, 0.52)
59. Graph: 3 2 31 22 ; 4 /y x y x= − =
Use INTERSECT to solve:
Chapter 11: Systems of Equations and Inequalities
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
4.7–4.7
3.1
–3.1 Solution: 1.65, 0.89x y= − = − or (–1.65, –0.89)
60. Graph: 3 31 2
23
2 ; 2 ;
4 /
y x y x
y x
= − = − −
=
Use INTERSECT to solve:
4.7–4.7
3.1
–3.1 Solution: 1.37, 2.14x y= − = or (–1.37, 2.14)
61. Graph: 4 44 41 2
3 4
12 ; 12 ;
2 / ; 2 /
y x y x
y x y x
= − = − −
= = −
Use INTERSECT to solve:
Solutions: 0.58, 1.86;x y= = 1.81, 1.05;x y= =
1.81, 1.05;x y= = − ; 0.58, 1.86x y= = − or
(0.58, 1.86), (1.81, 1.05), (1.81, –1.05), (0.58, –1.86)
62. Graph: 4 44 41 2
3
6 ; 6 ;
1/
y x y x
y x
= − = − −=
Use INTERSECT to solve:
Solutions: 0.64, 1.55;x y= = 1.55, 0.64;x y= =
0.64, 1.55;x y= − = − ; 1.55, 0.64x y= − = − or
(0.64, 1.55), (1.55, 0.64), (–0.64, –1.55), (–1.55, –0.64)
63. Graph: 1 22 / ; lny x y x= =
Use INTERSECT to solve:
4.7–4.7
3.1
–3.1 Solution: 2.35, 0.85x y= = or (2.35, 0.85)
Section 11.6: Systems of Nonlinear Equations
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
64. Graph: 2 21 2
3
4 ; 4 ;
ln
y x y x
y x
= − = − −=
Use INTERSECT to solve:
Solution: 1.90, 0.64;x y= = 0.14, 2.00x y= = −
or (1.90, 0.64), (0.14, –2.00)
65. Solve the first equation for x, substitute into the second equation and solve:
2 2
2 0 2
( 1) ( 1) 5
x y x y
x y
+ = = −
− + − =
2 2
2 2 2
( 2 1) ( 1) 5
4 4 1 2 1 5 5 2 3 0
(5 3)( 1) 0
3=0.6 or 1
56
= 1.2 or 25
y y
y y y y y y
y y
y y
x x
− − + − =
+ + + − + = + − =− + =
= = −
= − − =
The points of intersection are 6 3
, , (2, –1)5 5
−
.
66. Solve the first equation for x, substitute into the second equation and solve:
2 2
2 6 0 2 6
( 1) ( 1) 5
x y x y
x y
+ + = = − −
+ + + =
2 2
2 2
2
( 2 6 1) ( 1) 5
4 20 25 2 1 5
5 22 21 0
(5 7)( 3) 0
y y
y y y y
y y
y y
− − + + + =
+ + + + + =
+ + =+ + =
7
or 3516
or 05
y y
x x
= − = −
= − =
The points of intersection are 16 7
, , (0, 3)5 5
− − −
.
67. Complete the square on the second equation.
2
2
4 4 1 4
( 2) 3
y y x
y x
+ + = − +
+ = +
Substitute this result into the first equation. 2
2
2
( 1) 3 4
2 1 3 4
0
( 1) 0
0 or 1
x x
x x x
x x
x x
x x
− + + =
− + + + =
− =− =
= =2
2
If 0 : ( 2) 0 3
2 3 2 3
If 1: ( 2) 1 3
2 2 2 2
x y
y y
x y
y y
= + = +
+ = ± = − ±
= + = ++ = ± = − ±
The points of intersection are:
( ) ( ) ( ) ( )0, 2 3 , 0, 2 3 , 1, – 4 , 1, 0− − − + .
Chapter 11: Systems of Equations and Inequalities
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
68. Complete the square on the second equation, substitute into the first equation and solve:
2 2
2
( 2) ( 1) 4
2 5 0
x y
y y x
+ + − =
− − − =
2
2
2 1 5 1
( 1) 6
y y x
y x
− + = + +
− = +
2
2
2
( 2) 6 4
4 4 6 4
5 6 0
( 2)( 3) 0
2 or 3
x x
x x x
x x
x x
x x
+ + + =
+ + + + =
+ + =+ + =
= − = −
2
2
If 2 : ( 1) 2 6 1 2
1 or 3
If 3 : ( 1) 3 6 1 3
1 3
x y y
y y
x y y
y
= − − = − + − = ± = − =
= − − = − + − = ±
= ±
The points of intersection are:
( ) ( )3, 1 3 , 3,1 3 , ( 2, 1), ( 2, 3)− − − + − − − .
69. Solve the first equation for x, substitute into the second equation and solve:
2 2
4
3
6 1 0
yx
x x y
= − − + + =
4
34
3
43
yx
xy
xy
=−
− =
= +
22
22
22
4 43 6 3 1 0
16 24 249 18 1 0
168 0
yy y
yy yy
yy
+ − + + + =
+ + − − + + =
+ − =
4 2
4 2
2 2
2
2
16 8 0
8 16 0
( 4) 0
4 0
4
2
y y
y y
y
y
y
y
+ − =
− + =
− =
− =
== ±
4If 2 : 3 5
24
If 2 : 3 12
y x
y x
= = + =
= − = + =−
The points of intersection are: (1, –2), (5, 2).
Section 11.6: Systems of Nonlinear Equations
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
70. Substitute the first equation into the second equation and solve:
2 2
4
2
4 4 0
yx
x x y
= + + + − =
( )( )( )
( )
22
22
2 2
2 2
4 3 2
4 3 2
2 2
2 2
44 4 0
2
44 4
2
( 2) 4 4 16
4 4 4 4 16
8 16 16 16
8 16 0
8 16 0
( 4) 0
0 or 4
2 2
x xx
x xx
x x x
x x x x
x x x
x x x
x x x
x x
x x
y y
+ + − = +
+ − = − +
+ + − = −
+ + + − = −
+ + − = −
+ + =
+ + =
+ == = −= = −
The points of intersection are: (0, 2), (–4, –2).
71. Let and x y be the two numbers. The system of
equations is:
2 2
2 2
10
x y x y
x y
− = = +
+ =
Solve the first equation for x, substitute into the second equation and solve:
( )
( )( )
2 2
2 2
2
2 10
4 4 10
2 3 0
3 1 0 3 or 1
y y
y y y
y y
y y y y
+ + =
+ + + =
+ − =+ − = = − =
If 3 : 3 2 1
If 1: 1 2 3
y x
y x
= − = − + = −= = + =
The two numbers are 1 and 3 or –1 and –3.
72. Let and x y be the two numbers. The system of
equations is:
2 2
7 7
21
x y x y
x y
+ = = −
− =
Solve the first equation for x, substitute into the second equation and solve:
( )2 2
2 2
7 21
49 14 21
14 28
2 7 2 5
y y
y y y
y
y x
− − =
− + − =− = −
= = − =
The two numbers are 2 and 5.
73. Let and x y be the two numbers. The system of
equations is:
2 2
44
8
xy xy
x y
= = + =
Solve the first equation for x, substitute into the second equation and solve:
( )
22
22
4 2
4 2
22
2
48
168
16 8
8 16 0
4 0
4
2
yy
yy
y y
y y
y
y
y
+ =
+ =
+ =
− + =
− =
== ±
4 4If 2 : 2; If 2 : 2
2 2y x y x= = = = − = = −
−
The two numbers are 2 and 2 or –2 and –2.
74. Let and x y be the two numbers. The system of
equations is:
2 2
1010
21
xy xy
x y
= = − =
Solve the first equation for x, substitute into the second equation and solve:
Chapter 11: Systems of Equations and Inequalities
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( )( )
22
22
4 2
4 2
2 2
1021
10021
100 21
21 100 0
4 25 0
yy
yy
y y
y y
y y
− =
− =
− =
+ − =
− + =
2 4
2
y
y
== ±
or 2 25 (no real solution)y = −
10210
2
If 2 : 5
If 2 : 5
y x
y x −
= = =
= − = = −
The two numbers are 2 and 5 or –2 and –5.
75. Let and x y be the two numbers. The system of
equations is:
1 15
x y xy
x y
− = + =
Solve the first equation for x, substitute into the second equation and solve:
( )1 1
x xy y
yx y y x
y
− =
− = =−
1
1 13 3
1 23 3
1 15
1 15
25
2 5
6 2
1 1
3 21
yy
y
y
y y
y
y
y y
y
y x
−+ =
− + =
− =
− ==
= = = =−
The two numbers are 12 and 1
3 .
76. Let and x y be the two numbers. The system of
equations is:
1 13
x y xy
x y
+ = − =
Solve the first equation for x, substitute into the
second equation and solve:
( )1 1
xy x y
yx y y x
y
− =
− = =−
1
1 13
1 13
23
2 3
2 2
1 11
1 1 2
yy
y
y
y y
y
y
y y
y
y x
−− =
− − =
− =
− == −
−= − = =− −
The two numbers are 1− and 12 .
77. 2
310 10
a
ba b a b
= + = = −
Solve the second equation for a , substitute into the first equation and solve:
10 2
33(10 ) 2
30 3 2
30 5
6 4
b
bb b
b b
b
b a
− =
− =− =
== =
10; 2a b b a+ = − =
The ratio of a b+ to b a− is 102 5= .
78. 4
314 14
a
ba b a b
= + = = −
Solve the second equation for a , substitute into the first equation and solve:
( )
14 4
33 14 4
42 3 4
42 7
6 8
b
bb b
b b
b
b a
− =
− =− =
== =
2; 14a b a b− = + =
The ratio of a b− to a b+ is 2 1714 = .
Section 11.6: Systems of Nonlinear Equations
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79. Let x = the width of the rectangle. Let y = the length of the rectangle.
2 2 16
15
x y
xy
+ = =
Solve the first equation for y, substitute into the second equation and solve. 2 2 16
2 16 2
8
x y
y x
y x
+ == −= −
( )
( )( )
2
2
8 15
8 15
8 15 0
5 3 0
x x
x x
x x
x x
− =
− =
− + =− − =
5x = or 3x = The dimensions of the rectangle are 3 inches by 5 inches.
80. Let 2x = the side of the first square. Let 3x = the side of the second square.
( ) ( )2 2
2 2
2
2
2 3 52
4 9 52
13 52
4
2
x x
x x
x
x
x
+ =
+ =
=
== ±
Note that we must have 0x > . The sides of the first square are (2)(2) = 4 feet and the sides of the second square are (3)(2) = 6 feet.
81. Let x = the radius of the first circle. Let y = the radius of the second circle.
2 2
2 2 12
20
x y
x y
π + π = π
π + π = π
Solve the first equation for y, substitute into the second equation and solve: 2 2 12
6
6
x y
x y
y x
π + π = π+ =
= −
2 2
2 2
2 2
2 2 2
2
20
20
(6 ) 20
36 12 20 2 12 16 0
6 8 0 ( 4)( 2) 0
4 or 2
2 4
x y
x y
x x
x x x x x
x x x x
x x
y y
π + π = π
+ =
+ − =
+ − + = − + =
− + = − − == == =
The radii of the circles are 2 centimeters and 4 centimeters.
82. Let x = the length of each of the two equal sides in the isosceles triangle. Let y = the length of the base.
The perimeter of the triangle: 18x x y+ + =
Since the altitude to the base y is 3, the Pythagorean theorem produces another equation.
2 22 2 23 9
2 4
y yx x
+ = + =
Solve the system of equations:
22
2 18 18 2
94
x y y x
yx
+ = = −
+ =
Solve the first equation for y, substitute into the second equation and solve.
( )
( )
22
22
2 2
18 29
4
324 72 49
4
81 18 9
18 90
5 18 2 5 8
xx
x xx
x x x
x
x y
−+ =
− + + =
− + + =− = −
= = − =
The base of the triangle is 8 centimeters.
83. The tortoise takes 9 + 3 = 12 minutes or 0.2 hour longer to complete the race than the hare. Let r = the rate of the hare. Let t = the time for the hare to complete the race. Then t + 0.2 = the time for the tortoise and
0.5r − = the rate for the tortoise. Since the length of the race is 21 meters, the distance equations are:
( ) ( )
2121
0.5 0.2 21
r t rt
r t
= = − + =
Solve the first equation for r, substitute into the
Chapter 11: Systems of Equations and Inequalities
1178
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second equation and solve:
( )
( )
( )( )
2
2
210.5 0.2 21
4.221 0.5 0.1 21
4.210 21 0.5 0.1 10 21
210 42 5 210
5 42 0
5 14 3 0
tt
tt
t t tt
t t t t
t t
t t
− + =
+ − − =
+ − − = ⋅
+ − − =
+ − =− + =
5 14 0
5 14
142.8
5
t
t
t
− ==
= =
or 3 0
3
t
t
+ == −
3t = − makes no sense, since time cannot be negative. Solve for r:
217.5
2.8r = =
The average speed of the hare is 7.5 meters per hour, and the average speed for the tortoise is 7 meters per hour.
84. Let 1 2 3, ,v v v = the speeds of runners 1, 2, 3.
Let 1 2 3, ,t t t = the times of runners 1, 2, 3.
Then by the conditions of the problem, we have the following system:
1 1
2 1
3 1
2 2
5280
5270
5260
5280
v t
v t
v t
v t
= = = =
Distance between the second runner and the third runner after 2t seconds is:
2 23 2 3 1
2 1
5280 5280
52805280 5260
5270
10.02
v tv t v t
v t
− = −
= −
≈
The second place runner beats the third place runner by about 10.02 feet.
85. Let x = the width of the cardboard. Let y = the
length of the cardboard. The width of the box will be 4x − , the length of the box will be
4y − , and the height is 2. The volume is
( 4)( 4)(2)V x y= − − .
Solve the system of equations: 216
216
2( 4)( 4) 224
xy yx
x y
= = − − =
Solve the first equation for y, substitute into the second equation and solve.
( )
( )( )
2
2
2
2162 8 4 224
1728432 8 32 224
432 8 1728 32 224
8 240 1728 0
30 216 0
12 18 0
xx
xx
x x x x
x x
x x
x x
− − =
− − + =
− − + =
− + =
− + =− − =
12 0
12
x
x
− ==
or 18 0
18
x
x
− ==
The cardboard should be 12 centimeters by 18 centimeters.
86. Let x = the width of the cardboard. Let y = the
length of the cardboard. The area of the cardboard is: 216xy =
The volume of the tube is: 2 224V r h= π =
where and 2 or 2
xh y r x r= π = =
π.
Solve the system of equations:
2 2
216216
224 2242 4
xy yx
x x yy
= =
π = = π π
Solve the first equation for y, substitute into the second equation and solve.
( )2 216
2244216 896
89613.03
216
xx
x
x
πππ
=
=
= ≈
( )2
896216
216216 21616.57
896y
x π π= = = ≈
The cardboard should be about 13.03 centimeters by 16.57 centimeters.
Section 11.6: Systems of Nonlinear Equations
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87. Find equations relating area and perimeter: 2 2 4500
3 3 ( ) 300
x y
x y x y
+ =
+ + − =
Solve the second equation for y, substitute into the first equation and solve: 4 2 300
2 300 4
150 2
x y
y x
y x
+ == −= −
2 2
2 2
2
2
2
(150 2 ) 4500
22,500 600 4 4500
5 600 18,000 0
120 3600 0
( 60) 0
60 0
60
150 2(60) 30
x x
x x x
x x
x x
x
x
x
y
+ − =
+ − + =
− + =
− + =
− =− =
== − =
The sides of the squares are 30 feet and 60 feet.
88. Let x = the length of a side of the square. Let r = the radius of the circle.
The area of the square is 2x and the area of the
circle is 2rπ . The perimeter of the square is 4x
and the circumference of the circle is 2 rπ . Find
equations relating area and perimeter: 2 2 100
4 2 60
x r
x r
+ π =
+ π =
Solve the second equation for x, substitute into the first equation and solve: 4 2 60
4 60 2
115
2
x r
x r
x r
+ π == − π
= − π
22
2 2 2
2 2
115 100
2
1225 15 100
41
15 125 04
r r
r r r
r r
− π + π =
− π + π + π =
π + π − π + =
2 2 2
2 2
2
14 ( 15 ) 4 (125)
4
1225 500
4
100 500 0
b ac − = − π − π + π = π − π + π
= π − π <
Since the discriminant is less than zero, it is impossible to cut the wire into two pieces whose total area equals 100 square feet.
89. Solve the system for and l w : 2 2l w P
l w A
+ = =
Solve the first equation for l , substitute into the second equation and solve.
2
2
2 2
2
2
2
02
l P w
Pl w
Pw w A
Pw w A
Pw w A
= −
= −
− =
− =
− + =
22
2
2
2 42 4
2
164 4
2 2
16162
2 4
P PP P
P P
AA
w
AP P A
± −± −= =
−± ± −= =
2
2 2
2
2 2
16If then
4
16 16
2 4 4
16If then
4
16 16
2 4 4
P P Aw
P P P A P P Al
P P Aw
P P P A P P Al
+ −=
+ − − −= − =
− −=
− − + −= − =
If it is required that length be greater than width, then the solution is:
2 216 16and
4 4
P P A P P Aw l
− − + −= =
90. Solve the system for and l b :
22 2
2 2
4
P b l b P l
bh l
= + = −
+ =
Solve the first equation for b , substitute into the second equation and solve.
Chapter 11: Systems of Equations and Inequalities
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
( )
2 2 2
22 2
2 2 2 2
2 2
2 2
4 4
4 2 4
4 4 4 4
4 4
4
4
h b l
h P l l
h P Pl l l
h P Pl
h Pl
P
+ =
+ − =
+ − + =
+ =
+=
2 2 2 24 4
2 2
h P P hb P
P P
+ −= − =
91. Solve the equation: 2 4(2 4) 0m m− − =
( )
2
2
8 16 0
4 0
4
m m
m
m
− + =
− ==
Use the point-slope equation with slope 4 and the point (2, 4) to obtain the equation of the tangent line: 4 4( 2) 4 4 8 4 4y x y x y x− = − − = − = −
92. Solve the system: 2 2 10x y
y mx b
+ =
= +
Solve the system by substitution:
( )
( )
22
2 2 2 2
2 2 2
10
2 10 0
1 2 10 0
x mx b
x m x bmx b
m x bmx b
+ + =
+ + + − =
+ + + − =
Note that the tangent line passes through (1, 3). Find the relation between m and b: 3 (1) 3m b b m= + = −
There is one solution to the quadratic if the discriminant is zero.
( ) ( ) ( )2 2 2
2 2 2 2 2 2
2 2
2 4 1 10 0
4 4 40 4 40 0
40 4 40 0
bm m b
b m b m m b
m b
− + − =
− + − + =
− + =
Substitute for b and solve:
( )
( )
22
2 2
2
2
2
40 4 3 40 0
40 4 24 36 40 0
36 24 4 0
9 6 1 0
3 1 0
3 1
1
3
m m
m m m
m m
m m
m
m
m
− − + =
− + − + =
+ + =
+ + =
+ == −
= −
1 103 3
3 3b m
= − = − − =
The equation of the tangent line is 1 10
3 3y x= − + .
93. Solve the system: 2 2y x
y mx b
= +
= +
Solve the system by substitution: 2 22 2 0x mx b x mx b+ = + − + − =
Note that the tangent line passes through (1, 3). Find the relation between m and b: 3 (1) 3m b b m= + = −
Substitute into the quadratic to eliminate b: 2 22 (3 ) 0 ( 1) 0x mx m x mx m− + − − = − + − =
Find when the discriminant equals 0:
( ) ( )( )
( )
2
2
2
4 1 1 0
4 4 0
2 0
2 0
2
m m
m m
m
m
m
− − − =
− + =
− =− =
=
3 3 2 1b m= − = − = The equation of the tangent line is 2 1y x= + .
94. Solve the system: 2 5x y
y mx b
+ =
= +
Solve the system by substitution: 2 25 5 0x mx b x mx b+ + = + + − =
Note that the tangent line passes through (–2, 1). Find the relation between m and b:
( )1 2 2 1m b b m= − + = +
Substitute into the quadratic to eliminate b:
Section 11.6: Systems of Nonlinear Equations
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
( )
2
2
2 1 5 0
2 4 0
x mx m
x mx m
+ + + − =
+ + − =
Find when the discriminant equals 0:
( ) ( )( )
( )
2
2
2
4 1 2 4 0
8 16 0
4 0
4 0
4
m m
m m
m
m
m
− − =
− + =
− =− =
=
( )2 1 2 4 1 9b m= + = + =
The equation of the tangent line is 4 9y x= + .
95. Solve the system: 2 22 3 14x y
y mx b
+ =
= +
Solve the system by substitution:
( )
( )
22
2 2 2 2
2 2 2
2 3 14
2 3 6 3 14
3 2 6 3 14 0
x mx b
x m x mbx b
m x mbx b
+ + =
+ + + =
+ + + − =
Note that the tangent line passes through (1, 2). Find the relation between m and b: 2 (1) 2m b b m= + = −
Substitute into the quadratic to eliminate b: 2 2 2
2 2 2 2
(3 2) 6 (2 ) 3(2 ) 14 0
(3 2) (12 6 ) (3 12 2) 0
m x m m x m
m x m m x m m
+ + − + − − =
+ + − + − − = Find when the discriminant equals 0:
2 2 2 2
2
2
2
(12 6 ) 4(3 2)(3 12 2) 0
144 96 16 0
9 6 1 0
(3 1) 0
3 1 0
1
3
m m m m m
m m
m m
m
m
m
− − + − − =
+ + =
+ + =
+ =+ =
= −
1 72 2
3 3b m
= − = − − =
The equation of the tangent line is 1 7
3 3y x= − + .
96. Solve the system: 2 23 7x y
y mx b
+ =
= +
Solve the system by substitution:
( )
( )
22
2 2 2 2
2 2 2
3 7
3 2 7
3 2 7 0
x mx b
x m x mbx b
m x mbx b
+ + =
+ + + =
+ + + − =
Note that the tangent line passes through (–1, 2). Find the relation between m and b: 2 ( 1) 2m b b m= − + = +
There is one solution to the quadratic if the discriminant equals 0.
( ) ( )( )2 2 2
2 2 2 2 2 2
2 2
2 2
2 4 3 7 0
4 4 28 12 84 0
28 12 84 0
7 3 21 0
bm m b
b m b m m b
m b
m b
− + − =
− + − + =
− + =
− + =
Substitute for b and solve:
( )
( )
22
2 2
2
2
7 3 2 21 0
7 3 12 12 21 0
4 12 9 0
2 3 0
2 3
3
2
m m
m m m
m m
m
m
m
− + + =
− − − + =
− + =
− ==
=
3 72 2
2 2b m= + = + =
The equation of the tangent line is 3 7
2 2y x= + .
97. Solve the system: 2 2 3x y
y mx b
− =
= +
Solve the system by substitution:
( )
( )
22
2 2 2 2
2 2 2
3
2 3
1 2 3 0
x mx b
x m x mbx b
m x mbx b
− + =
− − − =
− − − − =
Note that the tangent line passes through (2, 1). Find the relation between m and b: 1 (2) 1 2m b b m= + = −
Substitute into the quadratic to eliminate b: 2 2 2
2 2 2 2
2 2 2 2
(1 ) 2 (1 2 ) (1 2 ) 3 0
(1 ) ( 2 4 ) 1 4 4 3 0
(1 ) ( 2 4 ) ( 4 4 4) 0
m x m m x m
m x m m x m m
m x m m x m m
− − − − − − =
− + − + − + − − =
− + − + + − + − =Find when the discriminant equals 0:
Chapter 11: Systems of Equations and Inequalities
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
( ) ( )( )
( )
22 2 2
2 3 4 4 3
2
2
2
2 4 4 1 4 4 4 0
4 16 16 16 16 16 16 0
4 16 16 0
4 4 0
2 0
2
m m m m m
m m m m m m
m m
m m
m
m
− + − − − + − =
− + − + − + =
− + =
− + =
− ==
The equation of the tangent line is 2 3y x= − .
