chapter system of equations and inequalities 5 solutions key to all chapter 5...system of equations...
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System of Equations and InequalitiesSolutions Key
ArE you rEAdy?
1. B 2. F
3. A 4. E
5. D
6. Plot (0, 1). Count 3 units up and 4 units right and plot another point. Draw the line connecting the two points.
2
-2
x
y
0 2
7. Plot (0, 5). Count 3 units down and 1 unit right and plot another point. Draw the line connecting the two points.
2
4
x
y
0 4
8. Plot (0, -6). Count 1 unit up and 1 unit right and plot another point. Draw the line connecting the two points.
-2
-4
-6
x y
0 2 4
9. x + y = 4 ______ -x ___ -x y = -x + 4 Plot (0, 4). Count 1 unit down and 1 unit right and
plot another point. Draw the line connecting the two points.
2
4
x
y
0 2 4
10. Plot (0, 4). Count 2 units down and 3 units right and plot another point. Draw the line connecting the two points.
2
x
y
0 2 4 6
11. Plot (0, -5). Count 0 units up and 1 unit right and plot another point. Draw the line connecting the two points.
-2
-4
x y
0 2 -2
12. -7x - 18 = 3 ________ + 18 ____ +18 -7x = 21
-7x ____ -7
= 21 ___ -7
x = -3
13. 12 = -3n + 6 ___ -6 _______ - 6 6 = -3n
6 ___ -3
= -3n ____ -3
-2 = n
14. 1 __ 2 d + 30 = 32
_______ - 30 ____ -30
1 __ 2 d = 2
2 ( 1 __ 2 d) = 2(2)
d = 4
15. -2p + 9 = -3 _______ - 9 ___ -9 -2p = -12
-2p
____ -2
= -12 ____ -2
p = 6
16. 33 = 5y + 8 ___ -8 ______ - 8 25 = 5y
25 ___ 5 =
5y ___
5
5 = y
17. -3 + 3x = 27 _______ +3 ___ +3 3x = 30
3x ___ 3
= 30 ___ 3
x = 10
18. 7x + y = 4 _______ -7x ____ -7x y = -7x + 4
19. y + 2 = -4x ____ - 2 ____ -2 y = -4x - 2
20. 8 = x - y ___ + y _____ + y 8 + y = x ______ -8 ___ -8 y = x - 8
21. x + 2 = y - 5 ____ + 5 _____ + 5 x + 7 = y y = x + 7
22. 2y - 3 = 12x _____ + 3 ___ +3 2y = 12x + 3
2y
___ 2 = 12x + 3 _______
2
y = 6x + 3 __ 2
23. y + 3 __ 4
x = 4
______
- 3 __ 4
x ____
- 3 __ 4
x
y = - 3 __ 4
x + 4
24. t - 5 = 7 - 5 = 2
25. 9 - 2a = 9 - 2(4) = 9 - 8 = 1
26. 1 __ 2 x - 2
= 1 __ 2 (14) - 2
= 7 - 2 = 5
27. n + 15 = 37 + 15 = 52
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28. 9c + 4
= 9 ( 1 __ 3 ) + 4
= 3 + 4 = 7
29. 16 + 3d = 16 + 3(5) = 16 + 15 = 31
30. b - 9 ≥ 1 _____ + 9 ___ +9 b ≥ 10
8 4 0 12 16 20 -4
31. -2x < 10
-2x ____ -2
> 10 ___ -2
x > -5
0 -4 -3 -2 -1 -5 -6
32. 3y ≤ -3
3y
___ 3 ≤ -3 ___
3
y ≤ -1
0 2 -6 -4 -2 -8 -10
33. 1 __ 3 y ≤ 5
3 ( 1 __ 3 y) ≤ 3(5)
y ≤ 15
5 0 10 15 20 -5 -10
SolvIng SyStEmS by grAphIng
CHECK IT OUT!
1a. _____________ 2x + y = 5 2(1) + (3) 5 2 + 3 5 5 5 3
_______________ -2x + y = 1 -2(1) + (3) 1 -2 + 3 1 1 1 3
(1, 3) is a solution of the system.
b. _____________ x - 2y = 4 2 - 2(-1) 4 2 + 2 4 4 4 3
_____________ 3x + y = 6 3(2) - (1) 6 6 - 1 6 5 6 7
(2, -1) is not a solution of the system.
2a. 7
-2
x
y
0 3 -8
y = - 2x - 1 y = x + 5
(-2, 3)
The solution appears to be at (-2, 3).
b. 2x + y = 4 _______ -2x ____ -2x y = -2x + 4
4
-2
x
y
0 12 -4
y = -2x + 4
(3, -2)
y = 1 _ 3
x - 3
The solution appears to be at (3, -2).
3. Let x represent the number of movie rentals and y represent the total cost.
Video club A: y = 3x + 10 Video club B: y = 2x + 15 Graph y = 3x + 10 and y = 2x + 15.
(5, 25)
Cost of Movie Rentals
0 1 3 5 2 4 6 7
5 10 15 20 25 30
Movies
Cost
( $) Club B
Club A
The lines appear to intersect at (5, 25). So, the cost at both places will be the same for 5 movie rentals and that cost will be $25.
THINK AND DISCUSS
1. Locate the point of intersection. The ordered pair is a solution of the system.
2. Substitute the x- and y-values of the ordered pair into all of the equations. If all equations are true, the solution is correct.
3.
1. Graph the first and second equation.
2. Identify the point of intersection.
3. Check the solution.
Solving a Linear System by Graphing
ExErCISESGUIDED PRACTICE
1. an ordered pair that satisfies both equations
2. ______________ 3x + y = 4 3(2) + (-2) 4 6 - 2 4 4 4 3
______________ x - 3y = 4 (2) - 3(-2) 4 2 + 6 4 8 4 7
(2, -2) is not a solution of the system.
3. ______________ x - 2y = 5 (3) - 2(-1) 5 3 + 2 5 5 5 3
______________ 2x - y = 7 2(3) - (-1) 7 6 + 1 7 7 7 3
(3, -1) is a solution of the system.
4. _______________ -x + y = 6 -(-1) + (5) 6 1 + 5 6 6 6 3
________________ 2x + 3y = 13 2(-1) + 3(5) 13 -2 + 15 13 13 13 3
(-1, 5) is a solution of the system.
5. 5
-3
x
y
0 6
y = -x + 3
(2, 1)
y = 1 _ 2 x
The solution appears to be at (2, 1).
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6. 2x + y = 1 _______ -2x ____ -2x y = -2x + 1
-5
x
y
0 6
y = -2x + 1
y = x - 2
(1, -1) -2
The solution appears to be at (1, -1).
7. x + y = 3 ______ -x ___ -x y = -x + 3
9
x
y
0 1 -9
y = -x + 3
y = -2x - 1
(-4, 7)
The solution appears to be at (-4, 7).
8. Let x represent the number of cubic yards and let y represent the total cost.
Lawn and Garden: y = 30x + 30 Yard Depot: y = 25x + 55 Graph y = 30x + 30 and y = 25x + 55. Mulch Delivery Charges
0 1 3 5 2 4 6
40 80
120 160 200
Yard Depot
Lawn and Garden
Mulch (yd3)
Cost
( $) (5, 180)
The lines appear to intersect at (5, 180). So, the cost at both places will be the same for 5 cubic yards and that cost will be $180.
PRACTICE AND PROBLEM SOLVING
9. ______________ x - 2y = 8 (1) - 2(-4) 8 1 + 8 8 9 8 7
______________ 4x - y = 8 4(1) - (-4) 8 4 + 4 8 8 8 3
(1, -4) is not a solution of the system.
10. _________________ 2x - 3y = -7 2(-2) - 3(1) -7 -4 - 3 -7 -7 -7 3
________________ 3x + y = -5 3(-2) + (1) -5 -6 + 1 -5 -5 -5 3
(-2, 1) is a solution of the system.
11. ______________ 2x + y = 12 2(5) + (2) 12 10 + 2 12 12 12 3
_________________ -3y - x = -11 -3(2) - (5) -11 -6 - 5 -11 -11 -11 3
(5, 2) is a solution of the system.
12. 5
-4
x
y
0 1 -8
y = -x - 1
(-2, 1)
y = 1 _ 2 x + 2
The solution appears to be at (-2, 1).
13.
5
x
y
0 6
y = x y = -x + 6
(3, 3)
The solution appears to be at (3, 3).
14. x = -y + 3 ___ + y ______ +y x + y = 3 ______ -x ___ -x y = -x + 3
9
x
y
0 1 -9
y = -x + 3
y = -2x - 1
(-4, 7)
The solution appears to be at (-4, 7).
15. x + y = 2 _______ -x ___ -x y = -x + 2
2
-4
x
y
0 6 -2
(3, -1)
The solution appears to be at (3, -1).
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16. Let x represent the number of weeks and let y represent the distance run each week.
Angelo: y = x + 7 Marc: y = 2x + 4 Graph y = x + 7 and y = 2x + 4. Running Distance
0 1 3 5 2 4
3
6
9
12
Weeks
Dis
tanc
e ( m
iles)
Marc
Angelo
The lines appear to intersect at (3, 10). So the distance run by Angelo and Marc will be the same in 3 weeks and that distance will be 10 mi.
17a. Let x represent the number of carnations and y represent the amount of money.
Selling: y = 2x Buying: y = 0.50x + 16
⎧
⎨
⎩ y = 2x
y = 0.50x + 16
b. Carnation Sales
0 4 8 12
8
16
24
Carnations
Cost
($) Florist’s price
School band’s price
The solution represents the number of carnations that need to be sold for the band will break even.
c. The solution appears to be at (10.6, 21.3); no; you cannot sell part of a carnation; 11 carnations.
18a. Let h represent the number of hats and let c represent the total cost.
Hats Off: c = 5h + 50 Top Stuff: c = 6h + 25
b. Hat Charges
0 8 16 24
75
150
Hats
Cost
of h
ats
($)
Hats Off
Top Stuff
25 hats; $175
c. fewer than 25 hats: Top Stuff; more than 25 hats: Hats Off
19. (-2.4, -9.3)
20. 4.8x + 0.6y = 4 ___________ -4.8x _____ -4.8x 0.6y = -4.8x + 4
0.6y
____ 0.6
= -4.8x + 4 _________ 0.6
y = -8x + 20 ___ 3
(0.8, 0.1)
21. 8 __ 3 x + y = 5 __
9
________
- 8 __ 3 x
____ - 8 __
3 x
y = - 8 __ 3 x + 5 __
9
(0.3, -0.3)
22. (-1.6, 1.3)
23. Let x represent the number of white hydrangeas in the first bed and let y represent the number of pink hydrangeas in the second bed.
First bed: x + y = 45 Second bed: 2x + 3y = 120 Graph x + y = 45 and 2x + 3y = 120. x + y = 45 ______ -x ___ -x y = -x + 45
2x + 3y = 120 ________ -2x ____ -2x 3y = -2x + 120
3y
___ 3 = -2x + 120 _________
3
y = - 2 __ 3
x + 40
Number of Hydrangeas
0 5 10 15
10 20 30 40 50 60
White Hydrangeas
Pink
Hyd
rang
eas
Second Bed
First Bed (15, 30)
The lines appear to intersect at (15, 30). So 15 white hydrangeas and 30 pink hydrangeas should be planted in the first bed. The gardener should buy 15 + 2(15) = 15 + 30 = 45 white hydrangeas and 30 + 3(30) = 30 + 90 = 120 pink hydrangeas.
24. Morning: y = 5x + 200 Afternoon: y = 11x Graph y = 5x + 200 and y = 11x.
Burning Calories
0 30 10 20
80 160 240 320 400
Time (min)
Calo
ries
(33.3, 366.7)
Morning
Afternoon
The lines appear to intersect at (33.3, 366.7). So, Rusty shoud jog for at least 34 min to burn off more calories in the afternoon.
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25. Let x represent the number of years and y represent the height.
First tree: y = x + 2 Second tree: y = 0.5x + 6 Graph y = x + 2 and y = 0.5x + 6
Height of Trees
0 12 4 8 10 2 6
2 4 6 8
10
Years
Hei
ght
( ft)
12 14
(8, 10) Second Tree
First Tree
The lines appear to intersect at (8, 10). So, the trees will be the same height in 8 years.
26. Possible answer: Store A rents carpet cleaners for a fee of $10, plus $3 per day. Store B rents carpet cleaners for a fee of $20, plus $5 per day. For how many days rental will the costs be the same?
27. The coordinates of the intersection point of a system give an ordered pair that satisfies both equations.
TEST PREP
28. B; If x represents the number of miles and y represents the total cost, the cost for Taxi company A is y = 0.50x + 4 and the cost for Taxi company B is y = 0.25x + 5, so B is correct.
29. H; One line has a positive slope and a y-intercept of 1, and the other has a positive slope and a y-intercept of -3. The only system of equations with two such lines is H.
30. Check (2, 6) is on y = 2x + 2. ___________ y = 2x + 2 (6) 2(2) + 2 6 4 + 2 6 6 3
Solve for b. y = 2.5x + b (6) = 2.5(2) + b 6 = 5 + b ___ -5 ______ -5 1 = b
ChALLENGE AND ExTEND
31. Let x represent the number of months after month 1 and let y represent the number sold.
VCRs: y = -10x + 500 DVD Players: y = 15x + 250 Graph y = -10x + 500 and y = 15x + 250.
VCR & DVD Sales
0 12 4 8 10 2 6
100 200 300 400 500
Months
Num
ber
Sold
(10, 400)
VCR
DVD
The solution appears to be at (10, 400). So, the total number sold will be the same for both items in month 10 + 1 = 11 and that will be 400 of each item.
32a. Let x represent the number of minutes and y represent the cost.
Long Distance Inc.: y = 0.03x + 1.45 Far Away Calls: y = 0.02x + 1.52 Graph y = 0.03x + 1.45 and y = 0.02x + 1.45. Cost of Long Distance Calls
0 3 6 9
1.4
1.5
1.6
1.7
Minutes
Cost
($)
Far Away Calls
Long Distance Inc.
(7, 1.66)
The solution appears to be at (7, 1.66). So, the cost for both companies will be the same for 7 min and that cost will be $1.66.
b. It is better to use Long Distance Inc. if the call is under 7 minutes because it costs less. If the call is over 7 minutes, it is better to use Far Away Calls because it costs less.
c. The line will intersect the y-axis at 1.5. Far Away Calls is cheaper because it costs less per minute.
SolvIng SyStEmS by SubStItutIon
CHECK IT OUT!
1a. y = x + 3 2x + 5 = x + 3 ______ -x ______ -x x + 5 = 3 _____ - 5 ___ -5 x = -2
y = x + 3 y = -2 + 3 y = 1 (-2, 1)
b. x + 8y = 16 (2y - 4) + 8y = 16 10y - 4 = 16 ______ + 4 ___ +4 10y = 20
10y
____ 10
= 20 ___ 10
y = 2
x = 2y - 4 x = 2(2) - 4 x = 4 - 4 x = 0 (0, 2)
c. 2x + y = -4 _______ -2x ____ -2x y = -2x - 4
x + y = -7 x + (-2x - 4) = -7 -x - 4 = -7 ______ + 4 ___ +4 -x = -3 -1(-x) = -1(-3) x = 3
x + y = -7 3 + y = -7 ______ -3 ___ -3 y = -10 (3, -10)
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2. -2x + y = 8 _______ +2x ____ +2x y = 2x + 8
3x + 2y = 9 3x + 2(2x + 8) = 9 3x + 2(2x) + 2(8) = 9 3x + 4x + 16 = 9 7x + 16 = 9 _______ - 16 ____ -16 7x = -7
7x ___ 7 = -7 ___
7
x = -1
-2x + y = 8 -2(-1) + y = 8 2 + y = 8 ______ -2 ___ -2 y = 6 (-1, 6)
3a. Let m represent the number of months and let t represent the total cost.
Option 1: t = 60 + 80m Option 2: t = 160 + 70m
t = 60 + 80m 160 + 70m = 60 + 80m _________ - 70m ________ - 70m 160 = 60 + 10m ____ -60 __________ -60 100 = 10m
100 ____ 10
= 10m ____ 10
10 = m
t = 60 + 80m = 60 + 80(10) = 60 + 800 = 860 In 10 months, the total cost for each option will be
the same, $860.
b. The first option; the first option is cheaper for the first 9 months; the second option is cheaper after 10 months.
THINK AND DISCUSS
1. at the point (5, 10)
2. The solution for each is (5, 3).
x + y = 8 x = - y + 8
( - y + 8) - y = 2 - 2 y + 8 = 2
- 2 y = - 6 y = 3
x + 3 = 8 x = 5
(5, 3)
x + y = 8 y = - x + 8
x - ( - x + 8) = 2 x + x - 8 = 2
2 x - 8 = 2 2 x = 10
x = 5
5 + y = 8 y = 3
(5, 3)
x - y = 2 x = y + 2
( y + 2) + y = 8 2 y + 2 = 8
2 y = 6 y = 3
x - 3 = 2 x = 5
(5, 3)
x - y = 2 - y = - x + 2
y = x - 2
x + ( x - 2) = 8 2 x - 2 = 8
2 x = 10 x = 5
5 - y = 2 - y = - 3
y = 3 (5, 3)
Solve x + y = 8 for x. Solve x + y = 8 for y.
Solve x - y = 2 for x. Solve x - y = 2 for y.
x + y = 8x - y = 2
ExErCISESGUIDED PRACTICE
1. y = 5x - 10 3x + 8 = 5x - 10 _______ -3x ________ -3x 8 = 2x - 10 ____ +10 _______ + 10 18 = 2x
18 ___ 2 = 2x ___
2
9 = x
y = 5x - 10 y = 5(9) - 10 y = 45 - 10 y = 35 (9, 35)
2. 3x + y = 2 _______ -3x ____ -3x y = -3x + 2
4x + y = 20 4x + (-3x + 2) = 20 x + 2 = 20 _____ - 2 ___ -2 x = 18
3x + y = 2 3(18) + y = 2 54 + y = 2 ________ -54 ____ -54 y = -52 (18, -52)
3. 4x + y = 20 4x + (x + 5) = 20 5x + 5 = 20 ______ - 5 ___ -5 5x = 15
5x ___ 5 = 15 ___
5
x = 3
y = x + 5 y = 3 + 5 y = 8 (3, 8)
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4. x - 2y = 10 _____ + 2y ____ +2y x = 2y + 10
1 __ 2 x - 2y = 4
1 __ 2 (2y + 10) - 2y = 4
1 __ 2 (2y) + 1 __
2 (10) - 2y = 4
y + 5 - 2y = 4 5 - y = 4 ______ -5 ___ -5 -y = -1 -(-y) = -1(-1) y = 1
x - 2y = 10 x - 2(1) = 10 x - 2 = 10 _____ + 2 ___ +2 x = 12 (12, 1)
5. y - 4x = 3 _____ + 4x ____ +4x y = 4x + 3
2x - 3y = 21 2x - 3(4x + 3) = 21 2x - 3(4x) - 3(3) = 21 2x - 12x - 9 = 21 -10x - 9 = 21 ________ + 9 ___ +9 -10x = 30
-10x _____ -10
= 30 ____ -10
x = -3
y - 4x = 3 y - 4(-3) = 3 y + 12 = 3 ______ - 12 ____ -12 y = -9 (-3, -9)
6. -x - y = 0 -(y - 8) - y = 0 -(y) - (-8) - y = 0 -y + 8 - y = 0 8 - 2y = 0 _______ -8 ___ -8 -2y = -8
-2y
____ -2
= -8 ___ -2
y = 4
x = y - 8 x = 4 - 8 x = -4 (-4, 4)
7a. Let m represent the number of months and let t represent the total cost.
Green Lawn: t = 49 + 29m Grass Team: t = 25 + 37m
t = 49 + 29m 25 + 37m = 49 + 29m ________ - 29m ________ - 29m 25 + 8m = 49 _________ -25 ____ -25 8m = 24
8m ___ 8 = 24 ___
8
m = 3
t = 49 + 29m = 49 + 29(3) = 49 + 87 = 136 In 3 months, the total cost for each company will be
the same, $136.
b. Green Lawn; for 6 months, Green Lawn’s service costs only 49 + 29(6) = 49 + 174 = $223, while Grass Team’s costs 25 + 37(6) = 25 + 222 = $247.
PRACTICE AND PROBLEM SOLVING
8. y = x + 3 2x + 4 = x + 3 ______ -x ______ -x x + 4 = 3 _____ - 4 ___ -4 x = -1
y = x + 3 y = -1 + 3 y = 2 (-1, 2)
9. y = 2x + 10 -2x - 6 = 2x + 10 _______ +2x ________ +2x -6 = 4x + 10 ____ -10 _______ - 10 -16 = 4x
-16 ____ 4
= 4x ___ 4
-4 = x
y = 2x + 10 y = 2(-4) + 10 y = -8 + 10 y = 2 (-4, 2)
10. x + 2y = 8 _____ - 2y ____ -2y x = -2y + 8
x + 3y = 12 (-2y + 8) + 3y = 12 y + 8 = 12 _____ - 8 ___ -8 y = 4
x + 2y = 8 x + 2(4) = 8 x + 8 = 8 _____ - 8 ___ -8 x = 0 (0, 4)
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11. 2x + 2y = 2 ______ - 2y ____ -2y 2x = -2y + 2
2x ___ 2 =
-2y + 2 _______
2
x = -y + 1
-4x + 4y = 12 -4(-y + 1) + 4y = 12 -4(-y) - 4(1) + 4y = 12 4y - 4 + 4y = 12 8y - 4 = 12 ______ + 4 ___ +4 8y = 16
8y
___ 8 = 16 ___
8
y = 2
2x + 2y = 2 2x + 2(2) = 2 2x + 4 = 2 ______ - 4 ___ -4 2x = -2
2x ___ 2
= -2 ___ 2
x = -1 (-1, 2)
12. -y = -2x + 4 -(0.5x + 2) = -2x + 4 -(0.5x) - (2) = -2x + 4 -0.5x - 2 = -2x + 4 _________ +2x _______ +2x 1.5x - 2 = 4 _______ + 2 ___ +2 1.5x = 6
1.5x ____ 1.5
= 6 ___ 1.5
x = 4
y = 0.5x + 2 y = 0.5(4) + 2 y = 2 + 2 y = 4 (4, 4)
13. -x + y = 4 ______ +x ___ +x y = x + 4
3x - 2y = -7 3x - 2(x + 4) = -7 3x - 2(x) - 2(4) = -7 3x - 2x - 8 = -7 x - 8 = -7 _____ + 8 ___ +8 x = 1
-x + y = 4 -(1) + y = 4 -1 + y = 4 ______ +1 ___ +1 y = 5 (1, 5)
14. 3x + y = -8 _______ -3x ____ -3x y = -3x - 8
-2x - y = 6 -2x - (-3x - 8) = 6 -2x - (-3x) - (-8) = 6 -2x + 3x + 8 = 6 x + 8 = 6 _____ - 8 ___ -8 x = -2
3x + y = -8 3(-2) + y = -8 -6 + y = -8 ______ +6 ___ +6 y = -2 (-2, -2)
15. x + 2y = -1 _____ - 2y ____ -2y x = -2y - 1
4x - 4y = 20 4(-2y - 1) - 4y = 20 4(-2y) + 4(-1) - 4y = 20 -8y - 4 - 4y = 20 -12y - 4 = 20 ________ + 4 ___ +4 -12y = 24
-12y
_____ -12
= 24 ____ -12
y = -2
x + 2y = -1 x + 2(-2) = -1 x - 4 = -1 _____ + 4 ___ +4 x = 3 (3, -2)
16. 4x = y - 1 ___ +1 _____ + 1 4x + 1 = y
6x - 2y = -3 6x - 2(4x + 1) = -3 6x - 2(4x) - 2(1) = -3 6x - 8x - 2 = -3 -2x - 2 = -3 _______ + 2 ___ +2 -2x = -1
-2x ____ -2
= -1 ___ -2
x = 1 __ 2
4x = y - 1
4 ( 1 __ 2 ) = y - 1
2 = y - 1 ___ +1 _____ + 1 3 = y
( 1 __ 2 , 3)
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17a. Let m represent the number of months and let t represent the total cost.
Option 1: t = 150 + 35m Option 2: t = 60m
t = 150 + 35m 60m = 150 + 35m _____ -35m _________ - 35m 25m = 150
25m ____ 25
= 150 ____ 25
m = 6
t = 60m = 60(6) = 360 In 6 months, the total cost for each option will be
the same, $360.
b. the second option; for 5 months, it will cost only 60(5) = $300, while the other option costs
150 + 35(5) = 150 + 175 = $325.
18. x = 5
x + y = 8 5 + y = 8 ______ -5 ___ -5 y = 3 (5, 3)
19. x = 2y + 6 x = 2(-3x + 4) + 6 x = 2(-3x) + 2(4) + 6 x = -6x + 8 + 6 x = -6x + 14 ____ +6x ________ +6x 7x = 14
7x ___ 7 = 14 ___
7
x = 2
y = -3x + 4 y = -3(2) + 4 y = -6 + 4 y = -2 (2, -2)
20. 3x - y = 11 _______ -3x ____ -3x -y = -3x + 11 -1(-y) = -1(-3x + 11) y = -1(-3x) - 1(11) y = 3x - 11
5y - 7x = 1 5(3x - 11) - 7x = 1 5(3x) + 5(-11) - 7x = 1 15x - 55 - 7x = 1 8x - 55 = 1 _______ + 55 ____ +55 8x = 56
8x ___ 8 = 56 ___
8
x = 7
3x - y = 11 3(7) - y = 11 21 - y = 11 _______ -21 ____ -21 -y = -10 -1(-y) = -1(-10) y = 10 (7, 10)
21. x - y = 2 ____ + y ___ +y x = y + 2
1 __ 2 x + 1 __
3 y = 6
1 __ 2 (y + 2) + 1 __
3 y = 6
1 __ 2 (y) + 1 __
2 (2) + 1 __
3 y = 6
1 __ 2 y + 1 + 1 __
3 y = 6
5 __ 6 y + 1 = 6
______ - 1 ___ -1
5 __ 6 y = 5
6 __ 5 ( 5 __
6 y) = 6 __
5 (5)
y = 6
x - y = 2 x - 6 = 2 ____ + 6 ___ +6 x = 8 (8, 6)
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22. 2x + y = 5 2(7 - 2y) + y = 5 2(7) + 2(-2y) + y = 5 14 - 4y + y = 5 14 - 3y = 5 ________ -14 ____ -14 -3y = -9
-3y
____ -3
= -9 ___ -3
y = 3
x = 7 - 2y x = 7 - 2(3) x = 7 - 6 x = 1 (1, 3)
23. 2.2x + 5 = y 2.2x + 5 = 1.2x - 4 _________ -1.2x _________ -1.2x x + 5 = -4 _____ - 5 ___ -5 x = -9
y = 1.2x - 4 y = 1.2(-9) - 4 y = -10.8 - 4 y = -14.8 (-9, -14.8)
24. Let x represent the first number and let y represent the second number.
Sum is 50: x + y = 50 First is 43 less than twice second: x = 2y - 43
⎧
⎨
⎩ x + y = 50
x = 2y - 43
x + y = 50 (2y - 43) + y = 50 3y - 43 = 50 _______ + 43 ____ +43 3y = 93
3y
___ 3 = 93 ___
3
y = 31
x = 2y - 43 x = 2(31) - 43 x = 62 - 43 x = 19 The two numbers are 19 and 31.