98. Solve the system: 2 22 14y x
y mx b
− =
= +
Solve the system by substitution:
( )
( )
2 2
2 2 2 2
2 2 2
2 14
2 4 2 14
2 1 4 2 14 0
mx b x
m x mbx b x
m x mbx b
+ − =
+ + − =
− + + − =
Note that the tangent line passes through (2, 3). Find the relation between m and b: 3 (2) 3 2m b b m= + = −
There is one solution to the quadratic if the discriminant equals 0.
( ) ( )( )2 2 2
2 2 2 2 2 2
2 2
2 2
4 4 2 1 2 14 0
16 16 112 8 56 0
112 8 56 0
14 7 0
bm m b
b m b m m b
m b
m b
− − − =
− + + − =
+ − =
+ − =
Substitute for b and solve:
( )
( )
22
2 2
2
2
14 3 2 7 0
14 4 12 9 7 0
18 12 2 0
2 3 1 0
3 1
1
3
m m
m m m
m m
m
m
m
+ − − =
+ − + − =
− + =
− ==
=
1 73 2 3 2
3 3b m
= − = − =
The equation of the tangent line is 1 7
3 3y x= + .
99. Solve for 1 2and r r :
1 2
1 2
br r
ac
r ra
+ = − =
Substitute and solve:
1 2
2 2
22 2 0
br r
ab c
r ra a
b cr r
a a
= − −
− − =
− − − =
22 2 0ar br c+ + =
2
2
1 2
2
2
2
4
2
4
2
4 2
2 2
4
2
b b acr
ab
r ra
b b ac b
a a
b b ac b
a a
b b ac
a
− ± −=
= − − =
− ± − = − −
−= −
− −=
The solutions are: 2 2
1 24 4
and 2 2
b b ac b b acr r
a a
− + − − − −= = .
Section 11.6: Systems of Nonlinear Equations
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100. Consider the circle with equation
( ) ( )2 2 2x h y k r− + − = and the third degree
polynomial with equation 3 2y ax bx cx d= + + + .
Substituting the second equation into the first equation yields
( ) ( )22 3 2 2x h ax bx cx d k r− + + + + − = .
In order to find the roots for this equation we can expand the terms on the left hand side of the
equation. Notice that ( )2x h− yields a 2nd degree
polynomial, and ( )23 2ax bx cx d k+ + + − yields a
6th degree polynomial. Therefore, we need to find the roots of a 6th degree equation, and the Fundamental Theorem of Algebra states that there will be at most six real roots. Thus, the circle and the 3rd degree polynomial will intersect at most six times. Now consider the circle with equation
( ) ( )2 2 2x h y k r− + − = and the polynomial of
degree n with equation 2 3
0 1 2 3 ... nny a a x a x a x a x= + + + + + .
Substituting the first equation into the first equation yields
( ) ( )22 2 3 20 1 2 3 ... n
nx h a a x a x a x a x k r− + + + + + + − =
In order to find the roots for this equation we can expand the terms on the left hand side of the equation.
Notice that ( )2x h− yields a 2nd degree polynomial,
and ( )22 30 1 2 3 ... n
na a x a x a x a x k+ + + + + −
yields a polynomial of degree 2n. Therefore, we need to find the roots of an equation of degree 2n, and the Fundamental Theorem of Algebra states that there will be at most 2n real roots. Thus, the circle and the nth degree polynomial will intersect at most 2n times.
101. Since the area of the square piece of sheet metal is 100 square feet, the sheet’s dimensions are 10 feet by 10 feet. Let x = the length of the cut.
10 – 2x 10 – 2x
x
10
10
xx
The dimensions of the box are: length 10 2 ;x= −
width 10 2 ; heightx x= − = . Note that each of
these expressions must be positive. So we must have 0 and 10 2 0 5,x x x> − > < that is, 0 5x< < . So the volume of the box is given by
( ) ( ) ( )( )( ) ( )( ) ( )2
length width height
10 2 10 2
10 2
V
x x x
x x
= ⋅ ⋅
= − −
= −
a. In order to get a volume equal to 9 cubic feet,
we solve ( ) ( )210 2 9.x x− =
( ) ( )( )
2
2
2 3
10 2 9
100 40 4 9
100 40 4 9
x x
x x x
x x x
− =
− + =
− + =
So we need to solve the equation 3 24 40 100 9 0x x x− + − = .
Graphing 3 21 4 40 100 9y x x x= − + − on a
calculator yields the graph
10−2
−40
80
10−2
−40
80
10−2
−40
80
The graph indicates that there are three real zeros on the interval [0, 6]. Using the ZERO feature of a graphing calculator, we find that the three roots shown occur at 0.093x ≈ ,
4.274x ≈ and 5.632x ≈ . But we’ve already noted that we must have 0< <5x , so the only practical values for the sides of the square base are 0.093x ≈ feet and 4.274x ≈ feet.
b. Answers will vary.
Chapter 11: Systems of Equations and Inequalities
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Section 11.7
1. 3 4 8
4 4
1
x x
x
x
+ < −<<
{ } ( )1 or ,1x x < −∞
2. 3 2 6x y− =
The graph is a line. x-intercept: ( )3 2 0 6
3 6
2
x
x
x
− ===
y-intercept: ( )3 0 2 6
2 6
3
y
y
y
− =− =
= −
3. 2 2 9x y+ =
The graph is a circle. Center: (0, 0) ; Radius: 3
4. 2 4y x= +
The graph is a parabola.
x-intercepts: 2
2
0 4
4, no intercepts
x
x x
= +
= − −
y-intercept: 20 4 4y = + =
The vertex has x-coordinate:
( )0
02 2 1
bx
a= − = − = .
The y-coordinate of the vertex is 20 4 4y = + = .
5. True
6. 2y x= ; right; 2
7. dashes; solid
8. half-planes
9. False, see example 7b.
10. unbounded
11. 0x ≥ Graph the line 0x = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (2, 0). Since 2 ≥ 0 is true, shade the side of the line containing (2, 0).
12. 0y ≥
Graph the line 0y = . Use a solid line since the
inequality uses ≥. Choose a test point not on the line, such as (0, 2). Since 2 ≥ 0 is true, shade the side of the line containing (0, 2).
Section 11.7: Systems of Inequalities
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13. 4x ≥ Graph the line 4x = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (5, 0). Since 5 ≥ 0 is true, shade the side of the line containing (5, 0).
14. 2y ≤
Graph the line 2y = . Use a solid line since the
inequality uses ≤. Choose a test point not on the line, such as (5, 0). Since 0 ≤ 2 is true, shade the side of the line containing (5, 0).
15. 2 6x y+ ≥
Graph the line 2 6x y+ = . Use a solid line since
the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 6+ ≥ is
false, shade the opposite side of the line from (0, 0).
16. 3 2 6x y+ ≤
Graph the line 3 2 6x y+ = . Use a solid line since
the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 3(0) 2(0) 6+ ≤ is
true, shade the side of the line containing (0, 0).
17. 2 2 1x y+ >
Graph the circle 2 2 1x y+ = . Use a dashed line
since the inequality uses >. Choose a test point
not on the circle, such as (0, 0). Since 2 20 0 1+ > is false, shade the opposite side of the circle from (0, 0).
18. 2 2 9x y+ ≤
Graph the circle 2 2 9x y+ = . Use a solid line
since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since
2 20 0 9+ ≤ is true, shade the same side of the circle as (0, 0).
Chapter 11: Systems of Equations and Inequalities
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19. 2 1y x≤ −
Graph the parabola 2 1y x= − . Use a solid line
since the inequality uses ≤ . Choose a test point not on the parabola, such as (0, 0). Since
20 0 1≤ − is false, shade the opposite side of the parabola from (0, 0).
20. 2 2y x> +
Graph the parabola 2 2y x= + . Use a dashed line
since the inequality uses >. Choose a test point not on the parabola, such as (0, 0). Since
20 0 2> + is false, shade the opposite side of the parabola from (0, 0).
21. 4xy ≥
Graph the hyperbola 4xy = . Use a solid line
since the inequality uses ≥ . Choose a test point not on the hyperbola, such as (0, 0). Since 0 0 4⋅ ≥ is false, shade the opposite side of the hyperbola from (0, 0).
22. 1xy ≤
Graph the hyperbola 1xy = . Use a solid line
since the inequality uses ≤ . Choose a test point not on the hyperbola, such as (0, 0). Since 0 0 1⋅ ≤ is true, shade the same side of the hyperbola as (0, 0).
23. 2
2 4
x y
x y
+ ≤ + ≥
Graph the line 2x y+ = . Use a solid line since
the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 4x y+ = . Use a solid line since the
inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 4+ ≥ is false,
shade the opposite side of the line from (0, 0). The overlapping region is the solution.
24. 3 6
2 2
x y
x y
− ≥ + ≤
Graph the line 3 6x y− = . Use a solid line since
the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 3(0) 0 6− ≥ is
false, shade the opposite side of the line from (0, 0). Graph the line 2 2x y+ = . Use a solid
line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 2+ ≤ is true, shade the side of the line
containing (0, 0). The overlapping region is the solution.
Section 11.7: Systems of Inequalities
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25. 2 4
3 2 6
x y
x y
− ≤ + ≥ −
Graph the line 2 4x y− = . Use a solid line since
the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 4− ≤ is true,
shade the side of the line containing (0, 0). Graph the line 3 2 6x y+ = − . Use a solid line since the
inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 3(0) 2(0) 6+ ≥ − is
true, shade the side of the line containing (0, 0). The overlapping region is the solution.
26. 4 5 0
2 2
x y
x y
− ≤ − ≥
Graph the line 4 5 0x y− = . Use a solid line since
the inequality uses ≤. Choose a test point not on the line, such as (2, 0). Since 4(2) 5(0) 0− ≤ is
false, shade the opposite side of the line from (2, 0). Graph the line 2 2x y− = . Use a solid line
since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2− ≥ is
false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
27. 2 3 0
3 2 6
x y
x y
− ≤ + ≤
Graph the line 2 3 0x y− = . Use a solid line since
the inequality uses ≤. Choose a test point not on the line, such as (0, 3). Since 2(0) 3(3) 0− ≤ is
true, shade the side of the line containing (0, 3). Graph the line 3 2 6x y+ = . Use a solid line since
the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 3(0) 2(0) 6+ ≤ is
true, shade the side of the line containing (0, 0). The overlapping region is the solution.
28. 4 2
2 2
x y
x y
− ≥ + ≥
Graph the line 4 2x y− = . Use a solid line since
the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 4(0) 0 2− ≥ is
false, shade the opposite side of the line from (0, 0). Graph the line 2 2x y+ = . Use a solid
line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 2+ ≥ is false, shade the opposite side of
the line from (0, 0). The overlapping region is the solution.
Chapter 11: Systems of Equations and Inequalities
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29. 2 6
2 4 0
x y
x y
− ≤ − ≥
Graph the line 2 6x y− = . Use a solid line since
the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 6− ≤ is true,
shade the side of the line containing (0, 0). Graph the line 2 4 0x y− = . Use a solid line since the
inequality uses ≥. Choose a test point not on the line, such as (0, 2). Since 2(0) 4(2) 0− ≥ is false,
shade the opposite side of the line from (0, 2). The overlapping region is the solution.
30. 4 8
4 4
x y
x y
+ ≤ + ≥
Graph the line 4 8x y+ = . Use a solid line since
the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 4(0) 8+ ≤ is true,
shade the side of the line containing (0, 0). Graph the line 4 4x y+ = . Use a solid line since the
inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 4(0) 4+ ≥ is false,
shade the opposite side of the line from (0, 0). The overlapping region is the solution.
31. 2 2
2 2
x y
x y
+ ≥ − + ≥
Graph the line 2 2x y+ = − . Use a solid line
since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2+ ≥ − is true, shade the side of the line
containing (0, 0). Graph the line 2 2x y+ = . Use
a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2+ ≥ is false, shade the opposite side of
the line from (0, 0). The overlapping region is the solution.
32. 4 4
4 0
x y
x y
− ≤ − ≥
Graph the line 4 4x y− = . Use a solid line since
the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 4(0) 4− ≤ is true,
shade the side of the line containing (0, 0). Graph the line 4 0x y− = . Use a solid line since the
inequality uses ≥. Choose a test point not on the line, such as (1, 0). Since 1 4(0) 0− ≥ is true,
shade the side of the line containing (1, 0). The overlapping region is the solution.
Section 11.7: Systems of Inequalities
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33. 2 3 6
2 3 0
x y
x y
+ ≥ + ≤
Graph the line 2 3 6x y+ = . Use a solid line since
the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 6+ ≥ is
false, shade the opposite side of the line from (0, 0). Graph the line 2 3 0x y+ = . Use a solid
line since the inequality uses ≤. Choose a test point not on the line, such as (0, 2). Since 2(0) 3(2) 0+ ≤ is false, shade the opposite side of
the line from (0, 2). Since the regions do not overlap, the solution is an empty set.
34. 2 0
2 2
x y
x y
+ ≥ + ≥
Graph the line 2 0x y+ = . Use a solid line since
the inequality uses ≥. Choose a test point not on the line, such as (1, 0). Since 2(1) 0 0+ ≥ is true,
shade the side of the line containing (1, 0). Graph the line 2 2x y+ = . Use a solid line since the
inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2+ ≥ is false,
shade the opposite side of the line from (0, 0). The overlapping region is the solution.
35. 2 2 9
3
x y
x y
+ ≤
+ ≥
Graph the circle 2 2 9x y+ = . Use a solid line
since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since
2 20 0 9+ ≤ is true, shade the same side of the circle as (0, 0). Graph the line 3x y+ = . Use a solid line since
the inequality uses ≥ . Choose a test point not on the line, such as (0, 0). Since 0 0 3+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
36. 2 2 9
3
x y
x y
+ ≥
+ ≤
Graph the circle 2 2 9x y+ = . Use a solid line
since the inequality uses ≥. Choose a test point not on the circle, such as (0, 0). Since
2 20 0 9+ ≥ is false, shade the opposite side of the circle as (0, 0). Graph the line 3x y+ = . Use a solid line since
the inequality uses ≤ . Choose a test point not on the line, such as (0, 0). Since 0 0 3+ ≤ is true, shade the same side of the line as (0, 0). The overlapping region is the solution.
Chapter 11: Systems of Equations and Inequalities
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37. 2 4
2
y x
y x
≥ −
≤ −
Graph the parabola 2 4y x= − . Use a solid line
since the inequality uses ≥ . Choose a test point not on the parabola, such as (0, 0). Since
20 0 4≥ − is true, shade the same side of the parabola as (0, 0). Graph the line 2y x= − . Use
a solid line since the inequality uses ≤ . Choose a test point not on the line, such as (0, 0). Since 0 0 2≤ − is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
38. 2y x
y x
≤
≥
Graph the parabola 2y x= . Use a solid line
since the inequality uses ≤ . Choose a test point
not on the parabola, such as (1, 2). Since 22 1≤ is false, shade the opposite side of the parabola from (1, 2). Graph the line y x= . Use a solid
line since the inequality uses ≥ . Choose a test point not on the line, such as (1, 2). Since 2 1≥ is true, shade the same side of the line as (1, 2). The overlapping region is the solution.
39. 2 2
2
16
4
x y
y x
+ ≤
≥ −
Graph the circle 2 2 16x y+ = . Use a sold line
since the inequality is not strict. Choose a test point not on the circle, such as (0,0) . Since
2 20 0 16+ ≤ is true, shade the side of the circle containing (0,0) . Graph the parabola
2 4y x= − . Use a solid line since the inequality
is not strict. Choose a test point not on the
parabola, such as (0,0) . Since 20 0 4≥ − is true,
shade the side of the parabola that contains (0,0) . The overlapping region is the solution.
40. 2 2
2
25
5
x y
y x
+ ≤
≤ −
Graph the circle 2 2 25x y+ = . Use a sold line
since the inequality is not strict. Choose a test point not on the circle, such as (0,0) . Since
2 20 0 25+ ≤ is true, shade the side of the circle containing (0,0) . Graph the parabola
2 5y x= − . Use a solid line since the inequality
is not strict. Choose a test point not on the
parabola, such as (0,0) . Since 20 0 5≤ − is
false, shade the side of the parabola opposite that which contains the point (0,0) . The overlapping
region is the solution.
Section 11.7: Systems of Inequalities
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41. 2
4
1
xy
y x
≥
≥ +
Graph the hyperbola 4xy = . Use a solid line
since the inequality uses ≥ . Choose a test point not on the parabola, such as (0, 0). Since 0 0 4⋅ ≥ is false, shade the opposite side of the hyperbola from (0, 0). Graph the parabola
2 1y x= + . Use a solid line since the inequality
uses ≥ . Choose a test point not on the parabola,
such as (0, 0). Since 20 0 1≥ + is false, shade the opposite side of the parabola from (0, 0).The overlapping region is the solution.
42. 2
2
1
1
y x
y x
+ ≤
≥ −
Graph the parabola 2 1y x+ = . Use a solid line
since the inequality uses ≤ . Choose a test point not on the parabola, such as (0, 0). Since
0 + 02 ≤ 1 is true, shade the same side of the parabola as (0, 0). Graph the parabola
2 1y x= − . Use a solid line since the inequality
uses ≥ . Choose a test point not on the parabola,
such as (0, 0). Since 20 0 1≥ − is true, shade the same side of the parabola as (0, 0). The overlapping region is the solution.
43.
0
0
2 6
2 6
x
y
x y
x y
≥ ≥ + ≤ + ≤
Graph 0; 0x y≥ ≥ . Shaded region is the first
quadrant. Graph the line 2 6x y+ = . Use a solid
line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 6+ ≤ is true, shade the side of the line
containing (0, 0). Graph the line 2 6x y+ = . Use
a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 6+ ≤ is true, shade the side of the line
containing (0, 0). The overlapping region is the solution. The graph is bounded. Find the vertices:
The x-axis and y-axis intersect at (0, 0). The intersection of 2 6x y+ = and the y-axis is (0, 3).
The intersection of 2 6x y+ = and the x-axis is
(3, 0). To find the intersection of 2 6x y+ = and
2 6x y+ = , solve the system:
2 6
2 6
x y
x y
+ = + =
Solve the first equation for x: 6 2x y= − .
Substitute and solve: 2(6 2 ) 6
12 4 6
12 3 6
3 6
2
y y
y y
y
y
y
− + =− + =
− =− = −
=
6 2(2) 2x = − =
The point of intersection is (2, 2). The four corner points are (0, 0), (0, 3), (3, 0), and (2, 2).
Chapter 11: Systems of Equations and Inequalities
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44.
0
0
4
2 3 6
x
y
x y
x y
≥ ≥ + ≥ + ≥
Graph 0; 0x y≥ ≥ . Shaded region is the first
quadrant. Graph the line 4x y+ = . Use a solid
line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 4+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 3 6x y+ = . Use
a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 6+ ≥ is false, shade the opposite side of
the line from (0, 0). The overlapping region is the solution. The graph is unbounded.
Find the vertices: The intersection of 4x y+ = and the y-axis is
(0, 4). The intersection of 4x y+ = and the x-
axis is (4, 0). The two corner points are (0, 4), and (4, 0).
=
45.
0
0
2
2 4
x
y
x y
x y
≥ ≥ + ≥ + ≥
Graph 0; 0x y≥ ≥ . Shaded region is the first
quadrant. Graph the line 2x y+ = . Use a solid
line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 + 0 ≥ 2 is false, shade the opposite side of the line from (0, 0). Graph the line 2 4x y+ = . Use
a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 4+ ≥ is false, shade the opposite side of
the line from (0, 0). The overlapping region is the solution. The graph is unbounded.
Find the vertices: The intersection of 2x y+ = and the x-axis is
(2, 0). The intersection of 2 4x y+ = and the y-
axis is (0, 4). The two corner points are (2, 0), and (0, 4).
46.
0
0
3 6
2 2
x
y
x y
x y
≥ ≥ + ≤ + ≤
Graph 0; 0x y≥ ≥ . Shaded region is the first
quadrant. Graph the line 3 6x y+ = . Use a solid
line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 3(0) 0 6+ ≤ is true, shade the side of the line
containing (0, 0). Graph the line 2 2x y+ = . Use
a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2+ ≤ is true, shade the side of the line
containing (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of 0 and 0x y= = is (0, 0).
The intersection of 2 2x y+ = and the x-axis is
(1, 0). The intersection of 2 2x y+ = and the y-
axis is (0, 2). The three corner points are (0, 0), (1, 0), and (0, 2).
Section 11.7: Systems of Inequalities
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47.
0
0
2
2 3 12
3 12
x
y
x y
x y
x y
≥ ≥ + ≥ + ≤
+ ≤
Graph 0; 0x y≥ ≥ . Shaded region is the first
quadrant. Graph the line 2x y+ = . Use a solid
line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 3 12x y+ = .
Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 12+ ≤ is true, shade the side of
the line containing (0, 0). Graph the line 3 12x y+ = . Use a solid line since the inequality
uses ≤. Choose a test point not on the line, such as (0, 0). Since 3(0) 0 12+ ≤ is true, shade the
side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of 2x y+ = and the y-axis is
(0, 2). The intersection of 2x y+ = and the x-
axis is (2, 0). The intersection of 2 3 12x y+ =
and the y-axis is (0, 4). The intersection of 3 12x y+ = and the x-axis is (4, 0).
To find the intersection of 2 3 12x y+ = and
3 12x y+ = , solve the system:
2 3 12
3 12
x y
x y
+ = + =
Solve the second equation for y: 12 3y x= − .
Substitute and solve: 2 3(12 3 ) 12
2 36 9 12
7 24
24
7
x x
x x
x
x
+ − =+ − =
− = −
=
24 72 1212 3 12
7 7 7y
= − = − =
The point of intersection is 24 12
,7 7
.
The five corner points are (0, 2), (0, 4), (2, 0),
(4, 0), and 24 12
,7 7
.
48.
0
0
2
8
2 10
x
y
x y
x y
x y
≥ ≥ + ≥ + ≤
+ ≥
Graph 0; 0x y≥ ≥ . Shaded region is the first
quadrant. Graph the line 2x y+ = . Use a solid
line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 8x y+ = . Use a
solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 8+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 10x y+ = .
Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 10+ ≥ is false, shade the opposite
side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of 2x y+ = and the y-axis is
(0, 2). The intersection of 2 10x y+ = and the y-
axis is (0, 10). To find the intersection of 2 10 and 8x y x y+ = + = , solve the system:
2 10
8
x y
x y
+ = + =
Solve the second equation for y: 8y x= − .
Substitute and solve: 2 8 10
2
x x
x
+ − ==
8 2 6y = − =
The point of intersection is (2, 6). The five corner points are (0, 2), (2, 0), (0, 8), (5, 0) and (2, 6).
Chapter 11: Systems of Equations and Inequalities
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
49.
0
0
2
8
2 10
x
y
x y
x y
x y
≥ ≥ + ≥ + ≤
+ ≤
Graph 0; 0x y≥ ≥ . Shaded region is the first
quadrant. Graph the line 2x y+ = . Use a solid
line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 8x y+ = . Use a
solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 8+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 10x y+ = . Use
a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 10+ ≤ is true, shade the side of the line
containing (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of 2x y+ = and the y-axis is
(0, 2). The intersection of 2x y+ = and the x-axis
is (2, 0). The intersection of 8x y+ = and the y-
axis is (0, 8). The intersection of 2 10x y+ = and
the x-axis is (5, 0). To find the intersection of 8 and 2 10x y x y+ = + = , solve the system:
8
2 10
x y
x y
+ = + =
Solve the first equation for y: 8y x= − .