25. 20 coins: n + d = 20 Value $1.40: 0.05n + 0.10d = 1.40
n + d = 20 _______ -n ___ -n d = -n + 20
0.05n + 0.10d = 1.40 0.05n + 0.10(-n + 20) = 1.40 0.05n + 0.10(-n) + 0.10(20) = 1.40 0.05n - 0.10n + 2.00 = 1.40 -0.05n + 2.00 = 1.40 ____________ - 2.00 _____ -2.00 -0.05n = -0.60
-0.05n _______ -0.05
= -0.60 ______ -0.05
n = 12
n + d = 20 12 + d = 20 _______ -12 ____ -12 d = 8 There are 12 nickels and 8 dimes in the jar.
26. Let p represent the price of the popcorn and d represent the price of the drinks.
Customer #3598: 3p + 2d = 21 Customer #3599: 2p + 4d = 22
⎧
⎨
⎩ 3p + 2d = 21
2p + 4d = 22
2p + 4d = 22 _______ - 4d ____ -4d 2p = -4d + 22
2p
___ 2 = -4d + 22 _________
2
p = -2d + 11
3p + 2d = 21 3(-2d + 11) + 2d = 21 3(-2d) + 3(11) + 2d = 21 -6d + 33 + 2d = 21 33 - 4d = 21 ________ -33 ____ -33 -4d = -12
-4d ____ -4
= -12 ____ -4
d = 3
3p + 2d = 21 3p + 2(3) = 21 3p + 6 = 21 ______ - 6 ___ -6 3p = 15
3p
___ 3 = 15 ___
3
p = 5 The price of a large bag of popcorn is $5 and the
price of a small drink is $3.
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27. Let x represent the amount in the 5% account and let y represent the amount in the 6% account.
Total of $1000: x + y = 1000 $58 in interest: 0.05x + 0.06y = 58
⎧
⎨
⎩ x + y = 1000
0.05x + 0.06y = 58
x + y = 1000 ____ - y _____ -y x = -y + 1000
0.05x + 0.06y = 58 0.05(-y + 1000) + 0.06y = 58 0.05(-y) + 0.05(1000) + 0.06y = 58 -0.05y + 50 + 0.06y = 58 50 + 0.01y = 58 ___________ -50 ____ -50 0.01y = 8
0.01y
_____ 0.01
= 8 ____ 0.01
y = 800
x + y = 1000 x + 800 = 1000 ______ - 800 _____ -800 x = 200 Helen invested $200 in the 5% account and $800 in
the 6% account.
28. x + y = 90 x + (4x - 10) = 90 5x - 10 = 90 _______ + 10 ____ +10 5x = 100
5x ___ 5 = 100 ____
5
m∠x = 20°
y = 4x - 10 y = 4(20) - 10 y = 80 - 10 m∠y = 70°
29. x + y = 90 2y + y = 90 3y = 90
3y
___ 3 = 90 ___
3
m∠y = 30°
x = 2y x = 2(30) m∠x = 60°
m∠y = 30°
30. x + y = 90 x + 2(x - 15) = 90 x + 2(x) + 2(-15) = 90 x + 2x - 30 = 90 3x - 30 = 90 _______ + 30 ____ +30 3x = 120
3x ___ 3 = 120 ____
3
m∠x = 40°
y = 2(x - 15) y = 2(40 - 15) y = 2(25) m∠y = 50°
31a. rate · Time = Distance
With Headwind p - w · 3 = 240
With Tailwind p + w · 2 = 240
b. (p - w)3 = 240 (p)3 + (-w)3 = 240 3p - 3w = 240
(p + w)2 = 240 (p)2 + (w)2 = 240 2p + 2w = 240
⎧
⎨
⎩ 3p - 3w = 240
2p + 2w = 240
c. 3p - 3w = 240 _______ + 3w ____ +3w 3p = 3w + 240
3p
___ 3 = 3w + 240 ________
3
p = w + 80
2p + 2w = 240 2(w + 80) + 2w = 240 2(w) + 2(80) + 2w = 240 2w + 160 + 2w = 240 4w + 160 = 240 ________ - 160 -160 4w = 80
4w ___ 4 = 80 ___
4
w = 20
2p + 2w = 240 2p + 2(20) = 240 2p + 40 = 240 _______ - 40 ____ -40 2p = 200
2p
___ 2 = 200 ____
2
p = 100 The speed of the plane is 100 mi/h and the speed
of the wind is 20 mi/h.
32. Possible answer: Solve one of the equations for either x or y. Then substitute the expression equal to x or y into the other equation. This creates a one-variable equation that can be solved. When you get the value of one variable, substitute it into one of the original equations and solve to find the value of the other variable.
33. The solution of a system solved by graphing is the same as the solution of a system solved by substitution.
34a. Let x represent the cost of a book and let y represent the cost of a backpack.
Total $26: 2x + y = 26 Book costs $8 less than backpack: x = y - 8
⎧
⎨
⎩ 2x + y = 26
x = y - 8
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b. 2x + y = 26 2(y - 8) + y = 26 2(y) + 2(-8) + y = 26 2y - 16 + y = 26 3y - 16 = 26 _______ + 16 ____ +16 3y = 42
3y
___ 3 = 42 ___
3
y = 14
x = y - 8 x = 14 - 8 x = 6 book: $6; backpack: $14
c. 2x + y = 26 _______ -2x ____ -2x y = -2x + 26
x = y - 8 ___ + 8 _____ + 8 x + 8 = y
Juanita’s Purchase
0 4 8 12 14 2 6 10
8
16
24
32
4
12
20
28
Cost of Book ($)
Cost
of B
ackp
ack
($)
(6, 14)
The solution appears to be at (6, 14). Possible answer: Substitution works well since x
is already isolated. Graphing requires solving both equations for y.
35. Possible estimate: (1.75, -2.5) x + y = -0.6 ______ -x _____ -x y = -x - 0.6
2x - y = 6 2x - (-x - 0.6) = 6 2x - (-x) - (-0.6) = 6 2x + x + 0.6 = 6 3x + 0.6 = 6 _______ - 0.6 ____ -0.6 3x = 5.4
3x ___ 3 = 5.4 ___
3
x = 1.8
x + y = -0.6 1.8 + y = -0.6 ________ -1.8 ____ -1.8 y = -2.4 (1.8, -2.4)
TEST PREP
36. D; Since the total number of cousins is 24, m + f = 24 or f = 24 - m. Since the number of males was 6 less than twice the number of females, m = 2f - 6, so D is correct.
37. F; If d is the number of dimes and n is the number of nickels, then d + n represents the number of coins, so Roger has 12 coins. Since d is 5 greater than n, there are 5 more dimes than nickels, so F is correct.
ChALLENGE AND ExTEND
38. Let n represent the number of new cars and let u represent the number of used cars.
Total of 378 cars: n + u = 378
Ratio of new to used is 5:4: 5 __ 4 = n __ u
4n = 5u
⎧
⎨ ⎩ n + u = 378
4n = 5u
n + u = 378 _______ -n ___ -n u = -n + 378
4n = 5u 4n = 5(-n + 378) 4n = 5(-n) + 5(378) 4n = -5n + 1890 ____ +5n __________ +5n 9n = 1890
9n ___ 9 = 1890 _____
9
n = 210
n + u = 378 210 + u = 378 _________ -210 _____ -210 u = 168 The car dealership has 210 new cars and 168 used
cars.
39. t = 4
s + 3t = 10 s + 3(4) = 10 s + 12 = 10 ______ - 12 ____ -12 s = -2
2r - 3s - t = 12 2r - 3(-2) - 4 = 12 2r + 6 - 4 = 12 2r + 2 = 12 _____ - 2 ___ -2 2r = 10
2r __ 2 = 10 ___
2
r = 5
r = 5; s = -2; t = 4
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40. y + z = 5 ______ -y ___ -y z = -y + 5
2y - 4z = -14 2y - 4(-y + 5) = -14 2y - 4(-y) - 4(5) = -14 2y + 4y - 20 = -14 6y - 20 = -14 _______ + 20 ____ +20 6y = 6
6y
___ 6 = 6 __
6
y = 1
y + z = 5 1 + z = 5 ______ -1 ___ -1 z = 4
x + y + z = 7 x + 1 + 4 = 7 x + 5 = 7 _____ - 5 ___ -5 x = 2
x = 2; y = 1; z = 4
41. -b + c = -5 ______ +b ___ +b c = b - 5
3b + 2c = 15 3b + 2(b - 5) = 15 3b + 2(b) + 2(-5) = 15 3b + 2b - 10 = 15 5b - 10 = 15 _______ + 10 ____ +10 5b = 25
5b ___ 5 = 25 ___
5
b = 5
-b + c = -5 -5 + c = -5 ______ +5 ___ +5 c = 0
a + 2b + c = 19 a + 2(5) + 0 = 19 a + 10 = 19 ______ - 10 ____ -10 a = 9
a = 9; b = 5; c = 0
SolvIng SyStEmS by ElImInAtIon
CHECK IT OUT!
1. y + 3x = -2 _____________ + 2y - 3x = 14 3y + 0 = 12 3y = 12
3y
___ 3 = 12 ___
3
y = 4
y + 3x = -2 4 + 3x = -2 _______ -4 ___ -4 3x = -6
3x ___ 3 = -6 ___
3
x = -2 (-2, 4)
2. 3x + 3y = 15 ________________ - (-2x + 3y = -5)
3x + 3y = 15 ____________ + 2x - 3y = 5 5x + 0 = 20 5x = 20
5x ___ 5
= 20 ___ 5
x = 4
3x + 3y = 15 3(4) + 3y = 15 12 + 3y = 15 ________ -12 ____ -12 3y = 3
3y
___ 3
= 3 __ 3
y = 1 (4, 1)
3a. 3x + 2y = 6 _______________ + 3(-x + y = -2)
3x + 2y = 6 ________________ + (-3x + 3y = -6) 0 + 5y = 0 5y = 0
5y
___ 5 = 0 __
5
y = 0
-x + y = -2 -x + 0 = -2 -x = -2 -1(-x) = -1(-2) x = 2 (2, 0)
b. 3(2x + 5y = 26) __________________ + 2(-3x - 4y = -25)
6x + 15y = 78 ________________ + (-6x - 8y = -50) 0 + 7y = 28 7y = 28
7y
___ 7
= 28 ___ 7
y = 4
2x + 5y = 26 2x + 5(4) = 26 2x + 20 = 26 _______ - 20 ____ -20 2x = 6
2x ___ 2
= 6 __ 2
x = 3 (3, 4)
4. Let ℓ represent the number of lilies and let t represent the number of tulips.
Total $14.85: 1.25ℓ + 0.90t = 14.85 13 flowers: ℓ + t = 13
1.25ℓ + 0.90t = 14.85 ____________________ + (-1.25)(ℓ + t = 13)
1.25ℓ + 0.90t = 14.85 ________________________ + (-1.25ℓ - 1.25t = -16.25) 0 - 0.35t = -1.40 -0.35t = -1.40
-0.35t ______ -0.35
= -1.40 ______ -0.35
t = 4
ℓ + t = 13 ℓ + 4 = 13 ____ - 4 ___ -4 ℓ = 9 Sally bought 9 lilies and 4 tulips.
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THINK AND DISCUSS
1. Multiplying the second equation by -1 and adding is like adding the opposite, which is the same as subtracting; possible answer: 3x + y = 5; 3x + 2y = 6
2. No matter which variable you solve for first, the answer will be the same.
3.
Substitution: Elimination using addition or subtraction:
Elimination using multiplication:
Solving Systems of Linear Equations
x = 5 y x + y = 10
4 x + y = 1 2 x - y = 7
2 x + 6 y = 10 x + y = - 2
ExErCISESGUIDED PRACTICE
1. -x + y = 5 _____________ + x - 5y = -9 0 - 4y = -4 -4y = -4
-4y
____ -4
= -4 ___ -4
y = 1
x - 5y = -9 x - 5(1) = -9 x - 5 = -9 _____ + 5 ___ +5 x = -4 (-4, 1)
2. x + y = 12 __________ + x - y = 2 2x + 0 = 14 2x = 14
2x ___ 2 = 14 ___
2
x = 7
x + y = 12 7 + y = 12 ______ -7 ___ -7 y = 5 (7, 5)
3. 2x + 5y = -24 _____________ + 3x - 5y = 14 5x + 0 = -10 5x = -10
5x ___ 5 = -10 ____
5
x = -2
2x + 5y = -24 2(-2) + 5y = -24 -4 + 5y = -24 _______ +4 ____ +4 5y = -20
5y
___ 5 = -20 ____
5
y = -4 (-2, -4)
4. x - 10y = 60 ______________ - (x + 14y = 12)
x - 10y = 60 _________________ + (-x - 14y = -12) 0 - 24y = 48 -24y = 48
-24y
_____ -24
= 48 ____ -24
y = -2
x - 10y = 60 x - 10(-2) = 60 x + 20 = 60 ______ - 20 ____ -20 x = 40 (40, -2)
5. 5x + y = 0 ______________ - (5x + 2y = 30)
5x + y = 0 _________________ + (-5x - 2y = -30) 0 - y = -30 -y = -30 -1(-y) = -1(-30) y = 30
5x + y = 0 5x + 30 = 0 ______ - 30 ____ -30 5x = -30
5x ___ 5 = -30 ____
5
x = -6 (-6, 30)
6. -5x + 7y = 11 ________________ - (-5x + 3y = 19)
-5x + 7y = 11 ______________ + 5x - 3y = -19 0 + 4y = -8 4y = -8
4y
___ 4 = -8 ___
4
y = -2
-5x + 3y = 19 -5x + 3(-2) = 19 -5x - 6 = 19 _______ + 6 ___ +6 -5x = 25
-5x ____ -5
= 25 ___ -5
x = -5 (-5, -2)
7. 2x + 3y = 12 ______________ + 3(5x - y = 13)
2x + 3y = 12 ______________ + 15x - 3y = 39 17x + 0 = 51 17x = 51
17x ____ 17
= 51 ___ 17
x = 3
2x + 3y = 12 2(3) + 3y = 12 6 + 3y = 12 _______ -6 ___ -6 3y = 6
3y
___ 3 = 6 __
3
y = 2 (3, 2)
8. -3x + 4y = 12 _________________ + (-4)(2x + y = -8)
-3x + 4y = 12 ________________ + (-8x - 4y = 32) -11x + 0 = 44 -11x = 44
-11x _____ -11
= 44 ____ -11
x = -4
2x + y = -8 2(-4) + y = -8 -8 + y = -8 ______ +8 ___ +8 y = 0 (-4, 0)
9. 3(2x + 4y = -4) __________________ + (-2)(3x + 5y = -3)
6x + 12y = -12 _______________ + (-6x - 10y = 6) 0 + 2y = -6 2y = -6
2y
___ 2
= -6 ___ 2
y = -3
2x + 4y = -4 2x + 4(-3) = -4 2x - 12 = -4 _______ + 12 ____ +12 2x = 8
2x ___ 2 = 8 __
2
x = 4 (4, -3)
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10. Let x represent the number of hamburger bun packages and let y represent the number of hot-dog bun packages.
$20 of buns: 2.00x + 1.50y = 20 12 packages: x + y = 12
2.00x + 1.50y = 20 __________________ + (-2.00)(x + y = 12)
2.00x + 1.50y = 20 ______________________ + (-2.00x - 2.00y = -24) 0 - 0.50y = -4 -0.50y = -4
-0.50y
______ -0.50
= -4 ______ -0.50
y = 8
x + y = 12 x + 8 = 12 ____ - 8 ___ -8 x = 4 The Harlin family bought 4 packages of hamburger
buns and 8 packages of hot dog buns.
PRACTICE AND PROBLEM SOLVING
11. -x + y = -1 ___________ + 2x - y = 0 x + 0 = -1 x = -1
-x + y = -1 -(-1) + y = -1 1 + y = -1 ______ -1 ___ -1 y = -2 (-1, -2)
12. -2x + y = -20 ____________ + 2x + y = 48 0 + 2y = 28 2y = 28
2y
___ 2 = 28 ___
2
y = 14
2x + y = 48 2x + 14 = 48 ______ - 14 ____ -14 2x = 34
2x ___ 2 = 34 ___
2
x = 17 (17, 14)
13. 3x - y = -2 _____________ + (-2x + y = 3) x + 0 = 1 x = 1
-2x + y = 3 -2(1) + y = 3 -2 + y = 3 ______ +2 ___ +2 y = 5 (1, 5)
14. x - y = 4 ______________ -(x - 2y) = -10
x - y = 4 ________________ + (-x + 2y = -10) 0 + y = -6 y = -6
x - y = 4 x - (-6) = 4 x + 6 = 4 _____ - 6 ___ -6 x = -2 (-2, -6)
15. x + 2y = 5 _______________ -(3x + 2y) = -17
x + 2y = 5 _________________ + (-3x - 2y = -17) -2x + 0 = -12 -2x = -12
-2x ____ -2
= -12 ____ -2
x = 6
x + 2y = 5 6 + 2y = 5 _______ -6 ___ -6 2y = -1
2y
___ 2 = -1 ___
2
y = - 1 __ 2
(6, - 1 __ 2 )
16. 3x - 2y = -1 ______________ -(3x - 4y) = -9
3x - 2y = -1 ________________ + (-3x + 4y = -9) 0 + 2y = -10 2y = -10
2y
___ 2
= -10 ____ 2
y = -5
3x - 2y = -1 3x - 2(-5) = -1 3x + 10 = -1 _______ - 10 ____ -10 3x = -11
3x ___ 3
= -11 ____ 3
x = - 11 ___ 3
(- 11 ___ 3
, -5)
17. 3(x - y = -3) ____________ + 5x + 3y = 1
3x - 3y = -9 ____________ + 5x + 3y = 1 8x + 0 = -8 8x = -8
8x ___ 8 = -8 ___
8
x = -1
5x + 3y = 1 5(-1) + 3y = 1 -5 + 3y = 1 _______ +5 ___ +5 3y = 6
3y
___ 3 = 6 __
3
y = 2 (-1, 2)
18. 9x - 3y = 3 ___________________ + (-3)(3x + 8y = -17)
9x - 3y = 3 ________________ + (-9x - 24y = 51) 0 - 27y = 54 -27y = 54
-27y
_____ -27
= 54 ____ -27
y = -2
3x + 8y = -17 3x + 8(-2) = -17 3x - 16 = -17 _______ + 16 ____ +16 3x = -1
3x ___ 3 = -1 ___
3
x = - 1 __ 3
(- 1 __ 3 , -2)
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19. 3(5x + 2y = -1) _________________ + (-5)(3x + 7y = 11)
15x + 6y = -3 ___________________ + (-15x - 35y = -55) 0 - 29y = -58 -29y = -58
-29y
_____ -29
= -58 ____ -29
y = 2
3x + 7y = 11 3x + 7(2) = 11 3x + 14 = 11 _______ - 14 ____ -14 3x = -3
3x ___ 3 = -3 ___
3
x = -1 (-1, 2)
20. Let c represent the number of candles and let v represent the number of vases.
Spent $31.50: 3.00c + 4.25v = 31.50 8 tables: c + v = 8
3.00c + 4.25v = 31.50 ___________________ + (-3.00)(c + v = 8)
3.00c + 4.25v = 31.50 ________________________ + (-3.00c - 3.00v = -24) 0 + 1.25v = 7.50 1.25v = 7.50
1.25v _____ 1.25
= 7.50 ____ 1.25
v = 6
c + v = 8 c + 6 = 8 ____ - 6 ___ -6 c = 2 Mrs. Gonzalez bought 2 candles and 6 vases.
21. Difference is 2: ℓ - w = 2 Perimeter is 40: 2ℓ + 2w = 40
⎧
⎨ ⎩ ℓ - w = 2
2ℓ + 2w = 40
2(ℓ - w = 2) _____________ + 2ℓ + 2w = 40
2ℓ - 2w = 4 _____________ + 2ℓ + 2w = 40 4ℓ + 0 = 44 4ℓ = 44
4ℓ ___ 4 = 44 ___
4
ℓ = 11
2ℓ + 2w = 40 2(11) + 2w = 40 22 + 2w = 40 _________ -22 ____ -22 2w = 18
2w ___ 2 = 18 ___
2
w = 9 The length of the rectangle is 11 units and the width
of the rectangle is 9 units.
22. A; the student did not distribute the negative sign to all values in the parentheses and so mistakenly added 3 to -3.
23a. 1% Solution
+5%
Solution=
4 % Solution
Amount of Solution (mL)
x + y = 100
Amount of Acid (mL)
0.01x + 0.05y = 0.04(100)
b. 0.04(100) = 4
⎧
⎨ ⎩ x + y = 100
0.01x + 0.05y = 4
c. 0.01x + 0.05y = 4 ___________________ + (-0.01)(x + y = 100)
0.01x + 0.05y = 4 _____________________ + (-0.01x - 0.01y = -1) 0 + 0.04y = 3 0.04y = 3
0.04y
_____ 0.04
= 3 ____ 0.04
y = 75
x + y = 100 x + 75 = 100 _____ - 75 ____ -75 x = 25 The chemist should use 25 mL of the 1% solution
and 75 mL of the 5% solution.
24. (40, -2); Possible answer: elimination using subtraction; the coefficient of the x-values are the same.
25. (3, 3); Possible answer: elimination using multiplication; none of the variables have the same coefficient so they cannot be eliminated by addition or subtraction.
26. (3, -1); Possible answer: elimination using subtraction; the y-terms have the same coefficients.
27. ( 46 ___ 7 , 8 __
7 ) ; Possible answer: substitution; the second
equation is already solved for a variable.
28. (- 1 __ 2 , 2) ; Possible answer: graphing; both equations
are in slope-intercept form.
29. ( 15 ___ 7 , 9 __
7 ) ; Possible answer: elimination using
multiplication; when the first equation is multiplied by -2 the x-coefficients are opposites.
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30. Since they did not sell any of the expensive mulch, they did not sell any cocoa.
Let h represent the number of bags of hardwood mulch and let p represent the number of bags of pine bark.
Sold 176 bags: h + p = 176 Total sales of $520: 3.50h + 2.75p = 520
3.50h + 2.75p = 520 ___________________ + (-3.50)(h + p = 176)
3.50h + 2.75p = 520 _______________________ + (-3.50h - 3.50p = -616) 0 - 0.75p = -96 -0.75p = -96
-0.75p
_______ -0.75
= -96 ______ -0.75
p = 128
h + p = 176 h + 128 = 176 _______ - 128 _____ -128 h = 48 They sold 48 bags of hardwood mulch and 128 bags
of pine bark mulch.
31a. Let A represent the price of an item on shelf A and let B represent the price of an item on shelf B.
Option 1: 3A + 2B = 16 Option 2: 2A + 3B = 14
⎧
⎨ ⎩ 3A + 2B = 16
2A + 3B = 14
b. 2(3A + 2B = 16) ___________________ + (-3)(2A + 3B = 14)
6A + 4B = 32 _________________ + (-6A - 9B = -42) 0 - 5B = -10 -5B = -10
-5B ____ -5
= -10 ____ -5
B = 2
2A + 3B = 14 2A + 3(2) = 14 2A + 6 = 14 ______ - 6 ___ -6 2A = 8
2A ___ 2 = 8 __
2
A = 4
A = 4; B = 2
c. Buying the first package will save 3(6 - 4) + 2(3 - 2) = 6 + 2 = $8; buying the
second package will save 2(6 - 4) + 3(3 - 2) = 4 + 3 = $7.
32. -4(3x + y = 1) _______________ + 2x + 4y = -6
-12x - 4y = -4 ______________ + 2x + 4y = -6 -10x + 0 = -10 -10x = -10
-10x _____ -10
= -10 ____ -10
x = 1
3x + y = 1 3(1) + y = 1 3 + y = 1 ______ -3 ___ -3 y = -2
(1, -2); possible answer: to check the solution, algebraically, substitute 1 for x and -2 for y in both equations. If both equations are true statements, the solution is correct. To check the solution graphically, graph both equations on the same coordinate plane and locate their intersection. If the intersection is (1, -2) then the solution is correct.
TEST PREP
33. A; Let x represent the number of 2-point problems and y represent the number of 3-point problems. So, since the total number of problems is 25, x + y = 25. Since the total number of points is 60, 2x + 3y = 60.
34. G; Notice that 15 + 11 (1 1 __ 4 ) = 15 + 13.75 = 28.75
and 10 + 15 (1 1 __ 4 ) = 10 + 18.75 = 28.75, so G is
correct.
35a. Let s represent the number of student tickets sold and let n represent the number of nonstudent tickets sold.
358 tickets sold: s + n = 358 School made $752.25: 1.50s + 3.25n = 752.25
⎧
⎨ ⎩ s + n = 358
1.50s + 3.25n = 752.25
b. 1.50s + 3.25n = 752.25 ______________________ + (-1.50)(s + n = 358)
1.50s + 3.25n = 752.25 _________________________ + (-1.50s - 1.50n = -537) 0 + 1.75n = 215.25 1.75n = 215.25
1.75n _____ 1.75
= 215.25 ______ 1.75
n = 123
s + n = 358 s + 123 = 358 ______ - 123 _____ -123 s = 235 s = 235; n = 123; 235 student tickets were sold
and 123 nonstudent tickets were sold.
169 Holt McDougal Algebra 1
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ChALLENGE AND ExTEND
36. x + 16 1 __ 2 = - 3 __
4 y
x + 16 1 __ 2 = - 3 __
4 ( 1 __
2 x)
x + 33 ___ 2 = - 3 __
8 x
________ -x ____ -x
33 ___ 2 = - 11 ___
8 x
- 8 ___ 11
( 33 ___ 2 ) = - 8 ___
11 (- 11 ___
8 x)
-12 = x
y = 1 __ 2 x
y = 1 __ 2 (-12)
y = -6
x = -12; y = -6
37. 1 __ 2 z = 5
2 ( 1 __ 2 z) = 2(5)
z = 10
2x + y + z = 17 2x + y + 10 = 17 __________ - 10 ____ -10 2x + y = 7
2x + y = 7 __________ + x - y = 5 3x + 0 = 12 3x = 12
3x ___ 3 = 12 ___
3
x = 4
x - y = 5 4 - y = 5 ______ -4 ___ -4 -y = 1 -1(-y) = -1(1) y = -1
x = 4; y = -1; z = 10
38. x - 2y - z = -1 _____________________ + (-x + 2y + 4z = -11) 0 + 0 + 3z = -12 3z = -12
3z ___ 3 = -12 ____
3
z = -4
x - 2y - z = -1 x - 2y - (-4) = -1 x - 2y + 4 = -1 _________ - 4 ___ -4 x - 2y = -5
2x + y + z = 1 2x + y + (-4) = 1 2x + y - 4 = 1 _________ + 4 ___ +4 2x + y = 5
x - 2y = -5 ____________ + 2(2x + y = 5)
x - 2y = -5 _____________ + 4x + 2y = 10 5x + 0 = 5 5x = 5
5x ___ 5 = 5 __
5
x = 1
2x + y + z = 1 2(1) + y + (-4) = 1 2 + y - 4 = 1 y - 2 = 1 _____ + 2 ___ +2 y = 3
x = 1; y = 3; z = -4
39. Let x represent the number in the tens’ column and let y represent the number in the ones’ column.
Sum is 5: x + y = 5 The number is 10x + y. Result of multiplying by 3: 3(10x + y) = 42
⎧
⎨ ⎩ x + y = 5
3(10x + y) = 42
-3(x + y = 5) _______________ + 3(10x + y) = 42
-3x - 3y = -15 ______________ + 30x + 3y = 42 27x + 0 = 27 27x = 27
27x ____ 27
= 27 ___ 27
x = 1
x + y = 5 1 + y = 5 ______ -1 ___ -1 y = 4 The number is 10(1) + 4 = 14.