Substitute and solve: 2 8 10
2
x x
x
+ − ==
8 2 6y = − =
The point of intersection is (2, 6). The five corner points are (0, 2), (0, 8), (2, 0), (5, 0), and (2, 6).
50.
0
0
2
8
2 1
x
y
x y
x y
x y
≥ ≥ + ≥ + ≤
+ ≥
Graph 0; 0x y≥ ≥ . Shaded region is the first
quadrant. Graph the line 2x y+ = . Use a solid
line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 8x y+ = . Use a
solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 8+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 1x y+ = . Use
a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 1+ ≥ is false, shade the opposite side of the
line from (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of 2x y+ = and the y-axis is
(0, 2). The intersection of 2x y+ = and the x-
axis is (2, 0). The intersection of 8x y+ = and
the y-axis is (0, 8). The intersection of 8x y+ =
and the x-axis is (8, 0). The four corner points are (0, 2), (0, 8), (2, 0), and (8, 0).
Section 11.7: Systems of Inequalities
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51.
0
0
2 1
2 10
x
y
x y
x y
≥ ≥ + ≥ + ≤
Graph 0; 0x y≥ ≥ . Shaded region is the first
quadrant. Graph the line 2 1x y+ = . Use a solid
line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 1+ ≥ is false, shade the opposite side of the
line from (0, 0). Graph the line 2 10x y+ = . Use a
solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 10+ ≤ is true, shade the side of the line
containing (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of 2 1x y+ = and the y-axis is
(0, 0.5). The intersection of 2 1x y+ = and the
x-axis is (1, 0). The intersection of 2 10x y+ =
and the y-axis is (0, 5). The intersection of 2 10x y+ = and the x-axis is (10, 0). The four
corner points are (0, 0.5), (0, 5), (1, 0), and (10, 0).
52.
0
0
2 1
2 10
2
8
x
y
x y
x y
x y
x y
≥ ≥ + ≥ + ≤ + ≥
+ ≤
Graph 0; 0x y≥ ≥ . Shaded region is the first
quadrant. Graph the line 2 1x y+ = . Use a solid
line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 1+ ≥ is false, shade the opposite side of the
line from (0, 0). Graph the line 2 10x y+ = . Use
a solid line since the inequality uses ≤. Choose a
test point not on the line, such as (0, 0). Since 0 2(0) 10+ ≤ is true, shade the side of the line
containing (0, 0). Graph the line 2x y+ = . Use
a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 8x y+ = . Use a
solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 8+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of 2x y+ = and the y-axis is
(0, 2). The intersection of 2x y+ = and the x-
axis is (2, 0). The intersection of 2 10x y+ = and
the y-axis is (0, 5). The intersection of 8x y+ =
and the x-axis is (8, 0). To find the intersection of 8x y+ = and 2 10x y+ = , solve the system:
8
2 10
x y
x y
+ = + =
Solve the first equation for x: 8x y= − .
Substitute and solve: (8 ) 2 10
2
y y
y
− + ==
8 2 6x = − = The point of intersection is (6, 2). The five corner points are (0, 2), (0, 5), (2, 0), (8, 0), and (6, 2).
53. The system of linear inequalities is: 4
6
0
0
x
x y
x
y
≤ + ≤ ≥ ≥
Chapter 11: Systems of Equations and Inequalities
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54. The system of linear inequalities is: 5
2
6
0
0
y
x y
x
x
y
≤ + ≥ ≤ ≥
≥
55. The system of linear inequalities is: 20
15
50
0
0
x
y
x y
x y
x
≤ ≥ + ≤ − ≤
≥
56. The system of linear inequalities is: 6
5
3 4 12
2 8
0
y
x
x y
x y
x
≤ ≤ + ≥ − ≤
≥
57. a. Let x = the amount invested in Treasury bills, and let y = the amount invested in
corporate bonds. The constraints are:
50,000x y+ ≤ because the total investment
cannot exceed $50,000. 35,000x ≥ because the amount invested in
Treasury bills must be at least $35,000. 10,000y ≤ because the amount invested in
corporate bonds must not exceed $10,000. 0, 0x y≥ ≥ because a non-negative amount
must be invested. The system is
50,000
35,000
10,000
0
0
x y
x
y
x
y
+ ≤ ≥ ≤ ≥
≥
b. Graph the system.
The corner points are (35,000, 0),
(35,000, 10,000), (40,000, 10,000), (50,000, 0).
58. a. Let x = the # of standard model trucks, and let y = the # of deluxe model trucks.
The constraints are: 0, 0x y≥ ≥ because a non-negative number
of trucks must be manufactured. 2 3 80x y+ ≤ because the total painting hours
worked cannot exceed 80. 3 4 120x y+ ≤ because the total detailing
hours worked cannot exceed 120. The system is
0
0
2 3 80
3 4 120
x
y
x y
x y
≥ ≥ + ≤ + ≤
b. Graph the system.
The corner points are ( ) ( )800,0 , 0. , 40,0
3
.
Section 11.7: Systems of Inequalities
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59. a. Let x = the # of packages of the economy blend, and let y = the # of packages of the
superior blend. The constraints are:
0, 0x y≥ ≥ because a non-negative # of
packages must be produced. 4 8 75 16x y+ ≤ ⋅ because the total amount of
“A grade” coffee cannot exceed 75 pounds. (Note: 75 pounds = (75)(16) ounces.) 12 8 120 16x y+ ≤ ⋅ because the total amount of
“B grade” coffee cannot exceed 120 pounds. (Note: 120 pounds = (120)(16) ounces.) Simplifying the inequalities, we obtain:
4 8 75 16
2 75 4
2 300
x y
x y
x y
+ ≤ ⋅+ ≤ ⋅+ ≤
12 8 120 16
3 2 120 4
3 2 480
x y
x y
x y
+ ≤ ⋅+ ≤ ⋅+ ≤
The system is: 0
0
2 300
3 2 480
x
y
x y
x y
≥ ≥ + ≤ + ≤
b. Graph the system.
The corner points are (0, 0), (0, 150), (90, 105), (160, 0).
60. a. Let x = the # of lower-priced packages, and let y = the # of quality packages.
The constraints are: 0, 0x y≥ ≥ because a non-negative # of
packages must be produced. 8 6 120 16x y+ ≤ ⋅ because the total amount of
peanuts cannot exceed 120 pounds. (Note: 120 pounds = (120)(16) ounces.) 4 6 90 16x y+ ≤ ⋅ because the total amount of
cashews cannot exceed 90 pounds. (Note: 90 pounds = (90)(16) ounces.) Simplifying the inequalities, we obtain:
8 6 120 164 3 120 84 3 960
x yx yx y
+ ≤ ⋅+ ≤ ⋅+ ≤
4 6 90 162 3 90 82 3 720
x yx yx y
+ ≤ ⋅+ ≤ ⋅+ ≤
The system is 4 3 9602 3 720
0; 0
x yx y
x y
+ ≤ + ≤ ≥ ≥
b. Graph the system.
The corner points are (0, 0), (0, 240), (120, 160), (240, 0).
61. a. Let x = the # of microwaves, and let y = the # of printers.
The constraints are: 0, 0x y≥ ≥ because a non-negative # of
items must be shipped. 30 20 1600x y+ ≤ because a total cargo
weight cannot exceed 1600 pounds. 2 3 150x y+ ≤ because the total cargo
volume cannot exceed 150 cubic feet. Note that the inequality 30 20 1600x y+ ≤ can be
simplified: 3 2 160x y+ ≤ .
The system is: 3 2 1602 3 150
0; 0
x yx y
x y
+ ≤ + ≤ ≥ ≥
b. Graph the system.
The corner points are (0, 0), (0, 50), (36, 26),
160
3,0
.
Chapter 11: Systems of Equations and Inequalities
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Section 11.8
1. objective function
2. True
3. z x y= +
Vertex Value of
(0, 3) 0 3 3
(0, 6) 0 6 6
(5, 6) 5 6 11
(5, 2) 5 2 7
(4, 0) 4 0 4
z x y
z
z
z
z
z
= += + == + == + == + == + =
The maximum value is 11 at (5, 6), and the minimum value is 3 at (0, 3).
4. 2 3z x y= +
Vertex Value of 2 3
(0, 3) 2(0) 3(3) 9
(0, 6) 2(0) 3(6) 18
(5, 6) 2(5) 3(6) 28
(5, 2) 2(5) 3(2) 16
(4, 0) 2(4) 3(0) 8
z x y
z
z
z
z
z
= += + == + == + == + == + =
The maximum value is 28 at (5, 6), and the minimum value is 8 at (4, 0).
5. 10z x y= +
Vertex Value of 10
(0, 3) 0 10(3) 30
(0, 6) 0 10(6) 60
(5, 6) 5 10(6) 65
(5, 2) 5 10(2) 25
(4, 0) 4 10(0) 4
z x y
z
z
z
z
z
= += + == + == + == + == + =
The maximum value is 65 at (5, 6), and the minimum value is 4 at (4, 0).
6. 10z x y= +
Vertex Value of 10
(0, 3) 10(0) 3 3
(0, 6) 10(0) 6 6
(5, 6) 10(5) 6 56
(5, 2) 10(5) 2 52
(4, 0) 10(4) 0 40
z x y
z
z
z
z
z
= += + == + == + == + == + =
The maximum value is 56 at (5, 6), and the minimum value is 3 at (0, 3).
7. 5 7z x y= +
Vertex Value of 5 7
(0, 3) 5(0) 7(3) 21
(0, 6) 5(0) 7(6) 42
(5, 6) 5(5) 7(6) 67
(5, 2) 5(5) 7(2) 39
(4, 0) 5(4) 7(0) 20
z x y
z
z
z
z
z
= += + == + == + == + == + =
The maximum value is 67 at (5, 6), and the minimum value is 20 at (4, 0).
8. 7 5z x y= +
Vertex Value of 7 5
(0, 3) 7(0) 5(3) 15
(0, 6) 7(0) 5(6) 30
(5, 6) 7(5) 5(6) 65
(5, 2) 7(5) 5(2) 45
(4, 0) 7(4) 5(0) 28
z x y
z
z
z
z
z
= += + == + == + == + == + =
The maximum value is 65 at (5, 6), and the minimum value is 15 at (0, 3).
9. Maximize 2z x y= + subject to 0,x ≥
0, 6,y x y≥ + ≤ 1x y+ ≥ . Graph the
constraints.
y
x(6,0)
(0,6)
(0,1)
(1,0)
x + y = 6
x + y = 1
The corner points are (0, 1), (1, 0), (0, 6), (6, 0). Evaluate the objective function:
Vertex Value of 2
(0, 1) 2(0) 1 1
(0, 6) 2(0) 6 6
(1, 0) 2(1) 0 2
(6, 0) 2(6) 0 12
z x y
z
z
z
z
= += + == + == + =
= + =
The maximum value is 12 at (6, 0).
Section 11.8: Linear Programming
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
10. Maximize 3z x y= + subject to 0,x ≥ 0y ≥ ,
3x y+ ≥ 5, 7x y≤ ≤ . Graph the constraints.
y
x
x = 5
y = 7
x + y = 3
(5,7)(0,7)
(5,0)
(0,3)
(3,0)
The corner points are (0, 3), (3, 0), (0, 7), (5, 0), (5, 7). Evaluate the objective function:
Vertex Value of 3
(0, 3) 0 3(3) 9
(3, 0) 3 3(0) 3
(0, 7) 0 3(7) 21
(5, 0) 5 3(0) 5
(5, 7) 5 3(7) 26
z x y
z
z
z
z
z
= += + == + =
= + == + =
= + =
The maximum value is 26 at (5, 7).
11. Minimize 2 5z x y= + subject to 0,x ≥
0, 2, 5,y x y x≥ + ≥ ≤ 3y ≤ . Graph the
constraints.
y
x
(5,3)
(5,0)
(0,3)
(2,0)
(0,2)
x = 5
y = 3
x + y = 2
The corner points are (0, 2), (2, 0), (0, 3), (5, 0), (5, 3). Evaluate the objective function:
Vertex Value of 2 5
(0, 2) 2(0) 5(2) 10
(0, 3) 2(0) 5(3) 15
(2, 0) 2(2) 5(0) 4
(5, 0) 2(5) 5(0) 10
(5, 3) 2(5) 5(3) 25
z x y
z
z
z
z
z
= += + == + == + == + == + =
The minimum value is 4 at (2, 0).
12. Minimize 3 4z x y= + subject to 0x ≥ ,
0, 2 3 6,y x y≥ + ≥ 8x y+ ≤ . Graph the
constraints.
y
x
x + y = 8
2x + 3y = 6
(0,8)
(0,2)
(3,0) (8,0)
The corner points are (0, 2), (3, 0), (0, 8), (8, 0). Evaluate the objective function:
Vertex Value of 3 4
(0, 2) 3(0) 4(2) 8
(3, 0) 3(3) 4(0) 9
(0, 8) 3(0) 4(8) 32
(8, 0) 3(8) 4(0) 24
z x y
z
z
z
z
= += + == + =
= + == + =
The minimum value is 8 at (0, 2).
13. Maximize 3 5z x y= + subject to 0,x ≥ 0,y ≥
2,x y+ ≥ 2 3 12,x y+ ≤ 3 2 12x y+ ≤ . Graph
the constraints.
y
x
(2.4,2.4)
(0,4)
(0,2)
(2,0) (4,0)
3x + 2y = 12
2x + 3y = 12
x + y = 2
To find the intersection of 2 3 12x y+ = and
3 2 12x y+ = , solve the system:
2 3 123 2 12
x yx y
+ = + =
Solve the second equation for y: 3
62
y x= −
Substitute and solve:
Chapter 11: Systems of Equations and Inequalities
1200
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
32 3 6 12
29
2 18 1225
62
12
5
x x
x x
x
x
+ − =
+ − =
− = −
=
3 12 18 126 6
2 5 5 5y
= − = − =
The point of intersection is ( )2.4, 2.4 .
The corner points are (0, 2), (2, 0), (0, 4), (4, 0), (2.4, 2.4). Evaluate the objective function:
Vertex Value of 3 5
(0, 2) 3(0) 5(2) 10
(0, 4) 3(0) 5(4) 20
(2, 0) 3(2) 5(0) 6
(4, 0) 3(4) 5(0) 12
(2.4, 2.4) 3(2.4) 5(2.4) 19.2
z x y
z
z
z
z
z
= += + == + == + == + =
= + =
The maximum value is 20 at (0, 4).
14. Maximize 5 3z x y= + subject to 0,x ≥
0, 2, 8, 2 10y x y x y x y≥ + ≥ + ≤ + ≤ . Graph
the constraints.
y
x
x + y = 8
2x + y = 10
x + y = 2
(2,0)
(0,2)
(0,8)
(5,0)
(2,6)
To find the intersection of 8x y+ = and
2 10x y+ = , solve the system:
8
2 10
x y
x y
+ = + =
Solve the first equation for y: 8y x= − .
Substitute and solve: 2 8 10
2
x x
x
+ − ==
8 2 6y = − =
The point of intersection is (2, 6). The corner points are (0, 2), (2, 0), (0, 8), (5, 0), (2, 6). Evaluate the objective function:
Vertex Value of 5 3
(0, 2) 5(0) 3(2) 6
(0, 8) 5(0) 3(8) 24
(2, 0) 5(2) 3(0) 10
(5, 0) 5(5) 3(0) 25
(2, 6) 5(2) 3(6) 28
z x y
z
z
z
z
z
= += + == + == + == + == + =
The maximum value is 28 at (2, 6).
15. Minimize 5 4z x y= + subject to 0,x ≥
0, 2, 2 3 12, 3 12y x y x y x y≥ + ≥ + ≤ + ≤ . Graph
the constraints.
y
x
(0,4)
(0,2)
(2,0) (4,0)
x + y = 2
2x + 3y = 12
3x + y = 12
24
7, 12
7( )
To find the intersection of 2 3 12x y+ = and
3 12x y+ = , solve the system:
2 3 12
3 12
x y
x y
+ = + =
Solve the second equation for y: 12 3y x= −
Substitute and solve:
247
2 3(12 3 ) 122 36 9 12
7 24
x xx x
x
x
+ − =+ − =
− = −=
( ) 7224 127 7 7
12 3 12y = − = − =
The point of intersection is 24 12,
7 7
.
The corner points are (0, 2), (2, 0), (0, 4), (4, 0), 24 12
,7 7
. Evaluate the objective function:
( ) ( ) ( )24 12 24 127 7 7 7
Vertex Value of 5 4
(0, 2) 5(0) 4(2) 8
(0, 4) 5(0) 4(4) 16
(2, 0) 5(2) 4(0) 10
(4, 0) 5(4) 4(0) 20
, 5 4 24
z x y
z
z
z
z
z
= += + == + == + == + =
= + =
The minimum value is 8 at (0, 2).
Section 11.8: Linear Programming
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16. Minimize 2 3z x y= + subject to 0,x ≥
0, 3, 9, 3 6y x y x y x y≥ + ≥ + ≤ + ≥ . Graph
the constraints.
y
x
x + y = 9
x + y = 3x + 3y = 6
(0,9)
(0,3)
(6,0)(9,0)
32 , 3
2( )
To find the intersection of 3x y+ = and
3 6x y+ = , solve the system:
3
3 6
x y
x y
+ = + =
Solve the first equation for y: 3y x= − .
Substitute and solve: 3(3 ) 6
9 3 6
2 3
3
2
x x
x x
x
x
+ − =+ − =
− = −
=
3 33
2 2y = − =
The point of intersection is 3 3
,2 2
.
The corner points are (0, 3), (6, 0), (0, 9), (9, 0), 3 3
,2 2
. Evaluate the objective function:
Vertex Value of 2 3
(0, 3) 2(0) 3(3) 9
(0, 9) 2(0) 3(9) 27
(6, 0) 2(6) 3(0) 12
(9, 0) 2(9) 3(0) 18
3 3 3 3 15, 2 4
2 2 2 2 2
z x y
z
z
z
z
z
= += + =
= + == + == + =
= + =
The minimum value is 15
2 at
3 3,
2 2
.
17. Maximize 5 2z x y= + subject to 0,x ≥
0, 10, 2 10, 2 10y x y x y x y≥ + ≤ + ≥ + ≥ .
Graph the constraints.
y
x
(0,10)
(10,0)
x + y = 10
2x + y = 10x + 2y = 10
103 , 10
3( )
To find the intersection of 2 10x y+ = and
2 10x y+ = , solve the system:
2 10
2 10
x y
x y
+ = + =
Solve the first equation for y: 10 2y x= − .
Substitute and solve: 2(10 2 ) 10
20 4 10
3 10
10
3
x x
x x
x
x
+ − =+ − =
− = −
=
10 20 1010 2 10
3 3 3y
= − = − =
The point of intersection is (10/3 10/3). The corner points are (0, 10), (10, 0), (10/3, 10/3). Evaluate the objective function:
13
Vertex Value of 5 2
(0, 10) 5(0) 2(10) 20
(10, 0) 5(10) 2(0) 50
10 10 10 10 70, 5 2 23
3 3 3 3 3
z x y
z
z
z
= += + == + =
= + = =
The maximum value is 50 at (10, 0).
Chapter 11: Systems of Equations and Inequalities
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
18. Maximize 2 4z x y= + subject to
0, 0, 2 4, 9x y x y x y≥ ≥ + ≥ + ≤ .
Graph the constraints.
y
x
x + y = 9
2x + y = 4
(0,9)
(9,0)(2,0)
(0,4)
The corner points are (0, 9), (9, 0), (0, 4), (2, 0). Evaluate the objective function:
Vertex Value of 2 4
(0, 9) 2(0) 4(9) 36
(9, 0) 2(9) 4(0) 18
(0, 4) 2(0) 4(4) 16
(2, 0) 2(2) 4(0) 4
z x y
z
z
z
z
= += + == + == + == + =
The maximum value is 36 at (0, 9).
19. Let x = the number of downhill skis produced, and let y = the number of cross-country skis
produced. The total profit is: 70 50P x y= + .
Profit is to be maximized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A positive number of skis must be
produced. 2 40x y+ ≤ Manufacturing time available.
32x y+ ≤ Finishing time available.
Graph the constraints.
y
x
(8,24)
(20,0)(0,0)
(0,32)
2x + y = 40
x + y = 32
To find the intersection of 32x y+ = and
2 40x y+ = , solve the system:
32
2 40
x y
x y
+ = + =
Solve the first equation for y: 32y x= − .
Substitute and solve: 2 (32 ) 40
8
x x
x
+ − ==
32 8 24y = − =
The point of intersection is (8, 24). The corner points are (0, 0), (0, 32), (20, 0), (8, 24). Evaluate the objective function:
Vertex Value of 70 50
(0, 0) 70(0) 50(0) 0
(0, 32) 70(0) 50(32) 1600
(20, 0) 70(20) 50(0) 1400
(8, 24) 70(8) 50(24) 1760
P x y
P
P
P
P
= += + =
= + == + == + =
The maximum profit is $1760, when 8 downhill skis and 24 cross-country skis are produced.
With the increase of the manufacturing time to 48 hours, we do the following: The constraints are:
0, 0x y≥ ≥ A positive number of skis must be
produced. 2 48x y+ ≤ Manufacturing time available.
32x y+ ≤ Finishing time available.
Graph the constraints.
y
x
(16,16)
(0,32)
(24,0)(0,0)
2x + y = 48
x + y = 32
To find the intersection of 32x y+ = and
2 48x y+ = , solve the system:
32
2 48
x y
x y
+ = + =
Solve the first equation for y: 32y x= − .
Substitute and solve: 2 (32 ) 48
16
x x
x
+ − ==
32 16 16y = − =
The point of intersection is (16, 16). The corner points are (0, 0), (0, 32), (24, 0), (16, 16). Evaluate the objective function:
Section 11.8: Linear Programming
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Vertex Value of 70 50
(0, 0) 70(0) 50(0) 0
(0, 32) 70(0) 50(32) 1600
(24, 0) 70(24) 50(0) 1680
(16, 16) 70(16) 50(16) 1920
P x y
P
P
P
P
= += + =
= + == + == + =
The maximum profit is $1920, when 16 downhill skis and 16 cross-country skis are produced.
20. Let x = the number of acres of soybeans planted , and let y = the number of acres of wheat planted.
The total profit is: 180 100P x y= + . Profit is to
be maximized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A non-negative number of acres
must be planted. 70x y+ ≤ Acres available to plant.
60 30 1800x y+ ≤ Money available for
preparation. 3 4 120x y+ ≤ Workdays available.
Graph the constraints.
y
x
x + y = 70
60x + 30y = 1800
3x + 4y = 120
(0,30)
(30,0)(0,0)
(24,12)
To find the intersection of 60 30 1800x y+ = and
3 4 120x y+ = , solve the system:
60 30 1800
3 4 120
x y
x y
+ = + =
Solve the first equation for y: 60 30 1800
2 60
60 2
x y
x y
y x
+ =+ =
= −
Substitute and solve: 3 4(60 2 ) 120
3 240 8 120
5 120
24
x x
x x
x
x
+ − =+ − =
− = −=
60 2(24) 12y = − =
The point of intersection is (24, 12). The corner points are (0, 0), (0, 30), (30, 0), (24, 12). Evaluate the objective function:
Vertex Value of 180 100
(0, 0) 180(0) 100(0) 0
(0, 30) 180(0) 100(30) 3000
(30, 0) 180(30) 100(0) 5400
(24, 12) 180(24) 100(12) 5520
P x y
P
P
P
P
= += + =
= + == + == + =
The maximum profit is $5520, when 24 acres of soybeans and 12 acres of wheat are planted.
With the increase of the preparation costs to $2400, we do the following: The constraints are:
0, 0x y≥ ≥ A non-negative number of acres
must be planted. 70x y+ ≤ Acres available to plant.