SolvIng SpEcIAl SyStEmS
CHECK IT OUT!
1. Possible answer: Substitute -2x + 5 for y in the second equation. 2x + (-2x + 5) = 1 2x - 2x + 5 = 1 0 + 5 = 07
2. Possible answer: Substitute x - 3 for y in the second equation. -x(x - 3) - 3 = 0 x - x + 3 - 3 = 0 0 + 0 = 0 0 = 03
3a. x + 2y = -4 → y = - 1 __ 2 x - 2
-2(y + 2) = x → y = - 1 __ 2 x - 2
This system is consistent and dependent. It has infinitely many solutions.
b. y = -2(x - 1) → y = -2x + 2 y = -x + 3 → y = -x + 3 This system is consistent and independent. It has
one solution.
c. 2x - 3y = 6 → y = 2 __ 3 x - 2
y = 2 __ 3 x → y = 2 __
3 x
This system is inconsistent. It has no solution.
4. Yes. Let x represent the number of months and let y represent the total amount in the account.
Matt: y = 20x + 100 Ben: y = 30x + 80 The graphs of the two equations have different
slopes so they intersect. If the pattern continues, the amount of money in the two accounts will eventually be equal.
170 Holt McDougal Algebra 1
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THINK AND DISCUSS
1. solving algebraically by substitution or elimination, comparing slopes and y-intercepts, or graphing
2.
Exactly one solution: independent;
possible graph:
Infinitely many solutions: dependent;
possible graph:
No solution: inconsistent;
possible graph:
Linear System of Equations
x
y
x
y
x
y
ExErCISESGUIDED PRACTICE
1. consistent
2. Possible answer: Substitute x + 1 for y in the second equation.-x + (x + 1) = 3 -x + x + 1 = 3 1 = 37
3. Possible answer: Substitute -3x + 2 for y in the first equation. 3x + (-3x + 2) = 6 3x - 3x + 2 = 6 2 = 67
4. Possible answer: Compare slopes and intercepts.-y = 4x + 1 → y = -4x - 14x + y = 2 → y = -4x - 2The lines have the same slope and different y-intercepts. Therefore the lines are parallel.
5. Possible answer: Substitute -x + 3 for y in the second equation. x + (-x + 3) - 3 = 0 x - x + 3 - 3 = 0 0 = 03
6. Possible answer: Substitute -2x - 4 for y in the second equation.2x - (2x - 4) - 4 = 0 2x - 2x + 4 - 4 = 0 0 = 03
7. Possible answer: Add the two equations. -7x + y = -2 ___________ + 7x - y = 2 0 = 03
8. y = 2(x + 3) → y = 2x + 6 -2y = 2x + 6 → y = -x - 3 This system is consistent and independent. It has
one solution.
9. y = -3x - 1 → y = -3x - 1 3x + y = 1 → y = -3x + 1 This system is inconsistent. It has no solution.
10. 9y = 3x + 18 → y = 1 __ 3 x + 2
1 __ 3 x - y = -2 → y = 1 __
3 x + 2
This system is consistent and dependent. It has infinitely many solutions.
11. Yes. Let x represent the number of hours since Luke started to run and let y represent the number of miles.
Micah: y = 4x + 2 Luke: y = 6x The graphs of the two equations have different
slopes so they intersect. If their rates continue, Micah’s and Luke’s distance will eventually be equal.
PRACTICE AND PROBLEM SOLVING
12. Possible answer: Substitute 2x - 2 for y in the second equation.-2x + (2x - 2) = 1 -2x + 2x - 2 = 1 -2 = 17
13. Possible answer: Substitute -x - 1 for y in the first equation.x + (-x - 1) = 3 x - x - 1 = 3 -1 = 37
14. Possible answer: Substitute - 1 __ 2
x - 4 for y in the first equation.
x + 2(- 1 __ 2 x - 4) = -4
x - x - 8 = -4 -8 = -47
15. Possible answer: Compare slopes and intercepts.-6 + y = 2x → y = 2x + 6 → y = 2x - 36The lines have the same slope and different y-intercepts. Therefore the lines are parallel.
16. Possible answer: Substitute -2x + 3 for y in the second equation.2x + (-2x + 3) - 3 = 0 2x - 2x + 3 - 3 = 0 0 = 03
17. Possible answer: Substitute x - 2 for y in the second equation.x - (x - 2) - 2 = 0 x - x + 2 - 2 = 0 0 = 03
18. Possible answer: Substitute -x - 4 for y in the first equation.x + (-x - 4) = -4 x - x - 4 = -4 -4 = -43
19. Possible answer: Compare slopes and intercepts.-9x - 3y = -18 → y = -3x + 6 3x + y = 6 → y = -3x + 6
The lines have the same slope and the same y-intercept. Therefore the graphs are one line.
171 Holt McDougal Algebra 1
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20. y = -x + 5 → y = -x + 5 x + y = 5 → y = -x + 5 This system is consistent and dependent. It has
infinitely many solutions.
21. y = -3x + 2 y = 3x This system is consistent and independent. It has
one solution.
22. y - 1 = 2x → y = 2x + 1 y = 2x - 1 → y = 2x - 1 This system is inconsistent. It has no solution.
23. Yes. Let x represent the number of hours and let y represent the number of miles.
Mandy: y = 5x Nikki: y = 6x - 1 The graphs of the two equations have different
slopes so they intersect. If the pattern continues, Nikki will eventually catch up to Mandy.
24. No. Let x represent the number of minutes after copier A started and let y represent the number of copies.
Copier A: y = 35x Copier B: y = 35x + 10 The graphs of the two equations are parallel lines,
so there is no solution. If the pattern continues, the number of copies made by each machine will never be equal.
25. Let x represent the number of DVDs rented and let y represent the number of video games rented.
First week: 4x + 2y = 18 → y = -2x + 9 Second week: 2x + y = 9 → y = -2x + 9 There are infinitely many answers for the cost
of each video game and DVD. The system is consistent and dependent.
26. Let x represent the number of pounds of cashews and let y represent the number of pounds of peanuts.
Rosa: x + 2y = 10 → y = - 1 __ 2 x + 5
Sabrina: 2x + y = 11 → y = -2x + 11 This system is consistent and independent. It has
one solution.
y = - 1 __ 2 x + 5
-2x + 11 = - 1 __ 2 x + 5
________ +2x _______ +2x
11 = 3 __ 2 x + 5
___ -5 ______ - 5
6 = 3 __ 2 x
2 __ 3 (6) = 2 __
3 ( 3 __
2 x)
4 = x
2x + y = 11 2(4) + y = 11 8 + y = 11 ______ -8 ___ -8 y = 3 The cashews cost $4/lb and the peanuts cost $3/lb.
27. Let x represent the number of years and let y represent the number of geodes.
Pam: y = 4x + 2 Tommy: y = 4x + 2 They will always have the same amount. Both
started with 2 and add 4 every year.
28a. 2(3) = 6 2(4) = 8 2(5) = 10 2(6) = 12
2(12) = 24 2(13) = 26 2(14) = 28 2(15) = 30
In both tables, the value of y is twice the value of x. y = 2x; y = 2x
b.
2
x
y
2 -2
It is one line in the coordinate plane.
c. The equations are the same.
d. The graphs will now be parallel lines. The equations will have the same slope but different y-intercepts, so you know they will be parallel when graphed.
29. The graph will be 2 parallel lines.
30a. Let b represent the number of buttons to be made and let c represent the total cost.
Buttons, etc.: c = 50 + 1.10b Logos: c = 40 + 1.10b
b. c = 50 + 1.10b → c = 1.10b + 50 c = 40 + 1.10b → c = 1.10b + 40 Never; the slope, or cost per button is the same,
and the initial cost, or y-intercepts, are different.
c. For the same price, the y-intercept would have to be changed to 40. Possible answer: The y-intercept may be lowered to a value less than 40, or the slope may be changed to a value less than 1.1.
31. Student A; the lines are not parallel, so they will intersect.
32. The graph of a system that is consistent and independent shows two lines that intersect at one point. The graph of a system that is consistent and dependent shows only one line.
TEST PREP
33. D; Since 2x - y = 3 → y = 2x - 3 and 6x - 3y = 9 → y = 2x -3, the system is consistent and dependent.
34. H; Since two graphs that have different slopes, they must intersect at only one point. So if the graphs have different slopes, the system must be consistent, and independent.
172 Holt McDougal Algebra 1
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ChALLENGE AND ExTEND
35. p = q; p ≠ q
36a. 3x + 4y = 0 → y = - 3 __ 4 x
4x + 3y = 0 → y = - 4 __ 3 x
This system is consistent and independent. Intersection is (0, 0).
b. 2x + 5y = 0 → y = - 2 __ 5 x
5x + 2y = 0 → y = - 5 __ 2 x
This system is consistent and independent. Intersection is (0, 0).
c. Those systems are consistent and independent, so they only have one solution.
Notice that: ax + by = 0 → y = - a __
b x
bx + ay = 0 → y = - b __ a x
Since both equations have a y-intercept of 0, they intersect at (0, 0). Therefore, systems of equations of this form only intersect at (0, 0).
rEAdy to go on? Section A Quiz
1. _______________ y = -2x - 3 (1) -2(-2) - 3 1 4 - 3 1 1 3
____________ y = x + 3 (1) (-2) + 3 1 1 3
(-2, 1) is a solution of the system.
2. _____________ x - 4y = 1 (9) - 4(2) 1 9 - 8 1 1 1 3
______________ 2x - 3y = 2 2(9) - 3(2) 2 18 - 6 2 12 2 7
(9, 2) is not a solution of the system.
3. ___________
y = - 1 __ 3 x
(-1) - 1 __ 3 (3)
-1 -1 3
_______________ y + 2x = 5 (-1) + 2(3) 5 -1 + 6 5 5 5 3
(3, -1) is a solution of the system.
4.
x
y
0
6
1 -2
-8
(- 2, 3)
y = x + 5
y = 1 _ 2 x + 4
The solution appears to be at (-2, 3).
5.
x
y
0
4
-5 -2 (0,-2)
2x - y = 2
y = -x - 2
3
The solution appears to be at (0, -2).
6.
x
y
0
7
-7
-5 (3,-5)
4x + y = 7
5
2 _ 3 x + y = - 3
The solution appears to be at (3, -5).
7. Let x represent the number of months and let y represent the amount of money.
Christiana: y = 10x + 50 Marlena: y = 15x + 30 Graph y = 10x + 50 and y = 15x + 30.
Money in Account
0 1 3 5 2 4 6
20 40 60 80
100 Christina
Marlena
Months
Am
ount
( $)
(4, 90)
The lines appear to intersect at (4, 90). So both accounts will have the same amount deposited in 4 months and that amount will be $90.
8. 2x + y = 11 2x + (-x + 5) = 11 x + 5 = 11 _____ - 5 ___ -5 x = 6
y = -x + 5 y = -(6) + 5 y = -1 (6, -1)
173 Holt McDougal Algebra 1
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9. 3x - y = -2 _______ -3x ____ -3x -y = -3x - 2 -1(-y) = -1(-3x - 2) y = 3x + 2
4x - 3y = -1 4x - 3(3x + 2) = -1 4x - 3(3x) - 3(2) = -1 4x - 9x - 6 = -1 -5x - 6 = -1 _______ + 6 ___ +6 -5x = 5
-5x ____ -5
= 5 ___ -5
x = -1
3x - y = -2 3(-1) - y = -2 -3 - y = -2 ______ +3 ___ +3 -y = 1 -1(-y) = -1(1) y = -1 (-1, -1)
10. y = -2x - 5 -x = -2x - 5 ____ +2x _______ +2x x = -5
y = -x y = -(-5) y = 5 (-5, 5)
11. x + 3y = 15 ______________ + 2x - 3y = -6 3x + 0 = 9 3x = 9
3x ___ 3 = 9 __
3
x = 3
x + 3y = 15 3 + 3y = 15 _______ -3 ___ -3 3y = 12
3y
___ 3 = 12 ___
3
y = 4 (3, 4)
12. x + y = 2 _____________ - (2x + y = -1)
x + y = 2 ______________ + (-2x - y = 1) -x + 0 = 3 -x = 3 -1(-x) = -1(3) x = -3
x + y = 2 -3 + y = 2 ______ +3 ___ +3 y = 5 (-3, 5)
13. 3(-2x + 5y = -1) _______________ + 2(3x + 2y = 11)
-6x + 15y = -3 ____________ + 6x + 4y = 22 0 + 19y = 19 19y = 19
19y
____ 19
= 19 ___ 19
y = 1
3x + 2y = 11 3x + 2(1) = 11 3x + 2 = 11 ______ - 2 ___ -2 3x = 9
3x ___ 3 = 9 __
3
x = 3 (3, 1)
14. Let x represent the number of black and white drawings and let y represent the number of color drawings. Note that 2 h = 120 min.
Drew for 2 hours: 10x + 25y = 120 9 drawings: x + y = 9
⎧
⎨ ⎩ 10x + 25y = 120
x + y = 9
10x + 25y = 120 _______________ + (-10)(x + y = 9)
10x + 25y = 120 ___________________ + (-10x - 10y = -90) 0 + 15y = 30 15y = 30
15y
____ 15
= 30 ___ 15
y = 2
x + y = 9 x + 2 = 9 ____ - 2 ___ -2 x = 7 Akira drew 7 black and white drawings and 2 color
drawings.
15. y = -2x - 6 → y = -2x - 6 2x + y = 5 → y = -2x + 5 This system has no solution so it is an inconsistent
system.
16. x + y = 2 → y = -x + 2 2x + 2y = -6 → y = -x - 3 This system has no solution so it is an inconsistent
system.
17. y = -2x + 4 → y = -2x + 4 2x + y = 4 → y = -2x + 4 If this system were graphed, the graphs would be
the same line. There are infinitely many solutions.
18. 3x = -6y + 3 → y = - 1 __ 2 x + 1 __
2
2y = -x + 1 → y = - 1 __ 2 x + 1 __
2
This system is consistent and dependent. It has infinitely many solutions.
19. y = -4x + 2 → y = -4x + 2 4x + y = -2 → y = -4x - 2 This system is inconsistent. It has no solution.
20. 4x - 3y = 8 → y = 4 __ 3 x - 8 __
3
y = 4(x + 2) → y = 4x + 8 This system is consistent and independent. It has
one solution.
SolvIng lInEAr InEQuAlItIES
CHECK IT OUT!
1a. ________ y < x + 1 5 4 + 1 5 < 5 7 (4, 5) is not a solution.
b. _________ y > x - 7 1 1 - 7 1 > -6 3 (1, 1) is a solution.
174 Holt McDougal Algebra 1
5-5
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2a. 4x - 3y > 12 ________ -4x ____ -4x -3y > -4x + 12
-3y
____ -3
< -4x + 12 ________ -3
y < 4 __ 3 x - 4
Graph the boundary line y = 4 __ 3 x - 4. Use a dashed
line for <. The inequality is <, so shade below the line.
2
-2
-4
x
y
0 2 -2
b. 2x - y - 4 > 0 _________ + y ___ +y 2x - 4 > y y < 2x - 4 Graph the boundary line y = 2x - 4. Use a dashed
line for <. The inequality is <, so shade below the line.
-2
-4
x y
0 2 -2
c. Graph the boundary line y = - 2 __ 3 x + 1. Use a solid
line for ≥. The inequality is ≥, so shade above the line.
2
-2
x
y
0 2 -2
3a. Let b represent the number of black olives and let g represent the number of green olives.
2.5b + 2g ≤ 6
b. 2.5b + 2g ≤ 6 __________ -2.5b _____ -2.5b 2g ≤ -2.5b + 6
2g
___ 2 ≤ -2.5b + 6 _________
2
g ≤ -1.25b + 3 Since Dirk cannot buy a negative number of pounds
of olives, the system is graphed only in Quadrant I. Graph the boundary line y = -1.25x + 3. Use a solid line for ≤. The inequality is ≤, so shade below the line.
Olive Combinations
0 1 2 3 4
1
2
3
4
Black olives
Gre
en o
lives
c. Possible answers: (1 lb black, 1 lb green), (0.5 lb black, 2 lb green)
4a. y-intercept: 0; slope: -1 y = mx + b → y = -x The graph is shaded below a dashed boundary line.
Replace = with < to write the inequality y < -x.
b. y-intercept: -3; slope -2 y = mx + b → y = -2x - 3 The graph is shaded above a solid boundary line.
Replace = with ≥ to write the inequality y ≥ -2x - 3.
THINK AND DISCUSS
1. Possible answer: In both cases you will always graph a line. They are different because the graph of a linear inequality can be a dashed or solid line, and the graph must be shaded on one side of the line.
2. To write a linear inequality from a graph, find the equation of the boundary line. Use ≥ or ≤ for a solid boundary line. Use > or < for a dashed boundary line. If the half-plane is shaded above the line, use > or ≥, and if it is shaded below, use < or ≤.
3. Inequality
Symbol
Boundary Line
Shading
<
Dashed
Below
>
Dashed
Above
≤
Solid
Below
≥
Solid
Above
y < 5 x + 2 y > 7 x - 3 y ≤ 9 x + 1 y ≥ - 3 x - 2
175 Holt McDougal Algebra 1
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ExErCISESGUIDED PRACTICE
1. No; a dashed line means that the ordered pairs are not solutions to the linear inequality.
2. ___________ y ≤ -x + 3 3 -(0) + 3 3 ≤ 3 3 (0, 3) is a solution.
3. _____________ y > -2x - 2 0 -2(2) - 2 0 -4 - 2 0 > -6 3 (2, 0) is a solution.
4. _____________ y < 2x + 4 1 2(-2) + 4 1 -4 + 4 1 < 0 7 (-2, 1) is not a solution.
5. Graph the boundary line y = -x. Use a solid line for ≤. The inequality is ≤, so shade below the line.
2
-2
x
y
0 2 -2
6. Graph the boundary line y = 3x + 1. Use a dashed line for >. The inequality is >, so shade above the line.
2
x
y
2 -2
7. -y < -x + 4 -1(-y) > -1(-x + 4) y > x - 4 Graph the boundary line y = x - 4. Use a dashed
line for >. The inequality is >, so shade above the line.
-2
-4
y
x 0 2 -2
8. -y ≥ x + 1 -1(-y) ≤ -1(x + 1) y ≤ -x - 1 Graph the boundary line y = -x - 1. Use a solid
line for ≤. The inequality is ≤, so shade below the line.
2
-2
-4
x
y
2 -2
9a. Let x represent the number of cups of orange juice and let y represent the number of cups of pineapple juice.
x + y ≤ 16
b. x + y ≤ 16 ______ -x ___ -x y ≤ -x + 16 Since Jack cannot make a negative number of cups,
the system is graphed only in Quadrant I. Graph the boundary line y = -x + 16. Use a solid line for ≤. The inequality is ≤ so shade below the line.
Punch Combinations
0 4 8 12 16
4
8
12
16
Orange juice (c)
Pine
appl
e ju
ice
(c)
c. Possible answer: (2 c orange , 2 c pineapple), (4 c orange, 10 c pineapple)
10. y-intercept: 3; slope: 0 y = mx + b → y = 3 The graph is shaded below a dashed boundary line.
Replace = with < to write the inequality y < 3.
11. y-intercept: 5; slope: 1 y = mx + b → y = x + 5 The graph is shaded above a solid boundary line.
Replace = with ≥ to write the inequality y ≥ x + 5.
PRACTICE AND PROBLEM SOLVING
12. ____________ y ≥ 2x + 3 3 2(3) + 3 3 6 + 3 3 ≥ 9 7 (2, 3) is not a solution.
13. ____________ y < 3x - 3 -1 3(1) - 3 -1 3 - 3 -1 < 0 3 (1, -1) is a solution.
14. ____________ y > 4x + 7 7 4(0) + 7 7 0 + 7 7 > 7 7 (0, 7) is not a solution.
15. Graph the boundary line y = -2x + 6. Use a dashed line for >. The inequality is >, so shade above the line.
4
-4
x
y
0 2
176 Holt McDougal Algebra 1
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16. -y ≥ 2x -1(-y) ≤ -1(2x) y ≤ -2x Graph the boundary line y = -2x. Use a solid line
for ≤. The inequality is ≤, so shade below the line.
-2
x
y
0 2 -2
17. x + y ≤ 2 ______ -x ___ -x y ≤ -x + 2 Graph the boundary line y = -x + 2. Use a solid
line for ≤. The inequality is ≤, so shade below the line.
2
-2
x
y
0 2 -2
18. x - y ≥ 0 ____ + y ___ +y x ≥ y y ≤ x Graph the boundary line y = x. Use a solid line for
≤. The inequality is ≤, so shade below the line.
2
-2
x
y
2 -2
19a. Let x represent the number of pounds of hamburgers and let y represent the number of pounds of hot dogs.
3x + 2y ≤ 30
b. 3x + 2y ≤ 30 ________ -3x ____ -3x 2y ≤ -3x + 30
2y
___ 2 ≤ -3x + 30 ________
2
y ≤ - 3 __ 2 x + 15
Since Beverly cannot buy a negative number of hamburgers or hot dogs, the system is graphed only in Quadrant I. Graph the boundary line
y = - 3 __ 2 x + 15. Use a solid line for ≤. The
inequality is ≤, so shade below the line.Food Combinations
0 4 8
6
12
Hamburger meat (lb)
Hot
dog
s (lb
)
c. Possible answer: (3, 2), (5, 6)
20. y-intercept: -1; slope: 2 __ 3
y = mx + b → y = 2 __ 3 x - 1
The graph is shaded above a dashed boundary line.
Replace = with > to write the inequality y > 2 __ 3
x - 1.
21. y-intercept: 3; slope: - 1 __ 5
y = mx + b → y = - 1 __ 5 x + 3
The graph is shaded below a solid boundary line. Replace = with ≤ to write the ineqaulity
y ≤ - 1 __ 5 x + 3.
22a. 125x + 100y ≥ 500
b. 125x + 100y ≥ 500 ____________ -125x ______ -125x 100y ≥ -125x + 500
100y
_____ 100
≥ -125x + 500 ___________ 100
y ≥ -1.25x + 5 Since the store cannot sell a negative number of
items, the system is graphed only in Quadrant I. Graph the boundary line y = -1.25x + 5. Use a solid line for ≥. The inequality is ≥ so shade above the line.
Electronics Sales
0 2 4 6 8
2
4
6
8
DVD players
CD p
laye
rs
c. x and y must be whole numbers.
d. Possible answer: (5 DVD, 1 CD), (7 DVD, 6 CD), (9 DVD, 2 CD)
23. y ≤ 2 - 3x y ≤ -3x + 2 Graph the boundary line y = -3x + 2. Use a solid
line for ≤. The inequality is ≤, so shade below the line.
2
-2
x
y
0 2 -2
177 Holt McDougal Algebra 1
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24. -y < 7 + x -1(-y) > -1(7 + x) y > -x - 7 Graph the boundary line y = -x - 7. Use a dashed
line for >. The inequality is >, so shade above the line.
4
x
y
0 4
-4
-4
25. 2x - y ≤ 4 _______ -2x ____ -2x -y ≤ -2x + 4 -1(-y) ≥ -1(-2x + 4) y ≥ 2x - 4 Graph the boundary line y = 2x - 4. Use a solid line
for ≥. The inequality is ≥, so shade above the line.
-2
-4
x y
0 2 -2
26. 3x - 2y > 6 ________ -3x ____ -3x -2y > -3x + 6
-2y
____ -2
< -3x + 6 _______ -2
y < 3 __ 2 x - 3
Graph the boundary line y = 3 __ 2 x - 3. Use a dashed
line for <. The inequality is <, so shade below the line.
-2
x y
0 2 -2
27a. Let x represent the length and let y represent the width.
2x + 2y ≤ 18
b. 2x + 2y ≤ 18 ________ -2x ____ -2x 2y ≤ -2x + 18
2y
___ 2 ≤ -2x + 18 ________
2
y ≤ -x + 9
Since Marvin cannot have an negative length or width, the system is graphed only in Quadrant I. Graph the boundary line y = -x + 9. Use a solid line for ≤. The inequality is ≤, so shade below the line.
Rectangular Garden
0 2 4 6 8
2
4
6
8
Length (yd)
Wid
th (y
d)
Possible answer: (3, 1), (1, 6), (6, 2)
c. 4 yd × 4 yd
28. 15x + 11y ≤ 77 __________ -15x _____ -15x 11y ≤ -15x + 77
11y
____ 11
≤ -15x + 77 _________ 11
y ≤ - 15 ___ 11
x + 7
Since Stephen can only buy a whole number of fish, the system is graphed only in Quadrant I. Graph the
boundary line y = - 15 ___ 11
x + 7. Use a solid line for ≤.
The inequality is ≤, so shade below the line.Fish Combinations
0 2 4 6
2
4
6
Yellow tangs
Clow
n fis
h
Possible answer: (2, 2.5); you cannot buy half a fish.
29. Graph the boundary line y = 1. Use a dashed line for >. The inequality is >, so shade above the line.
2
-2
x
y
0 2 -2
30. -2 < x x > -2 Graph the boundary line x = -2. Use a dashed line
for >. The inequality is >, so shade right of the line.
2
-2
x
y
0 2
178 Holt McDougal Algebra 1
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31. Graph the boundary line x = -3. Use a solid line for ≥. The inequality is ≥, so shade right of the line.
2
-2
x
y
0 2 -2
32. Graph the boundary line y = 0. Use a solid line for ≤. The inequality is ≤, so shade below the line.
2
-2
x
y
0 2 -2
33. 0 ≥ x x ≤ 0 Graph the boundary line x = 0. Use a solid line for
≤. The inequality is ≤, so shade left of the line.
2
-2
x
y
0 2 -2
34. -12 + y > 0 _______ +12 ____ +12 y > 12 Graph the boundary line y = 12. Use a dashed line
for >. The inequality is >, so shade above the line.
4
8
x
y
0 2 -2
35. x + 7 < 7 ____ - 7 ___ -7 x < 0 Graph the boundary line x = 0. Use a dashed line
for <. The inequality is <, so shade left of the line.
2
-2
x
y
0 2 -2
36. -4 ≥ x - y ___ + y _____ + y -4 + y ≥ x ______ +4 ___ +4 y ≥ x + 4 Graph the boundary line y = x + 4. Use a solid line
for ≥. The inequality is ≥, so shade above the line.
2
4
x
y
0 -2 -4
37. Let a represent the number of adult tickets and let s represent the number of student tickets.
7a + 4s ≥ 280 ________ -7a ____ -7a 4s ≥ -7a + 280
4s ___ 4 ≥ -7a + 280 __________
4
s ≥ -1.75a + 70 Since only a positive number of tickets can be sold,
the system is graphed only in Quadrant I. Graph the boundary line y = -1.75x + 70. Use a solid line for ≥. The inequality is ≥, so shade above the line.
Tickets Sold
0 20 40 60
20
40
60
Adult tickets
Stud
ent
tick
ets
Possible answer: (40, 10), (40, 20), (20, 100)
38. Possible answer: It is not possible to list an infinite number of solutions, so shading is used to show all the solutions in the coordinate plane.
39. Possible answer: Sammy volunteers at the hospital and the library. The total number of hours he volunteers is more than 5; h + ℓ > 5; Check students’ graphs and solutions.
40a. x + y ≤ 50
b. x + y ≤ 50 ______ -x ___ -x y ≤ -x + 50
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Since Gloria can only make a positive number of bears, the system is graphed only in Quadrant I. Graph the boundary line y = -x + 50. Use a solid line for ≤. The inequality is ≤, so shade below the line.