60 30 2400x y+ ≤ Money available for
preparation. 3 4 120x y+ ≤ Workdays available.
Graph the constraints.
y
x
x + y = 70
3x + 4y = 120
60x + 30y = 2400
(0,30)
(40,0)(0,0)
The corner points are (0, 0), (0, 30), (40, 0). Evaluate the objective function:
Vertex Value of 180 100
(0, 0) 180(0) 100(0) 0
(0, 30) 180(0) 100(30) 3000
(40, 0) 180(40) 100(0) 7200
P x y
P
P
P
= += + =
= + == + =
The maximum profit is $7200, when 40 acres of soybeans and 0 acres of wheat are planted.
21. Let x = the number of rectangular tables rented, and let y = the number of round tables rented.
The cost for the tables is: 28 52C x y= + . Cost is
to be minimized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A non-negative number of tables
must be used. 35x y+ ≤ Maximum number of tables.
6 10 250x y+ ≥ Number of guests.
15x ≤ Rectangular tables available.
Chapter 11: Systems of Equations and Inequalities
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Graph the constraints.
y
x
6 + 10 = 250x y
x y + = 35
(15, 20)
x = 15
(0, 25)
(0, 35)
(15, 16)
25
10
The corner points are (0, 25), (0, 35), (15, 20), (15, 16). Evaluate the objective function:
Vertex Value of 28 52
(0, 25) 28(0) 52(25) 1300
(0, 35) 28(0) 52(35) 1820
(15, 20) 28(15) 52(20) 1460
(15,16) 28(15) 52(16) 1252
C x y
C
C
C
C
= += + == + == + == + =
Kathleen should rent 15 rectangular tables and 16 round tables in order to minimize the cost. The minimum cost is $1252.00.
22. Let x = the number of buses rented, and let y =
the number of vans rented. The cost for the vehicles is: 975 350C x y= + . Cost is to be
minimized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A non-negative number of buses
and vans must be used. 40 8 320x y+ ≥ Number of regular seats.
3 36x y+ ≥ Number of handicapped seats.
Graph the constraints.
y
xx y + 3 = 36
40 + 8 = 320x y
(6, 10)(0, 12)
(0, 40)
(36, 0)
25
25
To find the intersection of 40 8 320x y+ = and
3 36x y+ = , solve the system:
40 8 320
3 36
x y
x y
+ = + =
Solve the second equation for x:
3 36
3 36
x y
x y
+ == − +
Substitute and solve: 40( 3 36) 8 320
120 1440 8 320
112 1120
10
y y
y y
y
y
− + + =− + + =
− = −=
3(10) 36 30 36 6x = − + = − + =
The point of intersection is (6, 10). The corner points are (0, 40), (6, 10), and (36, 0). Evaluate the objective function:
Vertex Value of 975 350
(0, 40) 975(0) 350(40) 14,000
(6,10) 975(6) 350(10) 9350
(36, 0) 975(36) 350(0) 35,100
C x y
C
C
C
= += + == + == + =
The college should rent 6 buses and 10 vans for a minimum cost of $9350.00.
23. Let x = the amount invested in junk bonds, and let y = the amount invested in Treasury bills.
The total income is: 0.09 0.07I x y= + . Income
is to be maximized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A non-negative amount must be
invested. 20,000x y+ ≤ Total investment cannot exceed
$20,000. 12,000x ≤ Amount invested in junk bonds
must not exceed $12,000. 8000y ≥ Amount invested in Treasury bills
must be at least $8000. a. y x≥ Amount invested in Treasury bills
must be equal to or greater than the amount invested in junk bonds.
Graph the constraints.
x + y = 20000
y = x
x = 12000
y = 8000
(0,20000)
(0,8000) (8000,8000)
(10000,10000)
Section 11.8: Linear Programming
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The corner points are (0, 20,000), (0, 8000), (8000, 8000), (10,000, 10,000). Evaluate the objective function:
Vertex Value of 0.09 0.07
(0, 20000) 0.09(0) 0.07(20000)
1400
(0, 8000) 0.09(0) 0.07(8000)
560
(8000, 8000) 0.09(8000) 0.07(8000)
1280
(10000, 10000) 0.09(10000) 0.07(10000)
1600
I x y
I
I
I
I
= += +== +== +== +=
The maximum income is $1600, when $10,000 is invested in junk bonds and $10,000 is invested in Treasury bills.
b. y x≤ Amount invested in Treasury bills
must not exceed the amount invested in junk bonds.
Graph the constraints.
x + y = 20000
y = x
x = 12000
y = 8000(8000,8000)
(10000,10000)
(12000,8000)
The corner points are (12,000, 8000), (8000, 8000), (10,000, 10,000). Evaluate the objective function:
Vertex Value of 0.09 0.07
(12000, 8000) 0.09(12000) 0.07(8000)
1640
(8000, 8000) 0.09(8000) 0.07(8000)
1280
(10000, 10000) 0.09(10000) 0.07(10000)
1600
I x y
I
I
I
= += +== +=
= +=
The maximum income is $1640, when $12,000 is invested in junk bonds and $8000 is invested in Treasury bills.
24. Let x = the number of hours that machine 1 is operated, and let y = the number of hours that
machine 2 is operated. The total cost is: 50 30C x y= + . Cost is to be minimized, so this
is the objective function. The constraints are:
0, 0x y≥ ≥ A positive number of hours must
be used. 10x ≤ Time used on machine 1. 10y ≤ Time used on machine 2.
60 40 240x y+ ≥ 8-inch pliers to be produced.
70 20 140x y+ ≥ 6-inch pliers to be produced.
Graph the constraints.
(10,10)(0,10)
(10,0)(4,0)
(0,7)
60x + 40y = 24070x + 20y = 140
12 , 21
4( )
To find the intersection of 60 40 240x y+ = and
70 20 140x y+ = , solve the system:
60 40 240
70 20 140
x y
x y
+ = + =
Divide the first equation by 2− and add the result to the second equation:
30 20 120
70 20 140
x y
x y
− − = −+ =
40 20
20 1
40 2
x
x
=
= =
Substitute and solve: 1
60 40 2402
30 40 240
40 210
210 21 15
40 4 4
y
y
y
y
+ =
+ ==
= = =
The point of intersection is 1 1
, 52 4
.
Chapter 11: Systems of Equations and Inequalities
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The corner points are (0, 7), (0, 10), (4, 0),
(10, 0), (10, 10), 1 1
, 52 4
. Evaluate the
objective function:
Vertex Value of 50 30
(0, 7) 50(0) 30(7) 210
(0, 10) 50(0) 30(10) 300
(4, 0) 50(4) 30(0) 200
(10, 0) 50(10) 30(0) 500
(10, 10) 50(10) 30(10) 800
1 1 1 1, 5 50 30 5 182.50
2 4 2 4
C x y
C
C
C
C
C
P
= += + == + == + == + == + =
= + =
The minimum cost is $182.50, when machine 1
is used for 1
2 hour and machine 2 is used for
15
4
hours.
25. Let x = the number of pounds of ground beef, and let y = the number of pounds of ground
pork. The total cost is: 0.75 0.45C x y= + . Cost
is to be minimized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A positive number of pounds
must be used. 200x ≤ Only 200 pounds of ground beef
are available. 50y ≥ At least 50 pounds of ground
pork must be used. 0.75 0.60 0.70( )x y x y+ ≥ + Leanness
condition (Note that the last equation will simplify to
1
2y x≤ .) Graph the constraints.
y
x
(200,100)
(200,50)(100,50)
The corner points are (100, 50), (200, 50), (200, 100). Evaluate the objective function:
Vertex Value of 0.75 0.45
(100, 50) 0.75(100) 0.45(50) 97.50
(200, 50) 0.75(200) 0.45(50) 172.50
(200, 100) 0.75(200) 0.45(100) 195
C x y
C
C
C
= += + == + == + =
The minimum cost is $97.50, when 100 pounds of ground beef and 50 pounds of ground pork are used.
26. Let x = the number of gallons of regular, and let y = the number of gallons of premium. The
total profit is: 0.75 0.90P x y= + . Profit is to be
maximized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A positive number of gallons
must be used.
1
4y x≥ At least one gallon of premium
for every 4 gallons of regular. 5 6 3000x y+ ≤ Daily shipping weight limit.
24 20 16(725)x y+ ≤ Available flavoring.
12 20 16(425)x y+ ≤ Available milk-fat
(Note: the last two inequalities simplify to 6 5 2900x y+ ≤ and 3 5 1700x y+ ≤ .)
Graph the constraints.
y
x
5 + 6 = 3000x y
(0, 340) (400, 100)
300
1501
4y x=
24 + 20 = 16(725)x y
12 + 20 = 16(425)x y
The corner points are (0, 0), (400, 100), (0, 340). Evaluate the objective function:
Vertex Value of 0.75 0.90
(0, 0) 0.75(0) 0.90(0) 0
(400,100) 0.75(400) 0.90(100) 390
(0, 340) 0.75(0) 0.90(340) 306
P x y
P
P
P
= += + == + == + =
Mom and Pop should produce 400 gallons of regular and 100 gallons of premium ice cream. The maximum profit is $390.00.
Section 11.8: Linear Programming
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27. Let x = the number of racing skates manufactured, and let y = the number of figure
skates manufactured. The total profit is: 10 12P x y= + . Profit is to be maximized, so
this is the objective function. The constraints are:
0, 0x y≥ ≥ A positive number of skates must
be manufactured. 6 4 120x y+ ≤ Only 120 hours are available for
fabrication. 2 40x y+ ≤ Only 40 hours are available for
finishing. Graph the constraints.
y
x
(10,15)
(20,0)(0,0)
(0,20)
To find the intersection of 6 4 120x y+ = and
+2 40x y = , solve the system:
6 4 120
2 40
x y
x y
+ = + =
Solve the second equation for x: 40 2x y= −
Substitute and solve: 6(40 2 ) 4 120
240 12 4 120
8 120
15
y y
y y
y
y
− + =− + =
− = −=
40 2(15) 10x = − =
The point of intersection is (10, 15). The corner points are (0, 0), (0, 20), (20, 0), (10, 15). Evaluate the objective function:
Vertex Value of 10 12
(0, 0) 10(0) 12(0) 0
(0, 20) 10(0) 12(20) 240
(20, 0) 10(20) 12(0) 200
(10, 15) 10(10) 12(15) 280
P x y
P
P
P
P
= += + =
= + == + == + =
The maximum profit is $280, when 10 racing skates and 15 figure skates are produced.
28. Let x = the amount placed in the AAA bond. Let y = the amount placed in a CD.
The total return is: 0.08 0.04R x y= + . Return is
to be maximized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A positive amount must be
invested in each. 50,000x y+ ≤ Total investment cannot exceed
$50,000. 20,000x ≤ Investment in the AAA bond cannot
exceed $20,000. 15,000y ≥ Investment in the CD must be at
least $15,000. y x≥ Investment in the CD must exceed or
equal the investment in the bond. Graph the constraints.
y
x
y = x
x + y = 50000
x = 20000
y = 15000
(0,50000)
(0,15000)
(15000,15000)
(20000,30000)
(20000,20000)
The corner points are (0, 50,000), (0, 15,000), (15,000, 15,000), (20,000, 20,000), (20,000, 30,000). Evaluate the objective function:
Vertex Value of 0.08 0.04
(0, 50000) 0.08(0) 0.04(50000)
2000
(0, 15000) 0.08(0) 0.04(15000)
600
(15000, 15000) 0.08(15000) 0.04(15000)
1800
(20000, 20000) 0.08(20000) 0.04(20000)
2400
(20000, 30000
R x y
R
R
R
R
= += +== +== +== +=
) 0.08(20000) 0.04(30000)
2800
R = +=
The maximum return is $2800, when $20,000 is invested in a AAA bond and $30,000 is invested in a CD.
Chapter 11: Systems of Equations and Inequalities
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29. Let x = the number of metal fasteners, and let y
= the number of plastic fasteners. The total cost is: 9 4C x y= + . Cost is to be minimized, so this
is the objective function. The constraints are: 2, 2x y≥ ≥ At least 2 of each fastener must be
made. 6x y+ ≥ At least 6 fasteners are needed.
4 2 24x y+ ≤ Only 24 hours are available.
Graph the constraints.
y
x
(2,8)
(5,2)(4,2)
(2,4)
The corner points are (2, 4), (2, 8), (4, 2), (5, 2). Evaluate the objective function:
Vertex Value of 9 4
(2, 4) 9(2) 4(4) 34
(2, 8) 9(2) 4(8) 50
(4, 2) 9(4) 4(2) 44
(5, 2) 9(5) 4(2) 53
C x y
C
C
C
C
= += + == + == + == + =
The minimum cost is $34, when 2 metal fasteners and 4 plastic fasteners are ordered.
30. Let x = the amount of “Gourmet Dog,” and let y = the amount of “Chow Hound.” The total
cost is: 0.40 0.32C x y= + . Cost is to be
minimized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A non-negative number of cans
must be purchased. 20 35 1175x y+ ≥ At least 1175 units of
vitamins per month. 75 50 2375x y+ ≥ At least 2375 calories per
month. 60x y+ ≤ Storage space for 60 cans.
Graph the constraints.
y
x
x + y = 60
75x+50y=2375
20x+35y=1175
(0,47.5)
(0,60)
(60,0)
(58.75,0)
(15,25)
The corner points are (0, 47.5), (0, 60), (60, 0), (58.75, 0), (15, 25). Evaluate the objective function:
Vertex Value of 0.40 0.32
(0, 47.5) 0.40(0) 0.32(47.5) 15.20
(0, 60) 0.40(0) 0.32(60) 19.20
(60, 0) 0.40(60) 0.32(0) 24.00
(58.75, 0) 0.40(58.75) 0.32(0) 23.50
(15, 25) 0.40(15) 0.32(25) 14.00
C x y
C
C
C
C
C
= += + == + == + =
= + == + =
The minimum cost is $14, when 15 cans of "Gourmet Dog" and 25 cans of “Chow Hound” are purchased.
31. Let x = the number of first class seats, and let y = the number of coach seats. Using the hint,
the revenue from x first class seats and y coach seats is ,Fx Cy+ where 0.F C> > Thus,
R Fx Cy= + is the objective function to be
maximized. The constraints are: 8 16x≤ ≤ Restriction on first class seats. 80 120y≤ ≤ Restriction on coach seats.
a. 1
12
x
y≤ Ratio of seats.
The constraints are: 8 16x≤ ≤ 80 120y≤ ≤
12x y≤
Graph the constraints.
Chapter 11 Review Exercises
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The corner points are (8, 96), (8, 120), and (10, 120). Evaluate the objective function:
Vertex Value of
(8, 96) 8 96
(8, 120) 8 120
(10, 120) 10 120
R Fx Cy
R F C
R F C
R F C
= += += += +
Since 0,C > 120 96 ,C C> so 8 120 8 96 .F C F C+ > + Since 0,F > 10 8 ,F F> so 10 120 8 120 .F C F C+ > + Thus, the maximum revenue occurs when the aircraft is configured with 10 first class seats and 120 coach seats.
b. 1
8
x
y≤
The constraints are: 8 16x≤ ≤ 80 120y≤ ≤
8x y≤
Graph the constraints.
The corner points are (8, 80), (8, 120), (15, 120), and (10, 80). Evaluate the objective function:
Vertex Value of
(8, 80) 8 80
(8, 120) 8 120
(15, 120) 15 120
(10, 80) 10 80
R Fx Cy
R F C
R F C
R F C
R F C
= += += += += +
Since 0F > and 0,C > 120 96 ,C C> the maximum value of R occurs at (15, 120). The maximum revenue occurs when the aircraft is configured with 15 first class seats and 120 coach seats.
c. Answers will vary.
32. Answers will vary.
Chapter 11 Review Exercises
1. 2 5
5 2 8
x y
x y
− = + =
Solve the first equation for y: 2 5y x= − .
Substitute and solve: 5 2(2 5) 8
5 4 10 8
9 18
2
x x
x x
x
x
+ − =+ − =
==
2(2) 5 4 5 1y = − = − = −
The solution is 2, 1x y= = − or (2, 1)− .
2. 2 3 2
7 3
x y
x y
+ = − =
Solve the second equation for y: 7 3y x= −
Substitute into the first equation and solve: 2 3(7 3) 2
2 21 9 2
23 11
11
23
x x
x x
x
x
+ − =+ − =
=
=
11 77 69 87 3
23 23 23 23y
= − = − =
The solution is 11 8
,23 23
x y= = or 11 8
,23 23
.
3. 3 4 4
13
2
x y
x y
− = − =
Solve the second equation for x: 1
32
x y= +
Substitute into the first equation and solve: 1
3 3 4 42
39 4 4
25
521
2
y y
y y
y
y
+ − =
+ − =
=
=
1 13 2
2 2x
= + =
The solution is 1
2,2
x y= = or 1
2,2
.
Chapter 11: Systems of Equations and Inequalities
1210
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
4. 2 0
135 4
2
x y
x y
+ = − = −
Solve the first equation for y: 2y x= −
Substitute into the second equation and solve: 13
5 4( 2 )2
135 8
213
1321
2
x x
x x
x
x
− − = −
+ = −
= −
= −
12 1
2y
= − − =
The solution is 1
, 12
x y= − = or 1
, 12
−
.
5. 2 4 0
3 2 4 0
x y
x y
− − = + − =
Solve the first equation for x: 2 4x y= +
Substitute into the second equation and solve: 3(2 4) 2 4 0
6 12 2 4 0
8 8
1
y y
y y
y
y
+ + − =+ + − =
= −= −
2( 1) 4 2x = − + =
The solution is 2, 1x y= = − or (2, 1)− .
6. 3 5 0
2 3 5 0
x y
x y
− + = + − =
Solve the first equation for x: 3 5x y= −
Substitute into the second equation and solve: 2(3 5) 3 5 0
6 10 3 5 0
9 15
5
3
y y
y y
y
y
− + − =− + − =
=
=
53 5 0
3x
= − =
The solution is 5
0,3
x y= = or 5
0,3
.
7. 2 5
3 4
y x
x y
= − = +
Substitute the first equation into the second equation and solve:
3(2 5) 4
6 15 4
5 11
11
5
x x
x x
x
x
= − += − +
− = −
=
11 32 5
5 5y
= − = −
The solution is 11 3
,5 5
x y= = − or 11 3
,5 5
−
.
8. 5 2
5 2
x y
y x
= + = +
Substitute the first equation into the second equation and solve:
5(5 2) 2
25 10 2
24 12
1
2
y y
y y
y
y
= + += + +
− =
= −
1 5 15 2 2
2 2 2x
= − + = − + = −
The solution is 1 1
,2 2
x y= − = − or 1 1
,2 2
− −
.
9. 3 4 0
1 3 40
2 2 3
x y
x y
− + = − + =
Multiply each side of the first equation by 3 and each side of the second equation by 6− and add:
3 9 12 0
3 9 8 0
4 0
x y
x y
− + =− + − =
=
There is no solution to the system. The system is inconsistent.
Chapter 11 Review Exercises
1211
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10. 1
24
4 2 0
x y
y x
+ = + + =
Solve the second equation for y: 4 2y x= − − .
Substitute into the first equation and solve: 1
( 4 2) 24
12
25
02
x x
x x
+ − − =
− − =
=
There is no solution to the system. The system of equations is inconsistent.
11. 2 3 13 0
3 2 0
x y
x y
+ − = − =
Multiply each side of the first equation by 2 and each side of the second equation by 3, and add to eliminate y:
4 6 26 0
9 6 0
13 26 0
13 26
2
x y
x y
x
x
x
+ − = − =
− ===
Substitute and solve for y: 3(2) 2 0
2 6
3
y
y
y
− =− = −
=
The solution is 2, 3x y= = or (2, 3).
12. 4 5 21
5 6 42
x y
x y
+ = + =
Multiply each side of the first equation by 5 and each side of the second equation by –4 and add to eliminate x:
20 25 105
20 24 168
63
x y
x y
y
+ = − − = −
= −
Substitute and solve for x: 4 5( 63) 21
4 315 21
4 336
84
x
x
x
x
+ − =− =
==
The solution is 84, 63x y= = − or (84, 63)− .
13. 3 2 8
212
3
x y
x y
− = − =
Multiply each side of the second equation by –3 and add to eliminate x:
3 2 8
3 2 36
0 28
x y
x y
− =− + = −
= −
There is no solution to the system. The system of equations is inconsistent.
14. 2 5 10
4 10 20
x y
x y
+ = + =
Multiply each side of the first equation by –2 and add to eliminate x:
4 10 20
4 10 20
0 0
x y
x y
− − = − + =
=
The system is dependent. 2 5 10
5 2 10
22
5
x y
y x
y x
+ == − +
= − +
The solution is 2
25
y x= − + , x is any real number
or 2
( , ) 2, is any real number5
x y y x x = − +
.
15.
2 6
2 3 13
3 2 3 16
x y z
x y z
x y z
+ − = − + = − − + = −
Multiply each side of the first equation by –2 and add to the second equation to eliminate x;
2 4 2 12
2 3 13
5 5 25
5
x y z
x y z
y z
y z
− − + = − − + = −
− + = −− =
Multiply each side of the first equation by –3 and add to the third equation to eliminate x:
3 6 3 18
3 2 3 16
8 6 34
x y z
x y z
y z
− − + = −− + = −
− + = −
Multiply each side of the first result by 8 and add to the second result to eliminate y:
Chapter 11: Systems of Equations and Inequalities
1212
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
8 8 40
8 6 34
2 6
3
y z
y z
z
z
− =− + = −
− == −
Substituting and solving for the other variables: ( 3) 5
2
y
y
− − ==
2(2) ( 3) 6
4 3 6
1
x
x
x
+ − − =+ + =
= −
The solution is 1, 2, 3x y z= − = = − or ( 1, 2, 3)− − .
16.
5 2
2 7
2 11
x y z
x y z
x y z
+ − = + + = − + =
Add the first equation and the second equation to eliminate z:
5 2
2 7
3 6 9
2 3
x y z
x y z
x y
x y
+ − = + + =
+ =− − = −
Multiply each side of the first equation by 2 and add to the third equation to eliminate z:
2 10 2 4
2 11
3 9 15
3 5
x y z
x y z
x y
x y
+ − =− + =
+ =+ =
Add the two results to eliminate x: 2 3
3 5
2
x y
x y
y
− − = −+ =
=
Substituting and solving for the other variables: 3(2) 5
6 5
1
x
x
x
+ =+ =
= −
2( 1) 2 7
2 2 7
7
z
z
z
− + + =− + + =
=
The solution is 1, 2, 7x y z= − = = or ( 1, 2, 7)− .
17.
2 4 15
2 4 27
5 6 2 3
x y z
x y z
x y z
− + = − + − = − − = −
Multiply the first equation by 1− and the second equation by 2, and then add to eliminate x:
2 4 15
2 4 8 54
8 9 69
x y z
x y z
y z
− + − = + − =
− =
Multiply the second equation by 5− and add to the third equation to eliminate x:
5 10 20 135
5 6 2 3
16 18 138
x y z
x y z
y z
− − + = −− − = −
− + = −
Multiply both sides of the first result by 2 and add to the second result to eliminate y: 16 18 138
16 18 138
0 0
y z
y z
− =− + = −
=
The system is dependent. 16 18 138
18 138 16
9 69
8 8
y z
z y
y z
− + = −+ =
= +
Substituting into the second equation and solving for x:
9 692 4 27
8 8
9 694 27
4 47 39
4 4
x z z
x z z
x z
+ + − =
+ + − =
= +
The solution is 7 39
4 4x z= + ,
9 69
8 8y z= + , z is
any real number or 7 39
( , , ) ,4 4
x y z x z = +
9 69, is any real number
8 8y z z
= +
.
18.