Teddy Bear Combinations
0 10 20 30 40
10
20
30
40
Girl bears
Boy
bear
s
c. Possible answer: (10, 20), (30, 10), (15, 15)
41. Student A; the graph is a solid line, so ≤ should be used, not <.
42. Possible answer: If the inequality is in slope-intercept form, then > or ≥ means you shade above the boundary line, and < or ≤ means you shade below the boundary line. The shading represents all the solutions of the inequality.
TEST PREP
43. B; Since 4 > 2 = -1 + 3, (1, 4) is a solution of y > -x + 3.
44. H; The boundary line is y = -2x + 3. Since the line is solid and the graph is shaded below the line, use ≤. So y ≤ -2x + 3, or 2x + y ≤ 3.
45. C; The equation is rewritten as x ≥ 3. The boundary line is solid because the inequaltiy is ≥. The graph is shaded to the right because all values of x greater than or equal to 3 are solutions.
ChALLENGE AND ExTEND
46. 0 ≥ -6 - 2x - 5y ____ +5y ____________ + 5y 5y ≥ -6 - 2x
5y
___ 5 ≥ -6 - 2x _______
5
y ≥ - 2 __ 5 x - 6 __
5
Graph the boundary line y = - 2 __ 5 x - 6 __
5 . Use a solid
line for ≥. The inequality is ≥, so shade above the line.
2
-2
x
y
0 -2 -4
47. Graph the boundary function y = |x|. Use a dashed line for >. The inequality is >, so shade above the function.
2
-2
x
y
0 2 -2
48. Graph the boundary function y = |x - 3|. Use a solid line for ≥. The inequality is ≥, so shade above the function.
2
-4
-2
x
y
0 6 4 2 -2
49. Find the boundary line.
m = y 2 - y 1
_______ x 2 - x 1 = 1.5 - 3 _______ -3 - 0
= -1.5 _____ -3
= 1 __ 2
y - y 1 = m(x - x 1 )
y - 1.5 = 1 __ 2 (x - (-3))
y - 1.5 = 1 __ 2 (x + 3)
y - 1.5 = 1 __ 2 x + 1.5
______ + 1.5 ________ + 1.5
y = 1 __ 2 x + 3
Since the points on the boundary line are solutions, use ≥ or ≤. Check which sign makes (1, 1) not a solution.
____________
y ≥ 1 __ 2 x + 3
1 1 __ 2 (1) + 3
1 1 __ 2 + 3
1 ≥ 7 __ 2 7
____________
y ≤ 1 __ 2 x + 3
1 1 __ 2 (1) + 3
1 1 __ 2 + 3
1 ≤ 7 __ 2 3
Since (1, 1) is not a solution of y ≥ 1 __ 2 x + 3, the
inequality is y ≥ 1 __ 2 x + 3.
50. The boundary lines must be x = 0 and y = 0. Since Quadrant I is not shaded, no points in that plane are solutions to either inequality, so the inequalities are x ≤ 0 and y ≤ 0.
180 Holt McDougal Algebra 1
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SolvIng SyStEmS of lInEAr InEQuAlItIES
CHECK IT OUT!
1a. _____________ y < -3x + 2 1 -3(0) + 2 1 < 2 3
__________ y ≥ x - 1 1 0 - 1 1 ≥ -1 3
(0, 1) is a solution of the system.
b. ___________ y > -x + 1 0 -0 + 1 0 > 1 7
__________ y > x - 1 0 0 - 1 0 > -1 3
(0, 0) is not a solution of the system.
2a.
-2
x
y
0 2
Possible answer: solutions: (3, 3), (4, 4); not solutions: (-3, 1), (-1, -4)
b. 3x + 6y ≤ 12 ________ -3x ____ -3x 6y ≤ -3x + 12
6y
___ 6 ≤ -3x + 12 ________
6
y ≤ - 1 __ 2 x + 2
4
-4
x
y
0 4 -4
Possible answer: solutions: (0, 0), (3, -2); not solutions: (4, 4), (1, -6)
3a.
2
-2
x
y
0
b.
2
x
y
2 -2
c.
-2
x
y
0 -2
4. Let x represent the number of pounds of pepper jack cheese and let y represent the number of pounds of cheddar cheese.
x ≥ 2 y ≥ 2 4x + 2y ≤ 20 The graph should only be in the first quadrant
because the number of pounds of cheese cannot be negative.
Cheese Combinations
0 2 4 6 8
2
4
6
8
Pepper jack cheese (lb)
Ched
dar
chee
se (l
b)
Alice could buy any combination represented by an ordered pair in the solution region. Answers do not need to be whole numbers because she can buy a real number of pounds of cheese.
Possible answer: (3 lb pepper jack, 2 lb cheddar), (2.5 lb pepper jack, 4 lb cheddar)
THINK AND DISCUSS
1. Possible answer: To write a system of linear inequalities from a graph, write the linear inequality for each graph that makes up the system.
2.
-4
y
0 -2 -4
Possible answer: (-4, -3)
Possible answer: (0, 0)
-4
x
y
0 -2 -4
y ≥ 2 x + 1
y > 1 __ 2 x - 2
y < 2 x + 1
y ≥ 1 __ 2 x - 2
ExErCISESGUIDED PRACTICE
1. all
2. ___________ y < -x + 3 0 -0 + 3 0 < 3 3
__________ y < x + 2 0 0 + 2 0 < 2 3
(0, 0) is a solution of the system.
3. _______ y < 3 0 < 3 3
__________ y > x - 2 0 0 - 2 0 > -2 3
(0, 0) is a solution of the system.
4. ________ y > 3x 0 3(1) 0 > 3 7
__________ y ≤ x + 1 0 1 + 1 2 ≤ 1 7
(1, 0) is not a solution of the system.
181 Holt McDougal Algebra 1
5-6
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5. 4
x
y
0 2 4 -2
Possible answer: solutions: (3, 3), (4, 3); not solutions: (0, 0), (2, 1)
6.
2
4
-2
x
y
0 2 4 -2
Possible answer: solutions: (0, 0), (1, 1); not solutions: (2, -1), (3, 1)
7. 3x + y ≥ 3 _______ -3x ____ -3x y ≥ -3x + 3
2
x
y
0 2 -2
Possible answer: solutions: (0, 4), (1, 4); not solutions: (2, -1), (3, 1)
8. 2x - 4y ≤ 8 ________ -2x ____ -2x -4y ≤ -2x + 8
-4y
____ -4
≥ -2x + 8 _______ -4
y ≥ 1 __ 2 x - 2
-4
x
y
0 4
Possible answer: solutions: (0, 0), (1, 1); not solutiosn: (3, 0), (-3, -4)
9. no solutions
x
y
0 2
10. no solutions
-2
x
y
0 2 -2
11. all points between the parallel lines and on the solid line
2
x
y
2 -2
12. all points between the parallel lines
2
x
y
0 2
13. same solutions as y > 2x - 1
2
-4
x
y
0 2 -2
14. same solutions as y ≤ -3x - 3
4
-4
x
y
-2
2
15. Let x represent the number of cups of lemonade and let y represent the number of cupcakes.
x ≥ 5 y ≥ 5 2x + y ≥ 25 The graph should only be in the first quadrant
because sales are not negative.Sales Goals
0 8 16 24
8
16
24
Lemonade (c)
Cupc
akes
(6, 13)
(10, 10)
To meet the sales goal, Sandy could sell any combination represented by an ordered pair of whole numbers in the solution region. Answer must be whole numbers because Sandy cannot sell a part of a cup of lemonade or cupcake.
Possible answer: (6, 13), (10, 10)
PRACTICE AND PROBLEM SOLVING
16. ___________ y ≥ -x - 1 0 -0 - 1 0 ≥ -1 3
____________ y ≤ 2x + 4 0 2(0) + 4 0 ≤ 4 3
(0, 0) is a solution of the system.
17. ___________ x + y < 3 0 + 0 3 0 < 3 3
____________ y > 3x - 4 0 3(0) - 4 0 > -4 3
(0, 0) is a solution of the system.
18. ________ y > 3x 0 3(1) 0 > 3 7
____________ y > 3x + 1 0 3(1) + 1 0 > 4 7
(1, 0) is not a solution of the system.
19. 2
-2
x
y
0 2 -2
Possible answer: solutions (-2, 0), (-3, 1); not solutions: (0, 0), (1, 4)
182 Holt McDougal Algebra 1
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20. 2
x
y
0
Possible answer: solutions (-2, -2), (-3, -2); not solutions: (0, 0), (1, -4)
21. 6x + 2y ≥ -2 ________ -6x ____ -6x 2y ≥ -6x - 2
2y
___ 2 ≥ -6x - 2 _______
2
y ≥ -3x - 1
4
x
y
0 2
Possible answer: solutions (-1, 3), (0, 5); not solutions: (0, 0), (1, 4)
22. 9x + 3y ≤ 6 ________ -9x ____ -9x 3y ≤ -9x + 6
3y
___ 3 ≤ -9x + 6 _______
3
y ≤ -3x + 2
2
-2
x
y
2 -2
Possible answer: solutions (-2, 0), (-3, 1); not solutions: (0, 0), (1, 4)
23. no solutions
2
4
x
y
0 2 -2
24. all points between the parallel lines
-4
x y
0 2 -2
25. all points are solutions
2
-2
x
y
0 -2
26. all points between the parallel lines
2
-2
x
y
2 -2
27. same solutions as y > 2
-2
x
y
0
28. same solutions as y ≤ 2x - 4
-4
x
y
0 2 4 -2
29. Let x represent the number of hours at the pharmacy and let y represent the number of hours baby-sitting.
15x + 10y ≥ 90 x + y ≤ 20 The graph should only be in the first quadrant
because time is not negative.
Linda’s Work Hours
0 6 12 18
6
12
18
Hours at pharmacy
Hou
rs b
aby-
sitt
ing
(8.5, 10)
(0, 9)
To meet her goals, Linda could work for any combination represented by an ordered pair in the solution region. Answers do not need to be whole numbers because she can work a real number of hours.
Possible answer: (0, 9), (8.5, 10)
30. Let x represent the number of acres of corn and let y represent the number of acres of soybeans.
x ≥ 40 y ≥ 50 x + y ≤ 200 The graph should only be in the first quadrant
because Tony cannot plant a negative number of acres.
Planting Combinations
0 50 100 150 200
50
100
150
200
Corn (acres)
Soyb
eans
(acr
es)
To meet his goals, Tony could plant any combination represented by an ordered pair in the solution region. Answers do not need to be whole numbers because he can plant a real number of acres.
Possible answer: (60, 80), (100, 60)
31.
-2
x
y
0 2 -2
32. 2
x
y
2 -2
183 Holt McDougal Algebra 1
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33. 2
-2
x
y
0 2 -2
34.
2
x
y
0 2 -2
35. ⎧
⎨ ⎩ y > x + 1
y < x + 3
36. ⎧
⎨ ⎩ y ≥ 1
y ≤ 3
37. ⎧
⎨ ⎩ y < 2
x ≥ -2
38. Let x represent the ages in years and let y represent the heights in inches.
x ≥ 17 x < 23 y ≥ 60 y ≤ 80 The graph should only be in the first quadrant
because time and height cannot be negative.
0 10 20 30 40
20
40
60
80
Age (yr)
Hei
ght
(in.)
Possible answer: (20, 70), (19, 75), (18, 69)
39. Student B; the inequality symbols are incorrect.
40. Let x represent the length of the area and let y represent the width of the area.
x ≥ 30 2x + 2y ≤ 150 The graph should only be in the first quadrant
because length is not negative.Dimensions for Dog Area
0 20 40 60
20
40
60
Width (ft)
Leng
th (f
t)
41. Yes; the solutions of the system ⎧
⎨ ⎩ y ≥ x + 4
y ≤ x + 4
are
represented by all ordered pairs on the line y = x + 4.
42a. Let x represent the number of girl bears and let y represent the number of boy bears.
x + y ≤ 40
b. 15x + 12y ≥ 540 c. Teddy Bear Combinations
0 12 24 36
12
24
36
Girl bears
Boy
bear
s
43. The boundary lines must be parallel. If the boundary lines are not parallel, there will be overlap.
TEST PREP
44. D; Since 2(1) + 1 = 3 ≥ 3 and 1 ≥ -1 = -2(1) + 1, (1, 1) is a solution of both inequalities and therefore, is a solution of the system.
45. G; The solution region is between the two lines. Therefore, the solutions lie below y = 2x + 1 and above y = 2x - 3.
46. y + x > 2 ____ - x ___ -x y > -x + 2
2
4
x
y
0 2 -2
Possible answer: (0, 3), (-2, 6)
ChALLENGE AND ExTEND
47.
2
x
y
0 2 -2
The solution region is a triangle. The height is about 3.5 units. The base is about 7 units.
A = 1 __ 2 bh
≈ 1 __ 2 (7)(3.5)
≈ 12 The answer is approximately 12 square units.
48. Possible answer: ⎧ ⎪
⎨
⎪ ⎩ y ≥ 3 __
2 x + 5 __
2
y > - 1 __ 2 x
184 Holt McDougal Algebra 1
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49. |y| < 1 y < 1 AND y > -1 Graph y < 1 and y > -1. The solutions of |y| < 1 lie
in the solution region.
2
-2
x
y
0
50. ⎧
⎨ ⎩ x < 0
y < 0
rEAdy to go on? Section b Quiz
1. _______________ y < -2x + 1 -2 -2(3) + 1 -2 -6 + 1 -2 < -5 7 (3, -2) is not a solution.
2. ____________ y ≥ 3x - 5 1 3(2) - 5 1 6 - 5 1 ≥ 1 3 (2, 1) is a solution.
3. _______________ y ≤ 4x - 10 -6 4(1) - 10 -6 4 - 10 -6 ≤ -6 3 (1, -6) is a solution.
4. Graph the boundary line y = 4x - 3. Use a solid line for ≥. The inequalitiy is ≥, so shade above the line.
-4
-2
x y
0 2 -2
5. 3x - y < 5 _______ -3x ____ -3x -y < -3x + 5 -1(-y) > -1(-3x + 5) y > 3x - 5 Graph the boundary line y = 3x - 5. Use a dashed
line for >. The inequalitiy is >, so shade above the line.