4 3 15
3 5 5
7 5 9 10
x y z
x y z
x y z
− + =− + − = −− − − =
Multiply the first equation by 3 and then add the second equation to eliminate x:
3 12 9 45
3 5 5
11 4 40
x y z
x y z
y z
− + =− + − = −
− + =
Multiply the first equation by 7 and add to the third equation to eliminate x: 7 28 21 105
7 5 9 10
33 12 115
115 11 4
3
x y z
x y z
y z
y z
− + =− − − =
− + =
− + =
Chapter 11 Review Exercises
1213
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Multiply the first result by 1− and adding it to the second result:
11 4 40
11511 4
3
5 0
3
y z
y z
− = −− + =
= −
The system has no solution. The system is inconsistent.
19. 3 2 8
4 1
x y
x y
+ = + = −
20.
2 5 2
5 3 8
2 0
x y z
x z
x y
+ + = − − = − =
21.
1 0 3 4
2 4 1 5
1 2 5 2
1 3 0 ( 4) 4 4
2 1 4 5 3 9
1 5 2 2 4 4
A C
− + = + −
+ + − − = + + = − + +
22.
1 0 3 4
2 4 1 5
1 2 5 2
1 3 0 ( 4) 2 4
2 1 4 5 1 1
1 5 2 2 6 0
A C
− − = − −
− − − − = − − = − − − − −
23.
1 0 6 1 6 0 6 0
6 6 2 4 6 2 6 4 12 24
1 2 6( 1) 6 2 6 12
A
⋅ ⋅ = ⋅ = ⋅ ⋅ = − − ⋅ −
24. 4 3 0
4 41 1 2
4 4 4( 3) 4 0
4 1 4 1 4( 2)
16 12 0
4 4 8
B−
− = − ⋅ − − ⋅ − − − ⋅
= − ⋅ − ⋅ − − −
= − −
25. 1 0
4 3 02 4
1 1 21 2
1(4) 0(1) 1( 3) 0(1) 1(0) 0( 2)
2(4) 4(1) 2( 3) 4(1) 2(0) 4( 2)
1(4) 2(1) 1( 3) 2(1) 1(0) 2( 2)
4 3 0
12 2 8
2 5 4
AB
− = ⋅ − −
+ − + + − = + − + + − − + − − + − + −
− = − − − −
26. 1 0
4 3 02 4
1 1 21 2
4(1) 3(2) 0( 1) 4(0) 3(4) 0(2)
1(1) 1(2) 2( 1) 1(0) 1(4) 2(2)
2 12
5 0
BA
− = ⋅ − − − + − − +
= + − − + − − −
=
27. 3 4
4 3 01 5
1 1 25 2
3(4) 4(1) 3( 3) 4(1) 3(0) 4( 2)
1(4) 5(1) 1( 3) 5(1) 1(0) 5( 2)
5(4) 2(1) 5( 3) 2(1) 5(0) 2( 2)
8 13 8
9 2 10
22 13 4
CB
− − = ⋅ −
− − − − − = + − + + − + − + + −
− = − − −
28. 3 4
4 3 01 5
1 1 25 2
4(3) 3(1) 0(5) 4( 4) 3(5) 0(2)
1(3) 1(1) 2(5) 1( 4) 1(5) 2(2)
9 31
6 3
BC
− − = ⋅ − − + − − +
= + − − + − −
= − −
Chapter 11: Systems of Equations and Inequalities
1214
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
29. 4 6
1 3A
=
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
( )
1 2
2 1 2
12 21 2 6
6 3
12
1 2 11 26 3
4 6 1 0
1 3 0 1
Interchange1 3 0 1
and 4 6 1 0
1 3 0 1 4
0 6 1 4
1 3 0 1
0 1
1 0 1 3
0 1
r r
R r r
R r
R r r
→ → = − + − −
→ = − − −
→ = − + −
Thus, 1
1 21 26 3
1A− −
= − .
30. 3 2
1 2A
− = −
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
( )
1 2
2 1 2
12 24
1 2 1314 4
1 12 2
314 4
Interchange3 2 1 0
and 1 2 0 1
1 2 0 1 3
3 2 1 0
1 2 0 1
0 4 1 3
1 2 0 1 2
0 1
1 0
0 1
r r
R r r
R r
R r r
− −
− → = + − − → = − − −
→ = + − − − −
→ − −
Thus, 1 1
1 2 231
4 4
A− − −= − −
.
31.
1 3 3
1 2 1
1 1 2
A
= −
Augment the matrix with the identity and use row operations to find the inverse:
( )
2 1 2
3 1 3
2 2
1 2 1
3 2 3
1 3 3 1 0 0
1 2 1 0 1 0
1 1 2 0 0 1
1 3 3 1 0 0
0 1 2 1 1 0
0 4 1 1 0 1
1 3 3 1 0 0
0 1 2 1 1 0
0 4 1 1 0 1
1 0 3 2 3 03
0 1 2 1 1 0 4
0 0 7 3 4 1
R r r
R r r
R r
R r r
R r r
−
= − + → − − − = − + − − −
→ − = − − − −
− − = − + → − = + −
( )13 37
3 4 17 7 7
5 9 37 7 7
1 3 11 1 27 7 7
2 3 23 4 17 7 7
1 0 3 2 3 0
0 1 2 1 1 0
0 0 1
1 0 03
0 1 0 2
0 0 1
R r
R r r
R r r
− −
→ − = − − = + → − = − + −
Thus,
5 9 37 7 7
1 1 1 27 7 73 4 17 7 7
A−
−
= − −
.
32.
3 1 2
3 2 1
1 1 1
A
= −
Augment the matrix with the identity and use row operations to find the inverse:
1 3
2 1 2
3 1 3
3 1 2 1 0 0
3 2 1 0 1 0
1 1 1 0 0 1
1 1 1 0 0 1Interchange
3 2 1 0 1 0 and
3 1 2 1 0 0
1 1 1 0 0 13
0 1 4 0 1 3 3
0 2 1 1 0 3
r r
R r r
R r r
−
→ −
= − + → − − − = − + − − −
( )2 2
1 1 1 0 0 1
0 1 4 0 1 3
0 2 1 1 0 3
R r
→ − = − − − −
Chapter 11 Review Exercises
1215
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
( )
1 2 1
3 2 3
13 37
31 27 7 7
3 517 7 7
1 3 194 17 7 7
2 3 231 2
7 7 7
1 0 3 0 1 2
0 1 4 0 1 3 2
0 0 7 1 2 3
1 0 3 0 1 2
0 1 4 0 1 3
0 0 1
1 0 03
0 1 0 4
0 0 1
R r r
R r r
R r
R r r
R r r
− − = − + → − = + − − −
→ − = − − = + → − = − + −
Thus,
3 517 7 7
1 94 17 7 7
31 27 7 7
A−
−
= − −
.
33. 4 8
1 2A
− = −
Augment the matrix with the identity and use row operations to find the inverse:
1 2
4 8 1 0
1 2 0 1
Interchange1 2 0 1
and 4 8 1 0 r r
− −
− → −
( )
( )
2 1 2
1 1
1 2 0 1 4
0 0 1 4
1 2 0 1
0 0 1 4
R r r
R r
− → = +
− −
→ = −
There is no inverse because there is no way to obtain the identity on the left side. The matrix is singular.
34. 3 1
6 2A
− = −
Augment the matrix with the identity and use row operations to find the inverse:
( )
( )
1 113 3
1 13
1 13 3
2 1 2
3 1 1 0
6 2 0 1
1 0
6 2 0 1
1 0 6
0 0 2 1
R r
R r r
− −
− −→ = −
− − −
→ = + −
There is no inverse because there is no way to obtain the identity on the left side. The matrix is singular.
35. 3 2 1
10 10 5
x y
x y
− = + =
Write the augmented matrix:
( )
( )
( )
2 1 2
1 2
2 1 2
12 2501
10
3 2 1
10 10 5
23 1 3161 2
Interchange161 2 and 23 1
161 2 30 50 5
161 2
0 1
R r r
r r
R r r
R r
−
−→ = − +
→ −
→ = − + − −
→ = −
( )25
1 2 11
10
01 160 1
R r r → = − +
The solution is 2 1
,5 10
x y= = or 2 1
,5 10
.
36. 3 2 6
1
2
x y
x y
+ = − = −
Write the augmented matrix:
12
3 2 6
1 1
− −
( )
( )
( )
12
1 2
12
2 1 2152
112
2 23 52
1 2 132
Interchange1 1
and 3 2 6
1 1 3
0 5
1 1
0 1
1 0 1
0 1
r r
R r r
R r
R r r
− − →
− −
→ = − + − −
→ =
→ = +
The solution is 3
1,2
x y= = or 3
1,2
.
Chapter 11: Systems of Equations and Inequalities
1216
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
37.
5 6 3 6
4 7 2 3
3 7 1
x y z
x y z
x y z
− − = − − = − + − =
Write the augmented matrix:
( )1 2 1
2 1 2
3 1 3
1392 2 211
11 111
3 32
5 6 3 6
27 34
3 71 1
91 1 1
27 34
3 71 1
91 1 14
0 39 11 23
4 260 2
91 1 1
0 1
0 131 2
R r r
R r r
R r r
R r
R r
− −
−− − −
−
−− −→ = − + − −
= − + → −− = − + − −−
− = −− →
= −
9 6011 11
1 2 139211 11
3 2 31042411 11
1 0
0 1
0 0
R r r
R r r
−= − + → − = − +
( )9 6011 11
392 113 311 11 24
133
91 3 11113
3 22 3 213 11
3
1 0
0 1
0 0 1
1 0 0 9
0 1 0
0 0 1
R r
R r r
R r r
− → − =
= + → = +
The solution is 13 13
9, ,3 3
x y z= = = or
13 139, ,
3 3
.
38.
2 5
4 3 1
8 5
x y z
x y z
x y z
+ + = − − = + − =
Write the augmented matrix:
52 1 1
34 1 1
8 51 1
−− −
2 1 2
3 1 3
51 112 2 2
1 1253 1
2 23
13 1
1 2 15 23
3 2 3
52 1 12
0 3 5 9 4
0 3 5 15
1
0 1 3
0 3 5 15
1 0 1
0 1 3 3
0 0 0 6
R r r
R r r
R r
R r
R r r
R r r
= − +
→ − − − = − + − − − = → = − − − − −
= − +→ = + −
There is no solution; the system is inconsistent.
39.
2 1
2 3 3
4 3 4 3
x z
x y
x y z
− = + = − − − =
Write the augmented matrix:
2 1 2
3 1 3
1 0 2 1
2 3 0 3
4 3 4 3
1 0 2 12
0 3 4 5 4
0 3 4 1
R r r
R r r
− − − −
− = − + → − = − + − −
( )54 12 23 3 3
1 0 2 1
0 1
0 3 4 1
R r
− → − = − −
( )543 2 33 3
1 0 2 1
0 1 3
0 0 8 6
R r r
− → − = + −
( )54 13 33 3 8
34
12
1 3 123 4
2 3 2334
1 0 2 1
0 1
0 0 1
1 0 02
0 1 0
0 0 1
R r
R r r
R r r
−
→ − = − − = +
→ − = − + −
The solution is 1 2 3
, ,2 3 4
x y z= − = − = − or
1 2 3, ,
2 3 4 − − −
.
Chapter 11 Review Exercises
1217
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
40.
2 2
2 2 1
6 4 3 5
x y z
x y z
x y z
+ − = − + = − + + =
Write the augmented matrix: 1 2 1 2
2 2 1 1
6 4 3 5
− − −
( )
2 1 2
3 1 3
51 12 22 6 6
13
1 2 1512 6
3 2 313
1351
2 61
15
1 2 1 22
0 6 3 5 6
0 8 9 7
1 2 1 2
0 1
0 8 9 7
1 0 02
0 1 8
0 0 5
1 0 0
0 1
0 0 1
R r r
R r r
R r
R r r
R r r
R
− = − + → − − = − + − −
− → − = − − − = − +
→ − = + −
→ − −
( )
( )
13 35
134 1
2 3 25 21
15
1 0 0
0 1 0
0 0 1
r
R r r
=
→ = + −
The solution is 1 4 1
, ,3 5 15
x y z= = = − or
1 4 1, ,
3 5 15 −
.
41.
0
5 6
2 2 1
x y z
x y z
x y z
− + = − − = − + =
Write the augmented matrix: 1 1 1 0
1 1 5 6
2 2 1 1
− − − −
2 1 2
3 1 3
1 1 1 0
0 0 6 6 2
0 0 1 1
R r r
R r r
− = − + → − = − + −
( )12 26
1 1 1 0
0 0 1 1
0 0 1 1
R r
− → − = − −
1 2 1
3 2 3
1 1 0 1
0 0 1 1
0 0 0 0
R r r
R r r
− = − + → − = +
The system is dependent. 1
1
x y
z
= + = −
The solution is 1x y= + , 1z = − , y is any real
number or {( , , ) 1, 1, is anyx y z x y z y= + = −
}real number .
42.
4 3 5 0
2 4 3 0
6 2 0
x y z
x y z
x y z
− + = + − = + + =
Write the augmented matrix:
( )3 54 4
11 14
4 3 5 0 1 0
2 4 3 0 2 4 3 0
6 2 1 0 6 2 1 0
R r
− − − → − =
3 54 4
2 1 211 112 2
3 1 313 132 2
3 524 4
2 211
23 313
132
1 2 14
3 2 3
1 02
0 0 6
0 0
1 0
0 1 1 0
0 1 1 0
1 0 0
0 1 1 0
0 0 0 0
R r r
R r r
R r
R r
R r r
R r r
−= − + → − = − + −
− = → − = −
= + → − = − +
The solution is 1
2x z= − , y z= , z is any real
number or
1( , , ) , , is any real number
2x y z x z y z z
= − =
.
Chapter 11: Systems of Equations and Inequalities
1218
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
43.
1
2 2 3
2 2 3 0
3 4 5 3
x y z t
x y z t
x y z t
x y z t
− − − = + − + = − − − = − + + = −
Write the augmented matrix: 1 1 1 1 1
2 1 1 2 3
1 2 2 3 0
3 4 1 5 3
− − − − − − − − −
2 1 2
3 1 3
4 1 4
2 3
2
1 1 1 1 12
0 3 1 4 1
0 1 1 2 13
0 1 4 8 6
1 1 1 1 1
Interchange0 1 1 2 1
and 0 3 1 4 1
0 1 4 8 6
1 1 1 1 1
0 1 1 2 1
0 3 1 4 1
0 1 4 8 6
R r r
R r r
R r r
r r
R
− − − = − + → = − + − − − − = − + − −
− − − − − − − → − −
− − − → = − − −
( )2r
1 2 1
3 2 3
4 2 4
13 32
14 45
2 3 2
4
1 0 0 1 2
0 1 1 2 1 3
0 0 2 2 2
0 0 5 10 5
1 0 0 1 2
0 1 1 2 1
0 0 1 1 1
0 0 1 2 1
1 0 0 1 2
=0 1 0 1 0
0 0 1 1 1 =
0 0 0 1 2
R r r
R r r
R r r
R r
R r
R r r
R r
= + → = − + − − − = + −
= − →
= − − + → − −
3 4r
+
1 4 1
2 4 2
3 4 3
1 0 0 0 4
0 1 0 0 2
0 0 1 0 3
0 0 0 1 2
R r r
R r r
R r r
= − + → = − + = − + −
The solution is 4, 2, 3, 2x y z t= = = = − or
(4, 2, 3, 2)− .
44.
3 3 42 3
3 2 35 6
x y z tx y zx z tx y z
− + − = + − = − + + =
+ + =
Write the augmented matrix: 1 3 3 1 41 2 1 0 31 0 3 2 31 1 5 0 6
− − − −
( )
2 1 2
3 1 3
4 1 4
74 115 5 5
2 25
3 2 15 5 5
74 1 15 5 5
1612 125 5 5
26 3815 5 5
1 3 3 1 40 5 4 1 7
0 3 0 3 10 4 2 1 2
1 3 3 1 4
0 1
0 3 0 3 10 4 2 1 2
1 0
0 1
0 0
0 0
R r rR r r
R r r
R r
R
− − = − + − − → = − + − = − + − −
− − → = − − − =− −
→
2 1
3 2 3
4 2 4
334
r r
R r r
R r r
+ = − + = − +
( )
3 2 15 5 5
74 15 5 5 5
3 31243
26 3815 5 5
31 3 11 5
3 42 3 24 5
3 264 3 42 5
3
13432
15
1 0
0 1
0 0 1 1
0 0
1 0 0 1 1
0 1 0 1
0 0 1 1
0 0 0 5
1 0 0 1 1
0 1 0 1
0 0 1 1
0 0 0 1
R r
R r r
R r r
R r r
− −
− − → =
− − = − + − → = + = − + −
− − − →
−
( )14 45
R r= −
1715
1 1 4 15
2 4 22215
3 4 32
15
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
R r rR r r
R r r
− = + − → = − + = − + −
The solution is 17 1 22 2
15 5 15 15, , ,x y z t= − = − = = −
or 17 1 22 2
, , ,15 5 15 15
− − −
.
Chapter 11 Review Exercises
1219
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
45. 3 4
3(3) 4(1) 9 4 51 3
= − = − =
46. 4 0
4(3) 1(0) 12 0 121 3
−= − − = − − = −
47.
01 46 62 1 1 261 2 1 4 03 31 4 4 134 1
1(6 6) 4( 3 24) 0( 1 8)
1(0) 4( 27) 0( 9) 0 108 0
108
− −− = − +
= − − − − + − −= − − + − = + +=
48.
3 1025 0 5 01 10 5 2 3 1013 32 1 1 231 2
2(3 10) 3(0 5) 10(0 1)
2( 7) 3(5) 10(1)
14 15 10
19
= − +− −−
= − − + + += − − += − − += −
49.
32 10 5 5 01 15 0 2 1 ( 3)16 0 0 62 26 02
2(0 6) 1(0 2) 3(30 0)
2( 6) 1( 2) 3(30)
12 2 90
100
−= − + −
= − − − − −= − − − −= − + −= −
50.
0123 32 1 1 231 2 2 1 0
4 2 1 2 1 41 4 2
2(4 12) 1(2 3) 0(4 2)
2( 8) 1(5) 0(6)
16 5 0
11
−= − − +
− −−
= − − − + + += − − − += − +=
51. 2 4
3 2 4x y
x y
− = + =
Set up and evaluate the determinants to use Cramer’s Rule:
1 2
3 21(2) 3( 2) 2 6 8D
−= = − − = + =
4 2
4 24(2) 4( 2) 8 8 16xD
−= = − − = + =
1 4
3 41(4) 3(4) 4 12 8yD = = − = − = −
The solution is 16
82xD
Dx = = = ,
8
81yD
Dy
−= = = −
or (2, 1)− .
52. 3 5
2 3 5x y
x y
− = − + =
Set up and evaluate the determinants to use Cramer’s Rule:
1 31(3) 2( 3) 3 6 9
2 3D
−= = − − = + =
5 35(3) 5( 3) 15 15 0
5 3xD− −
= = − − − = − + =
1 51(5) 2( 5) 5 10 15
2 5yD−
= = − − = + =
The solution is 0
90xD
Dx = = = ,
15 5
9 3yD
Dy = = =
or 5
30,
.
53. 2 3 13 0
3 2 0x y
x y
+ − = − =
Write the system is standard form: 2 3 133 2 0
x y
x y
+ = − =
Set up and evaluate the determinants to use Cramer’s Rule:
2 34 9 13
3 2D = = − − = −
−
13 326 0 26
0 2xD = = − − = −−
132 0 39 393 0
yD = = − = −
The solution is 26
132xD
Dx
−−
= = = ,
39
133yD
Dy
−−
= = = or (2, 3).
Chapter 11: Systems of Equations and Inequalities
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54. 3 4 12 0
5 2 6 0
x y
x y
− − = + + =
Write the system in standard form: 3 4 12
5 2 6
x y
x y
− = + = −
Set up and evaluate the determinants to use Cramer's Rule:
3 46 20 26
5 2D
−= = + =
12 424 24 0
6 2xD−
= = − =−
3 12 18 60 785 6
yD = = − − = −−
The solution is 0
026
xDx
D= = = ,
783
26yD
yD
−= = = − or (0, 3)− .
55. 2 6
2 3 13
3 2 3 16
x y z
x y z
x y z
+ − = − + = − − + = −
Set up and evaluate the determinants to use Cramer’s Rule:
( )
1 2 1
2 1 3
3 2 3
1 3 1 3 2 11 2 ( 1)
2 3 2 3 3 2
1 3 6 2( 3 6) ( 1)( 4 3)
3 6 1 10
D
−= −
−− − −
= − + −− − −
= − + − − + + − − += + + =
( )
6 2 1
13 1 3
16 2 3
1 3 13 3 13 16 2 ( 1)
2 3 16 3 16 2
6 3 6 2( 39 48) ( 1)(26 16)
18 18 10 10
xD
−= − −
− −− − − −
= − + −− − − −
= − + − − + + − −= − − = −
( )
1 6 1
2 13 3
3 16 3
13 3 2 3 2 131 6 ( 1)
16 3 3 3 3 16
1 39 48 6(6 9) ( 1)( 32 39)
9 18 7 20
yD
−= −
−− −
= − + −− −
= − + − − + − − += + − =
( )
1 2 6
2 1 13
3 2 16
1 13 2 13 2 11 2 6
2 16 3 16 3 2
1 16 26 2( 32 39) 6( 4 3)
10 14 6 30
zD = − −− −
− − − −= − +
− − − −= − − − + + − += − − − = −
The solution is 10
110
xDx
D
−= = = − ,
202
10yD
yD
= = = , 30
310
zDz
D
−= = = − or
( 1, 2, 3)− − .
56.
8
2 3 2
3 9 9
x y z
x y z
x y z
− + = + − = − − − =
Set up and evaluate the determinants to use Cramer’s Rule:
1 1 1
2 3 1
3 1 9
3 1 2 32 11 ( 1) 11 9 3 13 9
1( 27 1) 1( 18 3) 1( 2 9)
28 15 11
54
D
−= −
− −
− −= − − +− − −−
= − − + − + + − −= − − −= −
8 1 1
2 3 1
9 1 9
3 1 2 3128 ( 1) 11 9 9 19 9
8( 27 1) 1(18 9) 1(2 27)
224 27 25
222
xD
−= − −
− −
− −−−= − − +− − −−
= − − + + + −= − + −= −
1 8 1
2 2 1
3 9 9
1 2 1 22 21 8 19 9 3 9 3 9
1(18 9) 8( 18 3) 1(18 6)
27 120 24
171
yD = − −−
− −− −= − +− −
= + − − + + += + +=
Chapter 11 Review Exercises
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1 1 8
2 3 2
3 1 9
3 2 2 32 21 ( 1) 81 9 3 13 9
1(27 2) 1(18 6) 8( 2 9)
25 24 88
39
zD
−= −
−
− −= − − +− −
= − + + + − −= + −= −
The solution is 222 37
54 9xD
xD
−= = =−
,
171 19
54 6yD
yD
= = = −−
, 39 13
54 18zD
zD
−= = =−
or
37 19 13, ,
9 6 18 −
.
57. Let 8x y
a b= .
Then ( )2 2 8 162
x y
a b= = by Theorem (14).
The value of the determinant is multiplied by k when the elements of a column are multiplied by k.
58. Let 8x y
a b= .
Then 8y x
b a= − by Theorem (11). The
value of the determinant changes sign when any 2 columns are interchanged.
59. Find the partial fraction decomposition:
6( 4) ( 4)
( 4) 4
6 ( 4)
A Bx x x x
x x x x
A x Bx
− = − + − − = − +
Let x =4, then 6 (4 4) (4)
4 6
3
2
A B
B
B
= − +=
=
Let x = 0, then 6 (0 4) (0)
4 6
3
2
A B
A
A
= − +− =
= −
3 36 2 2
( 4) 4x x x x
−= +
− −
60. Find the partial fraction decomposition:
( 2)( 3) 2 3
x A B
x x x x= +
+ − + −
Multiply both sides by ( 2)( 3)x x+ − .
( 3) ( 2)x A x B x= − + +
Let 2x = − , then 2 ( 2 3) ( 2 2)
5 2
2
5
A B
A
A
− = − − + − +− = −
=
Let 3x = , then 3 (3 3) (3 2)
5 3
3
5
A B
B
B
= − + +=
=
2 35 5
( 2)( 3) 2 3x
x x x x= +
+ − + −
61. Find the partial fraction decomposition:
2 2
4
1( 1)
x A B C
x xx x x
− = + +−−
Multiply both sides by 2 ( 1)x x − 24 ( 1) ( 1)x Ax x B x Cx− = − + − +
Let 1x = , then 21 4 (1)(1 1) (1 1) (1)
3
3
A B C
C
C
− = − + − +− =
= −
Let 0x = , then 20 4 (0)(0 1) (0 1) (0)
4
4
A B C
B
B
− = − + − +− = −
=
Let 2x = , then 22 4 (2)(2 1) (2 1) (2)
2 2 4
2 2 4 4( 3)
2 6
3
A B C
A B C
A
A
A
− = − + − +− = + +
= − − − −==
2 2
4 3 4 3
1( 1)
x
x xx x x
− −= + +−−
Chapter 11: Systems of Equations and Inequalities
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62. Find the partial fraction decomposition:
2 2
2 6
2 1( 2) ( 1) ( 2)
x A B C
x xx x x
− = + +− −− − −
Multiply both sides by 2( 2) ( 1)x x− − . 22 6 ( 2)( 1) ( 1) ( 2)x A x x B x C x− = − − + − + −
Let 1x = , then 22(1) 6 (1 2)(1 1) (1 1) (1 2)
4
4
A B C
C
C
− = − − + − + −− =
= −
Let 2x = , then 22(2) 6 (2 2)(2 1) (2 1) (2 2)
2
A B C
B
− = − − + − + −− =
Let 0x = , then 22(0) 6 (0 2)(0 1) (0 1) (0 2)
6 2 4
6 2 ( 2) 4( 4)
2 8
4
A B C
A B C
A
A
A
− = − − + − + −− = − +− = − − + −
==
2 2
2 6 4 2 4
2 1( 2) ( 1) ( 2)
x
x xx x x
− − −= + +− −− − −
63. Find the partial fraction decomposition:
2 21( 9)( 1) 9
x A Bx C
xx x x
+= +++ + +
Multiply both sides by 2( 1)( 9)x x+ + . 2( 9) ( )( 1)x A x Bx C x= + + + +
Let 1x = − , then
( )( ) ( )( )( )21 1 9 1 1 1
1 (10) ( )(0)
1 10
1
10
A B C
A B C
A
A
− = − + + − + − +
− = + − +− =
= −
Let 0x = , then
( ) ( )( )( )20 0 9 0 0 1
0 9
10 9
10
9
10
A B C
A C
C
C
= + + + +
= +
= − +
=
Let 1x = , then ( ) ( )( )( )21 1 9 1 1 1
1 (10) ( )(2)
1 10 2 2
1 91 10 2 2
10 10
91 1 2
5
A B C
A B C
A B C
B
B
= + + + +
= + += + +
= − + +
= − + +
1
251
10
B
B
=
=
2 2
1 1 910 10 10
1( 9)( 1) 9
xx
xx x x
− += +
++ + +
64. Find the partial fraction decomposition:
2 2
3
2( 2)( 1) 1
x A Bx C
xx x x
+= +−− + +
Multiply both sides by 2( 2)( 1)x x− + .
23 ( 1) ( )( 2)x A x Bx C x= + + + −
Let 2x = , then 23(2) ((2) 1) ( (2) )(2 2)
6 5
6
5
A B C
A
A
= + + + −=
=
Let 0x = , then 23(0) (0 1) ( (0) )(0 2)
0= 2
60 2
56
253
5
A B C
A C
C
C
C
= + + + −−
= −
=
=
Let 1x = , then 23(1) (1 1) ( (1) )(1 2)
3 2
6 33 2
5 5
6
5
A B C
A B C
B
B
= + + + −= − −
= − −
= −
2 2
6 6 33 5 5 5
2( 2)( 1) 1
xx
xx x x
− += +
−− + +
Chapter 11 Review Exercises
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65. Find the partial fraction decomposition: 3
2 2 2 2 2( 4) 4 ( 4)
x Ax B Cx D
x x x
+ += ++ + +
Multiply both sides by 2 2( 4)x + . 3 2
3 3 2
3 3 2
( )( 4)
4 4
(4 ) 4
x Ax B x Cx D
x Ax Bx Ax B Cx D
x Ax Bx A C x B D
= + + + +
= + + + + +
= + + + + +
1; 0A B= = 4 0
4(1) 0
4
A C
C
C
+ =+ =
= −
4 0
4(0) 0
0
B D
D
D
+ =+ =
=
3
2 2 2 2 2
4
( 4) 4 ( 4)
x x x
x x x
−= ++ + +
66. Find the partial fraction decomposition: 3
2 2 2 2 2
1
( 16) 16 ( 16)
x Ax B Cx D
x x x
+ + += ++ + +
Multiply both sides by 2 2( 16)x + . 3 2
3 3 2
3 3 2
1 ( )( 16)
1 16 16
1 (16 ) 16
x Ax B x Cx D
x Ax Bx Ax B Cx D
x Ax Bx A C x B D
+ = + + + +
+ = + + + + +
+ = + + + + +
1; 0A B= = 16 0
16(1) 0
16
A C
C
C
+ =+ =
= −
16 1
16(0) 1
1
B D
D
D
+ =+ =
=
3
2 2 2 2 2
1 16 1
( 16) 16 ( 16)
x x x
x x x
+ − += ++ + +
67. Find the partial fraction decomposition: 2 2
2 2 2
2
( 1)( 1) ( 1)( 1)( 1)
1 1 1
x x
x x x x x
A B Cx D
x x x
=+ − + − +
+= + +− + +
Multiply both sides by 2( 1)( 1)( 1)x x x− + + . 2 2 2( 1)( 1) ( 1)( 1)
( )( 1)( 1)
x A x x B x x
Cx D x x
= + + + − ++ + − +
Let 1x = , then 2 2 21 (1 1)(1 1) (1 1)(1 1)
( (1) )(1 1)(1 1)
1 4
1
4
A B
C D
A
A
= + + + − ++ + − +
=
=
Let 1x = − , then 2 2
2
( 1) ( 1 1)(( 1) 1)
( 1 1)(( 1) 1)
( ( 1) )( 1 1)( 1 1)
1 4
1
4
A
B
C D
B
B
− = − + − +
+ − − − ++ − + − − − +
= −
= −
Let 0x = , then 2 2 20 (0 1)(0 1) (0 1)(0 1)
( (0) )(0 1)(0 1)
0
1 10
4 4
1
2
A B
C D
A B D
D
D
= + + + − ++ + − +
= − −
= − − −
=
Let 2x = , then 2 2 22 (2 1)(2 1) (2 1)(2 1)
( (2) )(2 1)(2 1)
4 15 5 6 3
1 1 14 15 5 6 3
4 4 2
15 5 36 4
4 4 26 0
0
A B
C D
A B C D
C
C
C
C
= + + + − ++ + − +
= + + +
= + − + +
= − + −
==
( )( )2
22 2
1 1 14 4 2
1 1 11 1
x
x x xx x
−= + +
− + ++ −
Chapter 11: Systems of Equations and Inequalities
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68. Find the partial fraction decomposition:
( ) ( ) ( )2 2 2
2
4 4
4 1 4 ( 1)( 1)
1 1 4
x x x x x
A B Cx D
x x x
=+ − + − +
+= + +− + +
Multiply both sides by ( )2 4 ( 1)( 1)x x x+ − + .
( ) ( )2 24 ( 1) 4 ( 1) 4
( )( 1)( 1)
A x x B x x
Cx D x x
= + + + − ++ + − +
Let 1x = , then
( ) ( )( )
2 24 (1 1) 1 4 (1 1) 1 4
(1) (1 1)(1 1)
4 10
4 2
10 5
A B
C D
A
A
= + + + − ++ + − +
=
= =
Let 1x = − , then
( ) ( )( )
2 24 ( 1 1) ( 1) 4 ( 1 1) ( 1) 4
( 1) ( 1 1)( 1 1)
4 10
4 2
10 5
A B
C D
B
B
= − + − + + − − − +
+ − + − − − += −
= − = −
Let 0x = , then
( ) ( )( )
2 24 (0 1) 0 4 (0 1) 0 4
(0) (0 1)(0 1)
4 4 4
2 24 4 4
5 5
8 84
5 54
5
A B
C D
A B D
D
D
D
= + + + − ++ + − +
= − −
= − − −
= + −
= −
Let 2x = , then
( ) ( )( )
2 24 (2 1) 2 4 (2 1) 2 4
(2) (2 1)(2 1)
4 24 8 6 3
2 2 44 24 8 6 3
5 5 5
48 16 124 6
5 5 56 0
0
A B
C D
A B C D
C
C
C
C
= + + + − ++ + − +
= + + +
= + − + + −
= − + −
==
( )( ) 22 2
2 2 45 5 54
1 1 44 1 x x xx x
− −= + +
− + ++ −
69. Solve the first equation for y, substitute into the second equation and solve:
2 2
2 3 0 2 3
5
x y y x
x y
+ + = → = − −
+ =
2 2 2 2
2
( 2 3) 5 4 12 9 5
5 12 4 0 (5 2)( 2) 0
2 or 2
511
15
x x x x x
x x x x
x x
y y
+ − − = → + + + =
+ + = → + + =
= − = −
= − =
Solutions: 2 11
, , ( 2,1)5 5
− − −
.
70. Add the equations to eliminate y, and solve: 2 2
2
2
16
2 8
2 8
x y
x y
x x
+ =
− = −
+ =
2 2 8 0 ( 4)( 2) 0
4 or 2
x x x x
x x
+ − = → + − == − =
2 2 2
2 2 2
If 4 :
( 4) 16 0 0
If 2 :
(2) 16 12 2 3
x
y y y
x
y y y
= −
− + = → = → =
=
+ = → = → = ±
Solutions: ( ) ( ) ( )– 4, 0 , 2, 2 3 , 2, 2 3− .
71. Multiply each side of the second equation by 2 and add the equations to eliminate xy:
2 2
22 2
2
2
2 10 2 10
3 2 2 6 4
7 14
2
2
xy y xy y
xy y xy y
y
y
y
+ = ⎯⎯→ + =
− + = ⎯⎯→ − + =
=
=
= ±
( ) ( )
( ) ( )
2
2
If 2 :
2 2 2 10 2 2 8 2 2
If 2 :
2 2 2 10 2 2 8
2 2
y
x x x
y
x x
x
=
+ = → = → =
= −
− + − = → − =
→ = −
Solutions: ( ) ( )2 2, 2 , 2 2, 2− −
Chapter 11 Review Exercises
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72. Multiply each side of the first equation by –2 and add the equations to eliminate y:
22 2 2 2
2 2 2 2
2
3 1 6 2 2
7 2 5 7 2 5
3
3
x y x y
x y x y
x
x
− − = ⎯⎯→ − + = −
− = ⎯⎯→ − =
=
= ±
( )
( )
2 2 2
2 2 2
If 3 :
3 3 1 8
8 2 2
If 3 :
3 3 1 8
8 2 2
x
y y
y
x
y y
y
=
− = → − = −
→ = ± = ±
= −
− − = → − = −
→ = ± = ±
Solutions:
( ) ( ) ( )( )
3, 2 2 , 3, 2 2 , 3, 2 2 ,
3, 2 2
− − − −
73. Substitute into the second equation into the first equation and solve:
2 2
2
6
3
x y y
x y
+ =
=
2
2
3 6
3 0
( 3) 0 0 or 3
y y y
y y
y y y y
+ =
− =− = → = =
2 2
2 2
If 0 : 3(0) 0 0
If 3 : 3(3) 9 3
y x x x
y x x x
= = → = → =
= = → = → = ±
Solutions: (0, 0), (–3, 3), (3, 3).
74. Multiply each side of the second equation by –1 and add the equations to eliminate y:
2 2 2 2
12 2 2 2
2
2 9 2 9
9 9
0 0
x y x y
x y x y
x x
−
+ = ⎯⎯→ + =
+ = ⎯⎯→ − − = −
= =
2 2 2If 0 : 0 9 9 3x y y y= + = → = → = ±
Solutions: (0, 3), (0, –3)
75. Factor the second equation, solve for x, substitute into the first equation and solve:
2 2
2 2
3 4 5 8
3 2 0
x xy y
x xy y
+ + =
+ + =
2 23 2 0
( 2 )( ) 0 2 or
x xy y
x y x y x y x y
+ + =+ + = → = − = −
Substitute 2x y= − and solve: 2 2
2 2
2 2 2
2
2
3 4 5 8
3( 2 ) 4( 2 ) 5 8
12 8 5 8
9 8
8 2 2
9 3
x xy y
y y y y
y y y
y
y y
+ + =
− + − + =
− + =
=
= = ±
Substitute x y= − and solve:
2 2
2 2
2 2 2
2
2
3 4 5 8
3( ) 4( ) 5 8
3 4 5 8
4 8
2 2
x xy y
y y y y
y y y
y
y y
+ + =
− + − + =
− + =
=
= = ±
2 2 2 2 4 2If : 2
3 3 3
2 2 2 2 4 2If : 2
3 3 3
y x
y x
−= = − = − −= = − =
If 2 : 2
If 2 : 2
y x
y x
= = −
= − =
Solutions:
( )( )
4 2 2 2 4 2 2 2, , , , 2, 2 ,
3 3 3 3
2, 2
− − −
−
76. 2 2
2
3 2 2 6
2 4
x xy y
xy y
+ − =
− = −
Multiply each side of the first equation by 2 and each side of the second equation by 3 and add to eliminate the constant:
2 2
2
2 2
6 4 4 12
3 6 12
6 7 10 0
x xy y
xy y
x xy y
+ − =
− = −
+ − =
Chapter 11: Systems of Equations and Inequalities
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
( ) ( )6 5 2 0
5 or 2
6
x y x y
x y x y
− + =
= = −
Substitute 56x y= and solve:
2
2 2
52 4
6
7 24 2 424
6 7 7
y y y
y y y
⋅ − = −
− = − → = → = ±
Substitute 2x y= − and solve:
2
2
2
2 2 4
4 4
1
1
y y y
y
y
y
− ⋅ − = −
− = −
== ±
2 42 5 2 42 5 42If :
7 6 7 21
2 42 5 2 42 5 42If :
7 6 7 21
y x
y x
= = ⋅ =
−= − = ⋅ = −
If 1: 2; If 1: 2y x y x= = − = − =
Solutions:
( )
( )
5 42 2 42 5 42 2 42, , , , 2,1 ,
21 7 21 7
2, 1
− − −
−
77.
2 2
2
3 2
1 0
x x y y
x xy
y
− + + = − − + + =
Multiply each side of the second equation by –y and add the equations to eliminate y:
2 2
2 2
3 2
0
2 2
1
x x y y
x x y y
x
x
− + + = −
− + − − =
− = −=
If 1:x = 2 2
2
1 3(1) 2
0
( 1) 0
0 or 1
y y
y y
y y
y y
− + + = −
+ =+ =
= = − Note that 0y ≠ because that would cause
division by zero in the original system. Solution: (1, –1)
78. Multiply each side of the second equation by –x and add the equations to eliminate x:
2 2 2 2
2
2
2
2 2
21 2
2
2 0
( 2) 0
0 or 2
x x y y x x y y
yx x x y
x
y y
y y
y y
y y
+ + = + → + + = + −+ = → − − = −
=
− =− =
= =
2 2 2
2 2 2
If 0 :
0 0 2 2 0
( 1)( 2) 0 1 or 2
If 2 :
2 2 2 0
( 1) 0 0 or 1
y
x x x x
x x x x
y
x x x x
x x x x
=
+ + = + → + − =→ − + = → = = −
=
+ + = + → + =→ + = → = = −
Note that 0x ≠ because that would cause division by zero in the original system. Solutions: (1, 0), (–2, 0), (–1, 2)
79. 3 4 12x y+ ≤
Graph the line 3 4 12x y+ = . Use a solid line
since the inequality uses ≤ . Choose a test point not on the line, such as (0, 0). Since
( ) ( )3 0 4 0 12+ ≤ is true, shade the side of the line
containing (0, 0).
y
x−5
3 + 4 x y 12≤
5
5
−5
80. 2 3 6x y− ≥
Graph the line 2 3 6x y− = . Use a solid line since
the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since ( ) ( )2 0 3 0 6− ≥ is
false, shade the side of the line opposite (0, 0). y
x−5 5
5
−5
62 − 3x y ≥
Chapter 11 Review Exercises
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81. 2y x≤
Graph the parabola 2y x= . Use a solid curve
since the inequality uses ≤ . Choose a test point
not on the parabola, such as (0, 1). Since 20 1≤ is false, shade the opposite side of the parabola from (0, 1).
y
x−5 5
5
−5
2y x≤
82. 2x y≥
Graph the circle 2x y= . Use a solid curve since
the inequality uses ≥ . Choose a test point not on
the parabola, such as (1, 0). Since 21 0≥ is true, shade the same side of the parabola as (1, 0).
y
x−5 5
5
−5
2x y≥
83. 2 2
2
x y
x y
− + ≤ + ≥
Graph the line 2 2x y− + = . Use a solid line
since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since
2(0) 0 2− + ≤ is true, shade the side of the line
containing (0, 0). Graph the line 2x y+ = . Use a
solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
y
–5
5
x–5 5
–2x + y = 2
x + y = 2
The graph is unbounded. Find the vertices: To find the intersection of 2x y+ = and
2 2x y− + = , solve the system:
2
2 2
x y
x y
+ =− + =
Solve the first equation for x: 2x y= − .
Substitute and solve: 2(2 ) 2
4 2 2
3 6
2
y y
y y
y
y
− − + =− + + =
==
2 2 0x = − = The point of intersection is (0, 2). The corner point is (0, 2).
84. 2 6
2 2
x y
x y
− ≤ + ≥
Graph the line 2 6x y− = . Use a solid line since
the inequality uses ≤ . Choose a test point not on the line, such as (0, 0). Since 0 2(0) 6− ≤ is true,
shade the side of the line containing (0, 0). Graph the line 2 2x y+ = . Use a solid line since the
inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2+ ≥ is false,
shade the opposite side of the line from (0, 0). The overlapping region is the solution.
y
–5
5
x–2 8
x – 2y = 6
(2, –2)
2x + y = 2 The graph is unbounded. Find the vertices: To find the intersection of 2 6x y− = and
2 2x y+ = , solve the system:
2 6
2 2
x y
x y
− = + =
Solve the first equation for x: 2 6x y= + .
Substitute and solve: 2(2 6) 2
4 12 2
5 10
2
y y
y y
y
y
+ + =+ + =
= −= −
2( 2) 6 2x = − + =
The point of intersection is (2, 2)− .
The corner point is (2, 2)− .
Chapter 11: Systems of Equations and Inequalities
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85.
0
0
4
2 3 6
x
y
x y
x y
≥ ≥ + ≤ + ≤
Graph 0; 0x y≥ ≥ . Shaded region is the first
quadrant. Graph the line 4x y+ = . Use a solid
line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 4+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 3 6x y+ = .
Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 6+ ≤ is true, shade the side of the
line containing (0, 0).
y
–2
8
x8
(0, 2)
(3, 0)(0, 0)
x + y = 4
2x + 3y = 6
The overlapping region is the solution. The graph is bounded. Find the vertices: The x-axis and y-axis intersect at (0, 0). The intersection of 2 3 6x y+ = and the y-axis is (0, 2). The
intersection of 2 3 6x y+ = and the x-axis is
(3, 0). The three corner points are (0, 0), (0, 2), and (3, 0).
86.
0
0
3 6
2 2
x
y
x y
x y
≥ ≥ + ≥ + ≥
Graph 0; 0x y≥ ≥ . Shaded region is the first
quadrant. Graph the line 3 6x y+ = . Use a solid
line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 3(0) + 0 ≥ 6 is false, shade the opposite side of the line from (0, 0). Graph the line 2 2x y+ = . Use a
solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) + 0 ≥ 2 is false, shade the opposite side of the line from (0, 0).
y
–2
8
x–2 8
(0, 6)
(2, 0)
3x + y = 6
2x + y = 2
The overlapping region is the solution. The graph is unbounded. Find the vertices: The intersection of 3 6x y+ = and the y-axis is
(0, 6). The intersection of 3 6x y+ = and the x-
axis is (2, 0). The two corner points are (0, 6), and (2, 0).
87.
0
0
2 8
2 2
x
y
x y
x y
≥ ≥ + ≤ + ≥
Graph 0; 0x y≥ ≥ . Shaded region is the first
quadrant. Graph the line 2 8x y+ = . Use a solid
line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) + 0 ≤ 8 is true, shade the side of the line containing (0, 0). Graph the line 2 2x y+ = . Use
a solid line since the inequality uses ≥ . Choose a test point not on the line, such as (0, 0). Since 0 + 2(0) ≥ 2 is false, shade the opposite side of the line from (0, 0).
y
–1
9
x–1 9
2x + y = 8
x + 2y = 2
(2, 0)
(4, 0)(0, 1)
(0, 8)
The overlapping region is the solution. The graph is bounded. Find the vertices: The intersection of
2 2x y+ = and the y-axis is (0, 1). The
intersection of 2 2x y+ = and the x-axis is (2, 0).
The intersection of 2 8x y+ = and the y-axis is
(0, 8). The intersection of 2 8x y+ = and the x-
axis is (4, 0). The four corner points are (0, 1), (0, 8), (2, 0), and (4, 0).
Chapter 11 Review Exercises
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88.
0
0
3 9
2 3 6
x
y
x y
x y
≥ ≥ + ≤ + ≥
Graph 0; 0x y≥ ≥ . Shaded region is the first
quadrant. Graph the line 3 9x y+ = . Use a solid
line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 3(0) 0 9+ ≤ is true, shade the side of the line
containing (0, 0).
y
–1
9
x–1 9
(0, 2)
(0, 9)
(3, 0)
2x + 3y = 6
3x + y = 9
Graph the line 2 3 6x y+ = . Use a solid line since
the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 6+ ≥ is
false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. The graph is bounded. Find the vertices: The intersection of 2 3 6x y+ = and the y-axis is
(0, 2). The intersection of 2 3 6x y+ = and the x-
axis is (3, 0). The intersection of 3 9x y+ = and
the y-axis is (0, 9). The intersection of 3 9x y+ =
and the x-axis is (3, 0). The three corner points are (0, 2), (0, 9), and (3, 0).
89. Graph the system of inequalities: 2 2 16
2
x y
x y
+ ≤
+ ≥
Graph the circle 2 2 16x y+ = .Use a solid line
since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since
2 20 0 16+ ≤ is true, shade the side of the circle containing (0, 0). Graph the line 2x y+ = . Use a solid line since
the inequality uses ≥ . Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
y
–5
5
x–5 5
x + y = 2
x2+ y2 = 16
90. Graph the system of inequalities: 2 1
3
y x
x y
≤ −
− ≤
Graph the parabola 2 1y x= − . Use a solid line
since the inequality uses ≤ . Choose a test point not on the parabola, such as (0, 0). Since
20 0 1≤ − is false, shade the opposite side of the parabola from (0, 0). Graph the line 3x y− = . Use a solid line since
the inequality uses ≤ . Choose a test point not on the line, such as (0, 0). Since 0 0 3− ≤ is true, shade the same side of the line as (0, 0). The overlapping region is the solution.
y
–5
5
x–2 8
x – y = 3
y2 = x – 1
91. Graph the system of inequalities: 2
4
y x
xy
≤
≤
Graph the parabola 2y x= . Use a solid line
since the inequality uses ≤ . Choose a test point
not on the parabola, such as (1, 2). Since 22 1≤ is false, shade the opposite side of the parabola from (1, 2). Graph the hyperbola 4xy = . Use a solid line
since the inequality uses ≤ . Choose a test point not on the hyperbola, such as (1, 2). Since 1 2 4⋅ ≤ is true, shade the same side of the hyperbola as (1, 2). The overlapping region is the solution.
Chapter 11: Systems of Equations and Inequalities
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y
–5
5
x
–5
5
xy = 4
y = x2
92. Graph the system of inequalities: 2 2
2 2
1
4
x y
x y
+ ≥
+ ≤
Graph the circle 2 2 1x y+ = . Use a solid line
since the inequality uses ≥ . Choose a test point not on the circle, such as (0, 0). Since
2 20 0 1+ ≥ is false, shade the opposite side of the circle from (0, 0).
Graph the circle 2 2 4x y+ = . Use a solid line
since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since
2 20 0 4+ ≤ is true, shade the same side of the circle as (0, 0). The overlapping region is the solution.
y
–5
5
x–5 5
x2 + y2 = 4
x2 + y2 = 1
93. Maximize 3 4z x y= + subject to 0x ≥ , 0y ≥ ,
3 2 6x y+ ≥ , 8x y+ ≤ . Graph the constraints.
y
x
(0,8)
(8,0)(2,0)
(0,3)
The corner points are (0, 3), (2, 0), (0, 8), (8, 0).
Evaluate the objective function:
Vertex Value of 3 4
(0, 3) 3(0) 4(3) 12
(0, 8) 3(0) 4(8) 32
(2, 0) 3(2) 4(0) 6
(8, 0) 3(8) 4(0) 24
z x y
z
z
z
z
= += + == + == + == + =
The maximum value is 32 at (0, 8).
94. Maximize 2 4z x y= + subject to 0x ≥ , 0y ≥ ,
6x y+ ≤ , 2x ≥ . Graph the constraints.
y
x
(2,4)
(6,0)(2,0)
The corner points are (2, 4), (2, 0), (6, 0). Evaluate the objective function:
Vertex Value of 2 4
(2, 4) 2(2) 4(4) 20
(2, 0) 2(2) 4(0) 4
(6, 0) 2(6) 4(0) 12
z x y
z
z
z
= += + == + == + =
The maximum value is 20 at (2, 4).
95. Minimize 3 5z x y= + subject to 0x ≥ , 0y ≥ ,
1x y+ ≥ , 3 2 12x y+ ≤ , 3 12x y+ ≤ .
Graph the constraints.
y
x(4,0)(1,0)
(0,1)
(0,4)127 , 24
7( )
To find the intersection of 3 2 12 and 3 12x y x y+ = + = , solve the system:
3 2 12
3 12
x y
x y
+ = + =
Solve the second equation for x: 12 3x y= −
Chapter 11 Review Exercises
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Substitute and solve: 3(12 3 ) 2 12
36 9 2 12
7 24
24
7
y y
y y
y
y
− + =− + =
− = −
=
24 72 1212 3 12
7 7 7x
= − = − =
The point of intersection is 12 24
,7 7
.
The corner points are (0, 1), (1, 0), (0, 4), (4, 0), 12 24
,7 7
.
Evaluate the objective function:
Vertex Value of 3 5
(0, 1) 3(0) 5(1) 5
(0, 4) 3(0) 5(4) 20
(1, 0) 3(1) 5(0) 3
(4, 0) 3(4) 5(0) 12
12 24 12 24 156, 3 5
7 7 7 7 7
z x y
z
z
z
z
z
= += + =
= + == + =
= + =
= + =
The minimum value is 3 at (1, 0).
96. Minimize 3z x y= + subject to 0x ≥ , 0y ≥ ,
8x ≤ , 6y ≤ , 2 4x y+ ≥ . Graph the
constraints.
y
x
(0,6)
(8,0)(2,0)
(0,4)
(8,6)
The corner points are (0, 4), (2, 0), (0, 6), (8, 0), (8, 6). Evaluate the objective function:
( )
Vertex Value of 3
(0, 6) 3(0) 6 6
(0, 4) 3(0) 4 4
(2, 0) 3(2) 0 6
(8, 0) 3(8) 0 24
8, 6 3(8) 6 30
z x y
z
z
z
z
z
= += + == + == + == + == + =
The minimum value is 4 at (0, 4).
97. 2 5 5
4 10
x y
x y A
+ = + =
Multiply each side of the first equation by –2 and eliminate x:
4 10 10
4 10
0 10
x y
x y A
A
− − = − + =
= −
If there are to be infinitely many solutions, the result of elimination should be 0 = 0. Therefore,
10 0 or 10A A− = = .
98. 2 5 5
4 10
x y
x y A
+ = + =
Multiply each side of the first equation by –2 and eliminate x:
4 10 10
4 10
0 10
x y
x y A
A
− − = − + =
= −
If the system is to be inconsistent, the result of elimination should be 0 = any number except 0. Therefore, 10 0 or 10A A− ≠ ≠ .
99. 2y ax bx c= + +
At (0, 1) the equation becomes: 21 (0) (0)
1
a b c
c
= + +=
At (1, 0) the equation becomes: 20 (1) (1)
0
0
a b c
a b c
a b c
= + += + +
+ + =
At (–2, 1) the equation becomes: 21 ( 2) ( 2)
1 4 2
4 2 1
a b c
a b c
a b c
= − + − += − +
− + =
The system of equations is: 0
4 2 1
1
a b c
a b c
c
+ + = − + = =
Substitute 1c = into the first and second equations and simplify:
1 0
1
1
a b
a b
a b
+ + =+ = −
= − −
4 2 1 1
4 2 0
a b
a b
− + =− =
Chapter 11: Systems of Equations and Inequalities
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Solve the first equation for a, substitute into the second equation and solve:
4( 1) 2 0
4 4 2 0
6 4
2
3
b b
b b
b
b
− − − =− − − =
− =
= −
2 11
3 3a = − = −
The quadratic function is 21 21
3 3y x x= − − + .
100. 2 2 0x y Dx Ey F+ + + + =
At (0, 1) the equation becomes: 2 20 1 (0) (1) 0
1
D E F
E F
+ + + + =+ = −
At (1, 0) the equation becomes: 2 21 0 (1) (0) 0
1
D E F
D F
+ + + + =+ = −
At (–2, 1) the equation becomes: 2 2( 2) 1 ( 2) (1) 0
2 5
D E F
D E F
− + + − + + =− + + = −
The system of equations is: 1
1
2 5
E F
D F
D E F
+ = − + = −− + + = −
Substitute 1E F+ = − into the third equation and solve for D:
2 ( 1) 5
2 4
2
D
D
D
− + − = −− = −
=
Substitute and solve: 2 1
3
F
F
+ = −= −
( 3) 1
2
E
E
+ − = −=
The equation of the circle is 2 2 2 2 3 0x y x y+ + + − = .
101. Let x = the number of pounds of coffee that costs $6.00 per pound, and let y = the number
of pounds of coffee that costs $9.00 per pound. Then 100x y+ = represents the total amount of
coffee in the blend. The value of the blend will be represented by the equation: 6 9 6.90(100)x y+ = . Solve the system of
equations: 100
6 9 690
x y
x y
+ = + =
Solve the first equation for y: 100y x= − .
Solve by substitution: 6 9(100 ) 690
6 900 9 690
3 210
70
x x
x x
x
x
+ − =+ − =
− = −=
100 70 30y = − =
The blend is made up of 70 pounds of the $6.00-per-pound coffee and 30 pounds of the $9.00-per-pound coffee.
102. Let x = the number of acres of corn, and let y
= the number of acres of soybeans. Then 1000x y+ = represents the total acreage on the
farm. The total cost will be represented by the equation: 65 45 54,325x y+ = . Solve the
system of equations: 1000
65 45 54,325
x y
x y
+ = + =
Solve the first equation for y: 1000y x= −
Solve by substitution: 65 45(1000 ) 54,325
65 45,000 45 54,325
20 9325
466.25
x x
x x
x
x
+ − =+ − =
==
1000 466.25 533.75y = − =
Corn should be planted on 466.25 acres and soybeans should be planted on 533.75 acres.
Chapter 11 Review Exercises
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103. Let x = the number of small boxes, let y = the
number of medium boxes, and let z = the number of large boxes. Oatmeal raisin equation: 2 2 15x y z+ + =
Chocolate chip equation: 2 10x y z+ + =
Shortbread equation: 3 11y z+ =
2 2 15
2 10
3 11
x y z
x y z
y z
+ + = + + = + =
Multiply each side of the second equation by –1 and add to the first equation to eliminate x:
2 2 15
2 10
3 11
5
x y z
x y z
y z
y
+ + =− − − = −
+ ==
Substituting and solving for the other variables: 5 3 11
3 6
2
z
z
z
+ ===
5 2(2) 10
9 10
1
x
x
x
+ + =+ =
=
Thus, 1 small box, 5 medium boxes, and 2 large boxes of cookies should be purchased.
104. a. Let x = the number of lower-priced packages, and let y = the number of quality
packages. Peanut inequality: 8 6 120(16)
4 3 960
x y
x y
+ ≤+ ≤
Cashew inequality: 4 6 72(16)
2 3 576
x y
x y
+ ≤+ ≤
The system of inequalities is:
0
0
4 3 960
2 3 576
x
y
x y
x y
≥ ≥ + ≤ + ≤
b. Graphing:
To find the intersection of 2 3 576x y+ =
and 4 3 960x y+ = , solve the system:
4 3 960
2 3 576
x y
x y
+ = + =
Subtract the second equation from the first:
4 3 960
2 3 576
2 384
192
x y
x y
x
x
+ =− − =
==
Substitute and solve: 2(192) 3 576
3 192
64
y
y
y
+ ===
The corner points are (0, 0), (0, 192), (240, 0), and (192, 64).
105. Let x = the speed of the boat in still water, and let y = the speed of the river current. The
distance from Chiritza to the Flotel Orellana is 100 kilometers.
Rate Time Distance
trip downstream 5 / 2 100
trip downstream 3 100
x y
x y
+−
The system of equations is: 5
( ) 10023( ) 100
x y
x y
+ = − =
Multiply both sides of the first equation by 6, multiply both sides of the second equation by 5, and add the results.
15 15 600
15 15 500
30 1100
x y
x y
x
+ =− =
=
1100 110
30 3x = =
1103 3 100
3
110 3 100
10 3
10
3
y
y
y
y
− =
− ==
=
The speed of the boat is 110 / 3 36.67 km/hr≈ ; the speed of the current is 10 / 3 3.33 km/hr≈ .
Chapter 11: Systems of Equations and Inequalities
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106. Let x = the speed of the jet stream, and let d = the distance from Chicago to Ft. Lauderdale. The jet stream flows from Chicago to Ft. Lauderdale because the time is shorter in that direction.
Rate Time Distance
Chicago to475 5 / 2
Ft. Lauderdale
Ft. Lauderdale475 17 / 6
to Chicago
x d
x d
+
−
The system of equations is:
( )( )( )( )
475 5 / 2
475 17 / 6
y d
y d
+ =
− =
Simplifying the system, we obtain: 2 5 2375
6 17 8075
d x
d x
− = + =
Multiply the first equation by 3− and add the result to the second equation:
6 15 7125
6 17 8075
d x
d x
− + = −+ =
32 950
950 475
32 16
x
x
=
= =
The speed of the jet stream is approximately 475 /16 29.69≈ miles per hour.
107. Let x = the number of hours for Bruce to do the job alone, let y = the number of hours for Bryce
to do the job alone, and let z = the number of hours for Marty to do the job alone. Then 1/x represents the fraction of the job that Bruce does in one hour. 1/y represents the fraction of the job that Bryce does in one hour. 1/z represents the fraction of the job that Marty does in one hour. The equation representing Bruce and Bryce working together is:
( )1 1 1 3
0.754 / 3 4x y
+ = = =
The equation representing Bryce and Marty working together is:
( )1 1 1 5
0.6258 / 5 6y z
+ = = =
The equation representing Bruce and Marty working together is:
( )1 1 1 3
0.3758 / 3 8x z
+ = = =
Solve the system of equations: 1 1
1 1
1 1
0.75
0.625
0.375
x y
y z
x z
− −
− −
− −
+ = + = + =
Let 1 1 1, ,u x v y w z− − −= = =
0.75
0.625
0.375
u v
v w
u w
+ = + = + =
Solve the first equation for u: 0.75u v= − . Solve the second equation for w: 0.625w v= − . Substitute into the third equation and solve: (0.75 ) (0.625 ) 0.375
2 1
0.5
v v
v
v
− + − =− = −
=
0.75 0.5 0.25u = − = 0.625 0.5 0.125w = − =
Solve for , , andx y z : 4, 2, 8 (reciprocals)x y z= = =
Bruce can do the job in 4 hours, Bryce in 2 hours, and Marty in 8 hours.
108. Let x = the number of dancing girls produced, and let y = the number of mermaids produced.
The total profit is: 25 30P x y= + . Profit is to be
maximized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A non-negative number of
figurines must be produced. 3 3 90x y+ ≤ 90 hours are available for
molding. 6 4 120x y+ ≤ 120 hours are available for
painting. 2 3 60x y+ ≤ 60 hours are available for glazing.
Graph the constraints.
y
x
(0,20)
(20,0)(0,0)
(12,12)
Chapter 11 Test
1235
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
To find the intersection of 6 4 120x y+ = and
2 3 60x y+ = , solve the system:
6 4 120
2 3 60
x y
x y
+ = + =
Multiply the second equation by 3, and subtract from the first equation:
6 4 120
6 9 180
x y
x y
+ =− − = −
5 60
12
y
y
− = −=
Substitute and solve: 2 3(12) 60
2 24
12
x
x
x
+ ===
The point of intersection is (12, 12). The corner points are (0, 0), (0, 20), (20, 0), (12, 12). Evaluate the objective function:
Vertex Value of 25 30
(0, 0) 25(0) 30(0) 0
(0, 20) 25(0) 30(20) 600
(20, 0) 25(20) 30(0) 500
(12, 12) 25(12) 30(12) 660
P x y
P
P
P
P
= += + =
= + == + == + =
The maximum profit is $660, when 12 dancing girl and 12 mermaid figurines are produced each day. To determine the excess, evaluate each constraint at 12x = and 12y = :
Molding: 3 3 3(12) 3(12) 36 36 72
Painting: 6 4 6(12) 4(12) 72 48 120
Glazing: 2 3 2(12) 3(12) 24 36 60
x y
x y
x y
+ = + = + =+ = + = + =+ = + = + =
Painting and glazing are at their capacity. Molding has 18 more hours available, since only 72 of the 90 hours are used.
109. Let x = the number of gasoline engines produced each week, and let y = the number of diesel
engines produced each week. The total cost is: 450 550C x y= + . Cost is to be minimized; thus,
this is the objective function. The constraints are: 20 60x≤ ≤ number of gasoline engines
needed and capacity each week. 15 40y≤ ≤ number of diesel engines needed
and capacity each week. 50x y+ ≥ number of engines produced to
prevent layoffs. Graph the constraints.
y
x
(60,40)
(60,15)
(20,40)
(35,15)
(20,30)
The corner points are (20, 30), (20, 40), (35, 15), (60, 15), (60, 40) Evaluate the objective function:
( ) ( ) ( )
Vertex Value of 450 550
(20, 30) 450(20) 550(30) 25,500
(20, 40) 450(35) 550(40) 31,000
(35, 15) 450(35) 550(15) 24,000
(60, 15) 450(60) 550(15) 35,250
60, 40 450 60 550 40 49,000
C x y
C
C
C
C
C
= += + == + == + == + =
= + =
The minimum cost is $24,000, when 35 gasoline engines and 15 diesel engines are produced. The excess capacity is 15 gasoline engines, since only 20 gasoline engines had to be delivered.
110. Answers will vary.
Chapter 11 Test
1. 2 7
4 3 9
x y
x y
− + = − + =
Substitution: We solve the first equation for y, obtaining
2 7y x= −
Next we substitute this result for y in the second equation and solve for x.
( )4 3 9
4 3 2 7 9
4 6 21 9
10 30
303
10
x y
x x
x x
x
x
+ =+ − =
+ − ==
= =
We can now obtain the value for y by letting 3x = in our substitution for y.
Chapter 11: Systems of Equations and Inequalities
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( )2 7
2 3 7 6 7 1
y x
y
= −= − = − = −
The solution of the system is 3x = , 1y = − or
(3, 1)− .
Elimination: Multiply each side of the first equation by 2 so that the coefficients of x in the two equations are negatives of each other. The result is the equivalent system
4 2 14
4 3 9
x y
x y
− + = − + =
We can replace the second equation of this system by the sum of the two equations. The result is the equivalent system
4 2 14
5 5
x y
y
− + = − = −
Now we solve the second equation for y. 5 5
51
5
y
y
= −−= = −
We back-substitute this value for y into the original first equation and solve for x.
( )2 7
2 1 7
2 6
63
2
x y
x
x
x
− + = −− + − = −
− = −−= =−
The solution of the system is 3x = , 1y = − or
(3, 1)− .
2. 1
2 13
5 30 18
x y
x y
− = − =
We choose to use the method of elimination and multiply the first equation by 15− to obtain the equivalent system
5 30 15
5 30 18
x y
x y
− + = − − =
We replace the second equation by the sum of the two equations to obtain the equivalent system
5 30 15
0 3
x y− + = − =
The second equation is a contradiction and has no solution. This means that the system itself has no solution and is therefore inconsistent.
3.
2 5 (1)
3 4 2 (2)
5 2 3 8 (3)
x y z
x y z
x y z
− + = + − = − + + =
We use the method of elimination and begin by eliminating the variable y from equation (2). Multiply each side of equation (1) by 4 and add the result to equation (2). This result becomes our new equation (2). 2 5 4 4 8 20
3 4 2 3 4 2
7 7 18 (2)
x y z x y z
x y z x y z
x z
− + = − + =+ − = − + − = −
+ =
We now eliminate the variable y from equation (3) by multiplying each side of equation (1) by 2 and adding the result to equation (3). The result becomes our new equation (3). 2 5 2 2 4 10
5 2 3 8 5 2 3 8
7 7 18 (3)
x y z x y z
x y z x y z
x z
− + = − + =+ + = + + =
+ =
Our (equivalent) system now looks like 2 5 (1)
7 7 18 (2)
7 7 18 (3)
x y z
x z
x z
− + = + = + =
Treat equations (2) and (3) as a system of two equations containing two variables, and eliminate the x variable by multiplying each side of equation (2) by 1− and adding the result to equation (3). The result becomes our new equation (3). 7 7 18 7 7 18
7 7 18 7 7 18
0 0 (3)
x z x z
x z x z
+ = − − = −+ = + =
=
We now have the equivalent system 2 5 (1)
7 7 18 (2)
0 0 (3)
x y z
x z
− + = + = =
This is equivalent to a system of two equations with three variables. Since one of the equations contains three variables and one contains only two variables, the system will be dependent. There are infinitely many solutions. We solve equation (2) for x and determine that
18
7x z= − + . Substitute this expression into
equation (1) to obtain y in terms of z.
Chapter 11 Test
1237
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
2 5
182 5
7
182 5
717
717
7
x y z
z y z
z y z
y z
y z
− + =
− + − + =
− + − + =
− + =
= −
The solution is 18 17
,7 7
x z y z= − + = − ,
z is any real number or 18
( , , ) ,7
x y z x z = − +
17, is any real number
7y z z
= −
.
4. 23
3 2 8 3 (1)
1 (2)
6 3 15 8 (3)
x y z
x y z
x y z
+ − = − − − + = − + =
We start by clearing the fraction in equation (2) by multiplying both sides of the equation by 3.
3 2 8 3 (1)
3 2 3 3 (2)
6 3 15 8 (3)
x y z
x y z
x y z
+ − = −− − + = − + =
We use the method of elimination and begin by eliminating the variable x from equation (2). The coefficients on x in equations (1) and (2) are negatives of each other so we simply add the two equations together. This result becomes our new equation (2). 3 2 8 3
3 2 3 3
5 0 (2)
x y z
x y z
z
+ − = −− − + =
− =
We now eliminate the variable x from equation (3) by multiplying each side of equation (1) by
2− and adding the result to equation (3). The result becomes our new equation (3). 3 2 8 3 6 4 16 6
6 3 15 8 6 3 15 8
7 31 14 (3)
x y z x y z
x y z x y z
y z
+ − = − − − + =− + = − + =
− + =
Our (equivalent) system now looks like 3 2 8 3 (1)
5 0 (2)7 31 14 (3)
x y zz
y z
+ − = − − = − + =
We solve equation (2) for z by dividing both
sides of the equation by 5− . 5 0
0
z
z
− ==
Back-substitute 0z = into equation (3) and solve for y.
7 31 147 31(0) 14
7 142
y zy
yy
− + =− + =
− == −
Finally, back-substitute 2y = − and 0z = into
equation (1) and solve for x. 3 2 8 3
3 2( 2) 8(0) 33 4 3
3 11
3
x y zx
xx
x
+ − = −+ − − = −
− = −=
=
The solution of the original system is 1
3x = , 2y = − , 0z = or
1, 2, 0
3 −
.
5. 4 5 0
2 6 19
5 5 10
x y z
x y
x y z
− + =− − + = − + − =
We first check the equations to make sure that all variable terms are on the left side of the equation and the constants are on the right side. If a variable is missing, we put it in with a coefficient of 0. Our system can be rewritten as
4 5 0
2 0 25
5 5 10
x y z
x y z
x y z
− + =− − + = − + − =
The augmented matrix is 4 5 1 02 1 0 25
1 5 5 10
− − − −
−
6. The matrix has three rows and represents a system with three equations. The three columns to the left of the vertical bar indicate that the system has three variables. We can let x, y, and z denote these variables. The column to the right of the vertical bar represents the constants on the right side of the equations. The system is
3 2 4 6
1 0 8 2
2 1 3 11
x y z
x y z
x y z
+ + = − + + =− + + = −
or
3 2 4 6
8 2
2 3 11
x y z
x z
x y z
+ + = − + =− + + = −
Chapter 11: Systems of Equations and Inequalities
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7. 1 1 4 6
2 2 0 4 1 3
3 2 1 8
2 2 4 6 6 4
0 8 1 3 1 11
6 4 1 8 5 12
A C
− + = − + − −
− = − + − = − −
8. 1 1 4 6
3 0 4 3 1 3
3 2 1 8
1 1 12 18 11 19
0 4 3 9 3 5
3 2 3 24 6 22
A C
− − = − − − −
− − − = − − − = − − −
9. Here we are taking the product of a 3 2× matrix and a 2 3× matrix. Since the number of columns in the first matrix is the same as the number of rows in the second matrix (2 in both cases), the operation can be performed and will result in a 3 3× matrix.
4 1 6 0 4( 2) 6 3 4 5 6 1
1 1 ( 3)0 1( 2) ( 3)3 1 5 ( 3)1
( 1)1 8 0 ( 1)( 2) 8 3 ( 1)5 8 1
4 10 26
1 11 2
1 26 3
4 61 2 5
1 30 3 1
1 8
CB
⋅ + ⋅ − + ⋅ ⋅ + ⋅
⋅ + − − + − ⋅ + −
− + ⋅ − − + ⋅ − + ⋅
−
−
− = − −
= =
10. Here we are taking the product of a 2 3× matrix and a 3 2× matrix. Since the number of columns in the first matrix is the same as the number of rows in the second matrix (3 in both cases), the operation can be performed and will result in a 2 2× matrix.
( ) ( ) ( ) ( )( ) ( )
1 1 2 0 5 3 1 1 2 4 5 2
0 1 3 0 1 3 0 1 3 4 1 2
1 11 2 5
0 40 3 1
3 2
16 17
3 10
BA
⋅ + − ⋅ + ⋅ ⋅ − + − ⋅ − + ⋅
⋅ + ⋅ + ⋅ ⋅ − + − + ⋅
− − = −
=
= −
11. We first form the matrix
[ ]2
3 2 1 0|
5 4 0 1A I
=
Next we use row operations to transform [ ]2|A I
into reduced row echelon form.
( )1 1
2 113 33
1 03 2 1 0
5 4 0 1 5 4 0 1R r
→ =
( )
( )
( )
2 1 2
2 2
1 2 1
2 13 3
523 3
2 13 3 3
25 32 2
25 3 32 2
1 0 5
0 1
1 0
0 1
1 0 2 1
0 1
R r r
R r
R rr
→ = − +
−
→ = −
− → = − + −
Therefore, 15 32 2
2 1A− −
= − .
12. We first form the matrix
[ ]3
1 1 1 1 0 0
| 2 5 1 0 1 0
2 3 0 0 0 1
B I
− = −
Next we use row operations to transform [ ]3|B I
into reduced row echelon form.
Chapter 11 Test
1239
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( )
2 1 2
3 1 3
2 2
1 2 1
3
3 2 1 17 7 7 7
54 17 7 73 2 17 7 7
51 47 7 7
1 1 1 1 0 0
2 5 1 0 1 0
2 3 0 0 0 1
1 1 1 1 0 02
0 7 3 2 1 0 2
0 5 2 2 0 1
1 1 1 1 0 0
0 1 0
0 5 2 2 0 1
1 0 0
0 1 0
0 0 1
R r r
R r r
R r
R r r
R
− −
− = − + → − − = − + − −
− → − − = − − = + → − −
− − 2 35r r
= − +
( )3 3
1 3 1
32 3 27
54 17 7 73 2 17 7 7
47
1 0 0
0 1 0 7
0 0 1 4 5 7
1 0 0 3 3 4
0 1 0 2 2 3
0 0 1 4 5 7
R r
R r r
R r r
→ − − =
− − − = − + → − − = + − −
Thus, 1
3 3 4
2 2 3
4 5 7
B−
− = − − − −
13. 6 3 12
2 2
x y
x y
+ = − = −
We start by writing the augmented matrix for the system.
6 3 12
2 1 2
− −
Next we use row operations to transform the augmented matrix into row echelon form.
( )
( )
1 2
2 1
11 12
2 1 2
12
12
6 3 12 2 1 2
2 1 2 6 3 12
1 1
6 3 12
1 1 6
0 6 18
R r
R r
R r
R r r
=− − → − − = − −
→ = − −
→ = − +
( )
( )
12 26
12 2 12
12
12
1 1
0 1 3
1 0
0 1 3
R r
R r r
− −→ =
→ = +
The solution of the system is 12x = , 3y = or
( )12 , 3
14. 1
74
8 2 56
x y
x y
+ = + =
We start by writing the augmented matrix for the system.
141 7
8 2 56
Next we use row operations to transform the augmented matrix into row echelon form.
14
2 1 2
14
1 7
88 2 56
1 7
0 0 0
R R r
= − +
The augmented matrix is now in row echelon form. Because the bottom row consists entirely of 0’s, the system actually consists of one equation in two variables. The system is dependent and therefore has an infinite number of solutions. Any ordered pair satisfying the
equation 1
74
x y+ = , or 4 28y x= − + , is a
solution to the system.
15. 2 4 3
2 7 15 12
4 7 13 10
x y z
x y z
x y z
+ + = − + + = − + + = −
We start by writing the augmented matrix for the system.
1 2 4 3
2 7 15 12
4 7 13 10
− − −
Next we use row operations to transform the augmented matrix into row echelon form.
Chapter 11: Systems of Equations and Inequalities
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( )
2 1 2
3 1 3
2 3
3 2
3 2 3
1 2 4 3
2 7 15 12
4 7 13 10
1 2 4 32
0 3 7 6 4
0 1 3 2
1 2 4 3
0 1 3 2
0 3 7 6
1 2 4 3
0 1 3 2 3
0 0 2 0
1 2 4 3
0 1 3 2
0 0 1 0
R r r
R r r
R r
R r
R r r
− − −
− = − + → − = − + − −
− = − → − = −
− → − = − + −
− → −
( )13 32 R r = −
The matrix is now in row echelon form. The last row represents the equation 0z = . Using 0z = we back-substitute into the equation 3 2y z+ = −
(from the second row) and obtain
( )3 2
3 0 2
2
y z
y
y
+ = −+ = −
= −
Using 2y = − and 0z = , we back-substitute into
the equation 2 4 3x y z+ + = − (from the first
row) and obtain
( ) ( )2 4 3
2 2 4 0 3
1
x y z
x
x
+ + = −+ − + = −
=
The solution is 1x = , 2y = − , 0z = or
(1, 2, 0)− .
16. 2 2 3 5
2 8
3 5 8 2
x y z
x y z
x y z
+ − = − + = + − = −
We start by writing the augmented matrix for the system.
2 2 3 5
1 1 2 8
3 5 8 2
− − − −
Next we use row operations to transform the augmented matrix into row echelon form.
( )
1 2
2 1
2 1 2
3 1 3
12 24
7 114 4
7 114 4
2 2 3 5
1 1 2 8
3 5 8 2
1 1 2 8
2 2 3 5
3 5 8 2
1 1 2 82
0 4 7 11 3
0 8 14 26
1 1 2 8
0 1
0 8 14 26
1 1 2 8
0 1
R r
R r
R r r
R r r
R r
− − − −
− = = − = − −
− = − + = − − = − + − −
−
= =− −
− −
−= − − ( )3 2 3 8
0 0 0 4
R r r
= − +
−
The last row represents the equation 0 4= − which is a contradiction. Therefore, the system has no solution and is be inconsistent.
17. ( ) ( ) ( )( )2 52 7 5 3 14 15 29
3 7
−= − − = − − = −
18.
2 4 6
1 4 0
1 2 4
−
− −
[ ] [ ][ ]
4 0 1 0 1 42 ( 4) 6
2 4 1 4 1 2
2 4( 4) 2(0) 4 1( 4) ( 1)(0)
6 1(2) ( 1)4
2( 16) 4( 4) 6(6)
32 16 36
12
= − − +− − − −
= − − + − − −
+ − −= − + − += − − += −
19. 4 3 23
3 5 19
x y
x y
+ = − − =
The determinant D of the coefficients of the variables is
( )( ) ( )( )4 34 5 3 3 20 9 29
3 5D = = − − = − − = −
−
Since 0D ≠ , Cramer’s Rule can be applied.
( )( ) ( )( )23 323 5 3 19 58
19 5xD−
= = − − − =−
Chapter 11 Test
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
( )( ) ( )( )4 234 19 23 3 145
3 19yD−
= = − − =
582
29xD
xD
= = = −−
1455
29yD
yD
= = = −−
The solution of the system is 2x = − , 5y = − or
( 2, 5)− − .
20. 4 3 2 15
2 3 15
5 5 2 18
x y z
x y z
x y z
− + =− + − = − − + =
The determinant D of the coefficients of the variables is
( )
( ) ( ) ( )( ) ( ) ( )
4 3 2
2 1 3
5 5 2
1 3 2 3 2 14 3 2
5 2 5 2 5 5
4 2 15 3 4 15 2 10 5
4 13 3 11 2 5
52 33 10
9
D
−= − −
−
− − − −= − − +
− −
= − + − + + −
= − + += − + += −
Since 0D ≠ , Cramer’s Rule can be applied.
( )
( ) ( ) ( )( ) ( ) ( )
15 3 2
15 1 3
18 5 2
1 3 15 3 15 115 3 2
5 2 18 2 18 5
15 2 15 3 30 54 2 75 18
15 13 3 24 2 57
9
xD
−= − −
−
− − − −= − − +
− −
= − + − + + −
= − + += −
( ) ( ) ( )( ) ( ) ( )
4 15 2
2 15 3
5 18 2
15 3 2 3 2 154 15 2
18 2 5 2 5 18
4 30 54 15 4 15 2 36 75
4 24 15 11 2 39
9
yD = − − −
− − − − − −= − +
= − + − − + + − +
= − += −
( )
( ) ( ) ( )( ) ( ) ( )
4 3 15
2 1 15
5 5 18
1 15 2 15 2 14 3 15
5 18 5 18 5 5
4 18 75 3 36 75 15 10 5
4 57 3 39 15 5
36
zD
−= − −
−
− − − −= − − +
− −
= − + − + + −
= − + += −
9
19
xDx
D
−= = =−
, 9
19
yDy
D= = = −
−,
364
9zD
zD
−= = =−
The solution of the system is 1x = , 1y = − ,
4z = or (1, 1, 4)− .
21. 2 2
2
3 12
9
x y
y x
+ =
=
Substitute 9x for 2y into the first equation and
solve for x:
( )2
2
2
3 9 12
3 9 12 0
3 4 0
( 1)( 4) 0
x x
x x
x x
x x
+ =
+ − =
+ − =− + =
1 or 4x x= = − Back substitute these values into the second equation to determine y:
1x = : 2 9(1) 9
3
y
y
= == ±
4x = − : 2 9( 4) 36
36 (not real)
y
y
= − = −
= ± −
The solutions of the system are (1, 3)− and
(1, 3) .
Chapter 11: Systems of Equations and Inequalities
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22. 2 22 3 5
1 1
y x
y x y x
− =
− = = +
Substitute 1x + for y into the first equation and solve for x:
( )( )
2 2
2 2
2 2
2
2
2 1 3 5
2 2 1 3 5
2 4 2 3 5
4 3 0
4 3 0
( 1)( 3) 0
x x
x x x
x x x
x x
x x
x x
+ − =
+ + − =
+ + − =
− + − =
− + =− − =
1 or 3x x= = Back substitute these values into the second equation to determine y:
1x = : 1 1 2y = + =
3x = : 3 1 4y = + =
The solutions of the system are (1, 2) and
(3, 4) .
23. 2 2 100
4 3 0
x y
x y
+ ≤
− ≥
Graph the circle 2 2 100x y+ = . Use a solid
curve since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since
2 20 0 100+ ≤ is true, shade the same side of the circle as (0, 0); that is, inside the circle.
Graph the line 4 3 0x y− = . Use a solid line since
the inequality uses ≥ . Choose a test point not on the line, such as (0, 1). Since 4(0) 3(1) 0− ≥ is
false, shade the opposite side of the line from (0, 1). The overlapping region is the solution.
24. ( )2
3 7
3
x
x
++
The denominator contains the repeated linear factor 3x + . Thus, the partial fraction decomposition takes on the form
( ) ( )2 2
3 7
33 3
x A B
xx x
+ = +++ +
Clear the fractions by multiplying both sides by
( )23x + . The result is the identity
( )3 7 3x A x B+ = + +
or
( )3 7 3x Ax A B+ = + +
We equate coefficients of like powers of x to obtain the system
3
7 3
A
A B
= = +
Therefore, we have 3A = . Substituting this result into the second equation gives
( )7 3
7 3 3
2
A B
B
B
= += +
− =
Thus, the partial fraction decomposition is
( ) ( )2 2
3 7 3 2
33 3
x
xx x
+ −= +++ +
.
25. ( )
2
22
4 3
3
x
x x
−
+
The denominator contains the linear factor x and
the repeated irreducible quadratic factor 2 3x + . The partial fraction decomposition takes on the form
( ) ( )2
2 2 22 2
4 3
33 3
x A Bx C Dx E
x xx x x
− + += + +++ +
We clear the fractions by multiplying both sides
by ( )22 3x x + to obtain the identity
( ) ( )( ) ( )2 2224 3 3 3x A x x x Bx C x Dx E− = + + + + + +
Collecting like terms yields
( ) ( )( ) ( )
2 4 3 24 3 6 3
3 9
x A B x Cx A B D x
C E x A
− = + + + + +
+ + +
Equating coefficients, we obtain the system
Chapter 11 Test
1243
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
0
0
6 3 4
3 0
9 3
A B
C
A B D
C E
A
+ = = + + = + =
= −
From the last equation we get 1
3A = − .
Substituting this value into the first equation
gives 1
3B = . From the second equation, we
know 0C = . Substituting this value into the fourth equation yields 0E = .
Substituting 1
3A = − and
1
3B = into the third
equation gives us
( ) ( )1 13 36 3 4
2 1 4
5
D
D
D
− + + =
− + + ==
Therefore, the partial fraction decomposition is
( ) ( ) ( )2
2 222 2
1 13 34 3 5
33 3
xx xx xx x x
−− = + +++ +
26. 0
0
2 8
2 3 2
x
y
x y
x y
≥ ≥ + ≥ − ≥
The inequalities 0x ≥ and 0y ≥ require that
the graph be in quadrant I. 2 8
14
2
x y
y x
+ ≥
≥ − +
Test the point ( )0,0 .
( )2 8
0 2 0 8 ?
0 8 false
x y+ ≥+ ≥≥
The point ( )0,0 is not a solution. Thus, the
graph of the inequality 2 8x y+ ≥ includes the
half-plane above the line 1
42
y x= − + . Because
the inequality is non-strict, the line is also part of the graph of the solution.
2 3 2
2 2
3 3
x y
y x
− ≥
≤ −
Test the point ( )0,0 .
( ) ( )2 3 2
2 0 3 0 2 ?
0 2 false
x y− ≥− ≥
≥
The point ( )0,0 is not a solution. Thus, the
graph of the inequality 2 3 2x y− ≥ includes the
half-plane below the line 2 2
3 3y x= − .
Because the inequality is non-strict, the line is also part of the graph of the solution. The overlapping shaded region (that is, the shaded region in the graph below) is the solution to the system of linear inequalities.
The graph is unbounded. The corner points are ( )4, 2 and ( )8,0 .
27. The objective function is 5 8z x y= + . We seek
the largest value of z that can occur if x and y are solutions of the system of linear inequalities
0
2 8
3 3
x
x y
x y
≥ + ≤ − ≤ −
2 8
2 8
x y
y x
+ == − +
3 3
3 3
11
3
x y
y x
y x
− = −− = − −
= +
The graph of this system (the feasible points) is shown as the shaded region in the figure below. The corner points of the feasible region are
( )0,1 , ( )3, 2 , and ( )0,8 .
Chapter 11: Systems of Equations and Inequalities
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4
8
y
x4 8
2 8x y+ =
3 3x y− = −
(0, 1)(3, 2)
(0, 8)
( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )
Corner point, , Value of obj. function,
0,1 5 0 8 1 8
3,2 5 3 8 2 31
0,8 5 0 8 8 64
x y z
z
z
z
= + == + == + =
From the table, we can see that the maximum value of z is 64, and it occurs at the point ( )0,8 .
28. Let j = unit price for flare jeans, c = unit price for camisoles, and t = unit price for t-shirts. The given information yields a system of equations with each of the three women yielding an equation.
2 2 4 90 (Megan)
3 42.5 (Paige)
3 2 62 (Kara)
j c t
j t
j c t
+ + = + = + + =
We can solve this system by using matrices.
( )11 12
2 2 4 90 1 1 2 451 0 3 42.5 1 0 3 42.5 1 3 2 62 1 3 2 62
R r = =
( )
( )
2 1 2
3 1 3
2 2
1 2 1
3 2 3
3 312
1 1 2 450 1 1 2.5 0 2 0 17
1 1 2 450 1 1 2.5 0 2 0 17
1 0 3 42.50 1 1 2.5
20 0 2 12
1 0 3 42.50 1 1 2.5 0 0 1 6
R r r
R r r
R r
R r r
R r r
R r
= − + = − − = − + = − = − = − + = − = − + = − =
The last row represents the equation 6z = . Substituting this result into 2.5y z− = (from the
second row) gives 2.5
6 2.5
8.5
y z
y
y
− =− =
=
Substituting 6z = into 3 42.5x z+ = (from the first row) gives
( )3 42.5
3 6 42.5
24.5
x z
x
x
+ =+ =
=
Thus, flare jeans cost $24.50, camisoles cost $8.50, and t-shirts cost $6.00.
Chapter 11 Cumulative Review
1. 22 0x x− =
( )2 1 0
0 or 2 1 0
2 1
1
2
x x
x x
x
x
− == − =
=
=
The solution set is 1
0,2
.
2. 3 1 4x + =
( )223 1 4
3 1 16
3 15
5
x
x
x
x
+ =
+ ===
Check:
3 5 1 4
15 1 4
16 4
4 4
⋅ + =
+ =
==
The solution set is { }5 .
3. 3 22 3 8 3 0x x x− − − =
The graph of 3 21 2 3 8 3Y x x x= − − − appears to
have an x-intercept at 3x = .
Using synthetic division:
3 2 3 8 3
6 9 3
2 3 1 0
− − −
Chapter 11 Cumulative Review
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Therefore,
( )( )( )( )( )
3 2
2
2 3 8 3 0
3 2 3 1 0
3 2 1 1 0
13 or or 1
2
x x x
x x x
x x x
x x x
− − − =
− + + =
− + + =
= = − = −
The solution set is 1
1, ,32
− −
.
4. 13 9x x+=
( ) 12
2 2
3 3
3 3
2 2
2
xx
x x
x x
x
+
+
=
== += −
The solution set is { }2− .
5. ( ) ( )3 3log 1 log 2 1 2x x− + + =
( )( )( )( )( )
( ) ( )
3
2
2
2
log 1 2 1 2
1 2 1 3
2 1 9
2 10 0
2 5 2 0
5 or 2
2
x x
x x
x x
x x
x x
x x
− + =
− + =
− − =
− − =− + =
= = −
Since 2x = − makes the original logarithms
undefined, the solution set is 5
2
.
6. 3x e=
( )ln 3 ln
ln 3 1
10.910
ln 3
x e
x
x
=
=
= ≈
The solution set is 1
0.910ln 3
≈
.
7. 3
4
2( )
1
xg x
x=
+
( )( )
( )3 3
4 4
2 2( )
11
x xg x g x
xx
− −− = = = −+− +
Thus, g is an odd function and its graph is
symmetric with respect to the origin.
8. 2 2 2 4 11 0x y x y+ − + − = 2 2
2 2
2 2
2 4 11
( 2 1) ( 4 4) 11 1 4
( 1) ( 2) 16
x x y y
x x y y
x y
− + + =− + + + + = + +
− + + =
Center: (1,–2); Radius: 4
9. 2( ) 3 1xf x −= +
Using the graph of 3xy = , shift the graph
horizontally 2 units to the right, then shift the graph vertically upward 1 unit.
Domain: ( , )−∞ ∞ Range: (1, )∞
Horizontal Asymptote: 1y =
10. 5
( )2
f xx
=+
5
25
Inverse2
( 2) 52 5
5 25 2 5
2
yx
xy
x yxy x
xy xx
yx x
=+
=+
+ =+ =
= −−= = −
Thus, 1 5( ) 2f x
x− = −
Domain of f = { | 2}x x ≠ −
Range of f = { | 0}y y ≠
Domain of 1f − = { | 0}x x ≠
Range of 1f − = { | 2}y y ≠ − .
Chapter 11: Systems of Equations and Inequalities
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11. a. 3 6y x= +
The graph is a line. x-intercept: y-intercept:
0 3 6
3 6
2
x
x
x
= += −= −
( )3 0 6
6
y = +=
b. 2 2 4x y+ =
The graph is a circle with center (0, 0) and radius 2.
c. 3y x=
d. 1
yx
=
e. y x=
f. xy e=
g. lny x=
Chapter 11 Cumulative Review
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h. 2 22 5 1x y+ =
The graph is an ellipse. 2 2
2 2
1 12 5
2 52 5
1
1
x y
x y
+ =
+ =
i. 2 23 1x y− =
The graph is a hyperbola
2 2
2 2
13
11 3
3
11
x y
x y
− =
− =
j. 2
2
2
2
2 4 1 0
2 1 4
4 ( 1)
1( 1)
4
x x y
x x y
y x
y x
− − + =
− + =
= −
= −
12. 3( ) 3 5f x x x= − +
a. Let 31 3 5Y x x= − + .
−3
−6
9
6
The zero of f is approximately 2.28− .
b.
−3
−6
9
6
−3
−6
9
6
f has a local maximum of 7 at 1x = − and a local minimum of 3 at 1x = .
c. f is increasing on the intervals ( , 1)−∞ − and
(1, )∞ .
Chapter 11: Systems of Equations and Inequalities
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Chapter 11 Projects
Project I – Internet Based Project
1. 80% = 0.80 18% = 0.18 2% = 0.02 40% = 0.40 50% = 0.50 10% = 0.10 20% = 0.20 60% = 0.60 20% = 0.20
2.
0.80 0.18 0.02
0.40 0.50 0.10
0.20 0.60 0.20
3. 0.80 0.18 0.02 1.00
0.40 0.50 0.10 1.00
0.20 0.60 0.20 1.00
+ + =+ + =+ + =
The sum of each row is 1 (or 100%). These represent the three possibilities of educational achievement for a parent of a child, unless someone does not attend school at all. Since these are rounded percents, chances are the other possibilities are negligible.
4.
2
2
0.8 0.18 0.02
0.4 0.5 0.1
0.2 0.6 0.2
0.716 0.246 0.038
0.54 0.382 0.078
0.44 0.456 0.104
P
= =
Grandchild of a college graduate is a college graduate: entry (1, 1): 0.716. The probability is 71.6%
5. Grandchild of a high school graduate finishes college: entry (2,1): 0.54. The probability is 54%.
6. grandchildren → k = 2. (2) (0) 2
0.716 0.246 0.038
[0.288 0.569 0.143] 0.54 0.382 0.078
0.44 0.456 0.104
[0.576388 0.353414 0.070198]
v v P=
=
=College: 57.6%≈ High School: 35%≈ Elementary: 7%≈
7. The matrix totally stops changing at
30
0.64885496 0.29770992 0.05343511
0.64885496 0.29770992 0.05343511
0.64885496 0.29770992 0.05343511
P
≈
Project II
a. 2 2 2 2 16× × × = codewords.
b. v uG= u will be the matrix representing all of the 4-digit information bit sequences.
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1
u
=
(Remember, this is mod two. That means that you only write down the remainder when dividing by 2. )
Chapter 11 Projects
1249
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
0 0 0 0 0 0 0
0 0 0 1 1 1 0
0 0 1 0 1 0 1
0 0 1 1 0 1 1
0 1 0 0 0 1 1
0 1 0 1 1 0 1
0 1 1 0 1 1 0
0 1 1 1 0 0 0
1 0 0 0 1 1 1
1 0 0 1 0 0 1
1 0 1 0 0 1 0
1 0 1 1 1 0 0
1 1 0 0 1 0 0
1 1 0 1 0 1 0
1 1 1 0 0 0 1
1 1 1 1 1 1 1
v uG
v
=
=
c. Answers will vary, but if we choose the 6th row and the 10th row: 0101101 1001001 1102102 → 1100100 (13th row)
d. v uG
VH uGH
==
0 0 0
0 0 0
0 0 0
0 0 0
GH
=
e.
1 1 1
0 1 1
1 0 1
[0 1 0 1 0 0 0] 1 1 0
1 0 0
0 1 0
0 0 1
[1 0 1]
rH
=
=
error code: 0010 000 r : 0101 000 0111 000 This is in the codeword list.
Project III
a. 1 1 1 1 1 1 1
3 4 5 6 7 8 9TA
=
b. 1( )
2.357
2.0357
T TB A A A Y
B
−=−
=
c. 2.0357 2.357y x= −
d. 2.0357 2.357y x= −
Project IV
Answers will vary.