-4
-2
x y 0 2 -2
6. 2x + 3y < 9 ________ -2x ____ -2x 3y < -2x + 9
3y
___ 3 < -2x + 9 _______
3
y < - 2 __ 3 x + 3
Graph the boundary line y = - 2 __ 3
x + 3. Use a
dashed line for <. The inequalitiy is <, so shade below the line.
2
-2
x
y
0 4 2
7. Graph the boundary line y = - 1 __ 2
x. Use a solid line
for ≤. The inequalitiy is ≤, so shade below the line.
4
2
x
y
0 4
8. Let x represent the number of pairs of pants and let y represent the number of shirts.
30x + 15y ≤ 150 __________ -30x _____ -30x 15y ≤ -30x + 150
15y
____ 15
≤ -30x + 150 __________ 15
y ≤ -2x + 10 Since Theo cannot buy a negative number of
clothes, the system is graphed only in Quadrant I. Graph the boundary line y = -2x + 10. Use a solid line for ≤. The inequality is ≤, so shade below the line.
Possible Clothing Combinations
Pants
0
4
8
Shir
ts
12
6 4 2
Possible answer: (4, 2), (3, 2), (2, 6)
9. y-intercept: -5; slope: 2 y = mx + b → y = 2x - 5 The graph is shaded below a dashed boundary line.
Replace = with < to write the inequality y < 2x - 5.
10. y-intercept: none; slope: undefined The graph is a vertical line at x = -3. The graph is shaded on the right side of a dashed
boundary line. Replace = with > to write the inequality x > -3.
185 Holt McDougal Algebra 1
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11. y-intercept: 0; slope: -1 y = mx + b → y = -x The graph is shaded above a solid boundary line.
Replace = with ≥ to write the inequality y ≥ -x.
12. ___________ y > -2 -1 > -2 3
_____________ y < x + 4 -1 -3 + 4 -1 < 1 3
(-3, -1) is a solution of the system.
13. ___________ y ≤ x + 4 0 -3 + 4 0 ≤ 1 3
_______________ y ≥ -2x - 6 0 -2(-3) - 6 0 6 - 6 0 ≥ 0 3
(-3, 0) is a solution of the system.
14. ________ y ≥ 3x 0 3(0) 0 ≥ 0 3
_____________ 2x + y < -1 2(0) + 0 -1 0 + 0 -1 0 < -1 7
(0, 0) is not a solution of the system.
15.
-3
x
y
0 -2 -6
Possible answer: solutions: (0, 0), (2, 2); not solutions: (-6, 0), (-4, 4)
16. x + y ≤ 2 ______ -x ___ -x y ≤ -x + 2
2x + y ≥ -1 _______ -2x ____ -2x y ≥ -2x - 1
6
-6
x
y
0 6 -6
Possible answer: solutions: (2, 0), (2, -2); not solutions: (6, 0), (-4, 0)
17. 2x - 5y ≤ -5 ________ -2x ____ -2x -5y ≤ -2x - 5
-5y
____ -5
≥ -2x - 5 _______ -5
y ≥ 2 __ 5 x + 1
3x + 2y < 10 ________ -3x ____ -3x 2y < -3x + 10
2y
___ 2 < -3x + 10 ________
2
y < - 3 __ 2 x + 5
4
2
x
y
0 2 -2
Possible answer: solutions: (-6, 6), (-5, 0); not solutions: (2, 0), (4, 8)
18.
2
-2
x
y
2 -3
The solutions of the system are the solutions of y ≥ x + 1.
19. 2
x
y
2 -2
The system has no solutions.
20.
2
-2
x
y
3 -3
The solutions are between the graphs of y = -3x + 5 and y = -3x - 2.
21. Let x represent the number of pounds of mangos and let y represent the number of pounds of apples.
x ≤ 45 y ≤ 50 4x + 3y ≥ 300 The graph should only be in the first quadrant
because sales cannot be negative.
Fruit Combinations
Mangos (lb)
0
28
56
Appl
es (l
b) 84
36 24 12
To meet his goals, the grocer could sell any combination represented by an ordered pair in the solution region. Answers do not need to be whole numbers because he can sell real number pounds of fruit.
Possible answer: (45, 40), (42, 50)
Study guIdE: rEvIEw
1. independent system 2. system of linear equations
3. solution of a system of linear inequalities
4. inconsistent system 5. independent system
186 Holt McDougal Algebra 1
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SOLvINg SySTEmS by grApHINg
6. ________________ y = -6x + 5 (-5) -6(0) + 5 -5 5 7
______________ x - y = 5 (0) - (-5) 5 5 5 3
(0, -5) is not a solution of the system.
7. ________________ x - 2y = -2 (4) - 2(3) -2 4 - 6 -2 -2 -2 3
____________
y = 1 __ 2 x + 1
(3) 1 __ 2 (4) + 1
3 2 + 1 3 3 3
(4, 3) is a solution of the system.
8. ________________ x + y = 9
(1 3 __ 4 ) + (7 1 __
4 ) 9
9 9 3
________________ 2y = 6x + 4
2 (7 1 __ 4 ) 6 (1 3 __
4 ) + 4
2 ( 29 ___ 4 ) 6 ( 7 __
4 ) + 4
29 ___ 2 21 ___
2 + 8 __
2
29 ___ 2 29 ___
2 3
(1 3 __ 4 , 7 1 __
4 ) is a solution of the system.
9. ________________ y = -2x + 5 (-1) -2(-1) + 5 -1 2 + 5 -1 7 8
________________ 3y = 6x + 3 3(-1) 6(-1) + 3 -3 -6 + 3 -3 -3 3
(-1, -1) is not a solution of the system.
10. 5
-4
x
y
0 4 -5
y = 3x + 2
y = -2x - 3
(-1, -1)
The solution appears to be at (-1, -1).
11. 2x - 2y = -2 ________ -2x ____ -2x -2y = -2x - 2
-2y
____ -2
= -2x - 2 _______ -2
y = x + 1
7
7
x
y
0
y = - 1 _ 3
x + 5
y = x + 1
(3, 4)
The solution appears to be at (3, 4).
12. Let x represent the number of hours and let y represent the cost.
Garage A: y = 0.50x + 6 Garage B: y = x + 2 Graph y = 0.5x + 6 and y = x + 2. Parking Cost
0 2 4 6
4
8
12
16
2
6
10
14
Time (hours)
Cost
($) Garage A
Garage B
(8, 10)
The lines appear to intersect at (8, 10). The cost at both places will be the same for 8 hours of parking and that cost will be $10.
SOLvINg SySTEmS by SUbSTITUTION
13. y = x + 3 2x + 12 = x + 3 _______ -x ______ -x x + 12 = 3 ______ - 12 ____ -12 x = -9
y = x + 3 y = -9 + 3 y = -6 (-9, -6)
14. y = -4x 2x - 3 = -4x _______ -2x ____ -2x -3 = -6x
-3 ___ -6
= -6x ____ -6
1 __ 2
= x
y = -4x
y = -4 ( 1 __ 2
)
y = -2
( 1 __ 2 , -2)
15. 2x + y = 4 _______ -2x ____ -2x y = -2x + 4
3x + y = 3 3x + (-2x + 4) = 3 x + 4 = 3 _____ - 4 ___ -4 x = -1
2x + y = 4 2(-1) + y = 4 -2 + y = 4 ______ +2 ___ +2 y = 6 (-1, 6)
187 Holt McDougal Algebra 1
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16. x + y = -1 x + (-2x + 3) = -1 -x + 3 = -1 ______ - 3 ___ -3 -x = -4 -1(-x) = -1(-4) x = 4
y = -2x + 3 y = -2(4) + 3 y = -8 + 3 y = -5 (4, -5)
17. -y - 2x = 8 -y - 2(y - 7) = 8 -y - 2(y) - 2(-7) = 8 -y - 2y + 14 = 8 -3y + 14 = 8 ________ - 14 ____ -14 -3y = -6
-3y
____ -3
= -6 ___ -3
y = 2
x = y - 7 x = 2 - 7 x = -5 (-5, 2)
18. 1 __ 2 x + y = 9
________
- 1 __ 2 x
____ - 1 __
2 x
y = - 1 __ 2 x + 9
3x - 4y = -6
3x - 4 (- 1 __ 2 x + 9) = -6
3x - 4 (- 1 __ 2 x) - 4(9) = -6
3x + 2x - 36 = -6 5x - 36 = -6 _______ + 36 ____ +36 5x = 30
5x ___ 5 = 30 ___
5
x = 6
1 __ 2 x + y = 9
1 __ 2 (6) + y = 9
3 + y = 9 ______ -3 ___ -3 y = 6 (6, 6)
19. Let x represent the number of hours and let y represent the total cost.
Motor Works: y = 70x + 650 Jim’s Car Care: y = 55x + 800
y = 70x + 650 55x + 800 = 70x + 650 __________ -55x __________ -55x 800 = 15x + 650 _____ -650 ________ - 650 150 = 15x
150 ____ 15
= 15x ____ 15
10 = x
y = 70x + 650 y = 70(10) + 650 y = 700 + 650 y = 1350 In 10 hours, the total cost for each garage will be the
same, $1350. Motor Works; 8 hours will cost $30 less at Motor
Works.
SOLvINg SySTEmS by ELImINATION
20. 4x + y = -1 _____________ + 2x - y = -5 6x + 0 = -6 6x = -6
6x ___ 6 = -6 ___
6
x = -1
4x + y = -1 4(-1) + y = -1 -4 + y = -1 ______ +4 ___ +4 y = 3 (-1, 3)
21. x + 2y = -1 _____________ -(x + y) = -2
x + 2y = -1 ______________ + (-x - y = -2) 0 + y = -3 y = -3
x + y = 2 x - 3 = 2 ____ + 3 ___ +3 x = 5 (5, -3)
22. (-2)(x + y = 12) _____________ + 2x + 5y = 27
-2x - 2y = -24 _____________ + 2x + 5y = 27 0 + 3y = 3 3y = 3
3y
___ 3 = 3 __
3
y = 1
x + y = 12 x + 1 = 12 ____ - 1 ___ -1 x = 11 (11, 1)
188 Holt McDougal Algebra 1
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23. 3x - 2y = -6
___________________
+ (-9) ( 1 __ 3 x + 3y = 9)
3x - 2y = -6 __________________ + (-3x - 27y = -81) 0 - 29y = -87 -29y = -87
-29y
_____ -29
= -87 ____ -29
y = 3
3x - 2y = -6 3x - 2(3) = -6 3x - 6 = -6 ______ + 6 ___ +6 3x = 0
3x ___ 3 = 0 __
3
x = 0 (0, 3)
24. 3x + y = 2 3x + (-4x) = 2 -x = 2 -1(-x) = -1(2) x = -2
y = -4x y = -4(-2) y = 8 (-2, 8); possible answer: substitution; the second
equation is already solved for y, and y has a coefficient of 1 in the first equation.
25.
1
- 7
7 x y
y = -2x + 1
y = 1 _ 3
x - 6 (3, -5)
(3, -5); possible answer: graphing; both equations are already in slope-intercept form.
26. 2y = -3x 2(-2x + 2) = -3x 2(-2x) + 2(2) = -3x -4x + 4 = -3x _______ +4x ____ +4x 4 = x
y = -2x + 2 y = -2(4) + 2 y = -8 + 2 y = -6 (4, -6); possible answer: substitution; the second
equation is already solved for y.
27. x - y = 0 ___________ + 3x + y = 8 4x + 0 = 8 4x = 8
4x ___ 4 = 8 __
4
x = 2
3x + y = 8 3(2) + y = 8 6 + y = 8 ______ -6 ___ -6 y = 2 (2, 2); possible answer: elimination; the coefficients
of the y-terms are opposites.
SOLvINg SpECIAL SySTEmS
28. This system is inconsistent. It has no solution.
29. y = -x + 4 → y = -x + 4 x + y = 4 → y = -x + 4 This system is consistent and dependent. It has
infinitely many solutions.
30. This system is consistent and independent. It has one solution.
y = 2x 3x + 2 = 2x _______ -3x ____ -3x 2 = -x -1(2) = -1(-x) -2 = x
y = 2x y = 2(-2) y = -4 (-2, -4)
31. -4x - y = 6 → y = -4x - 6
1 __ 2 y = -2x - 3 → y = -4x - 6
This system is consistent and dependent. It has infinitely many solutions.
32. x + 2y = 8 → y = - 1 __ 2 x + 4
y = - 1 __ 2 x + 4 → y = - 1 __
2 x + 4
This system is consistent and dependent. It has infinitely many solutions.
189 Holt McDougal Algebra 1
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33. y - 2x = -1 → y = 2x - 1 y + 2x = -5 → y = -2x - 5 This system is consistent and independent. It has
one solution. y - 2x = -1 _____________ + y + 2x = -5 2y + 0 = -6 2y = -6
2y
___ 2
= -6 ___ 2
y = -3
y + 2x = -5 -3 + 2x = -5 _______ +3 ___ +3 2x = -2
2x ___ 2 = -2 ___
2
x = -1 (-1, -3)
34. Yes. Let x represent the number of months after month 1 and let y represent the number of DVDs.
Tristan: y = 5x + 2 Marco: y = 4x + 8 The graphs of the two equations have different
slopes so they intersect. If the pattern continues, Tristan will eventually have the same number of DVDs as Marco. In fact, at month 7 (x = 6), Tristan and Marco will have 32 DVDs each.
35. This system is consistent and independent. It has one solution.
36. This system is inconsistent. It has no solution.
37. 2x + y = 2 → y = -2x + 2 y - 2 = -2x → y = -2x + 2 This system is consistent and dependent. It has
infinitely many solutions.
38. -3x - y = -5 → y = -3x + 5 y = -3x - 5 → y = -3x - 5 This system is inconsistent. It has no solution.
39. 2x + 3y = 1 → y = - 2 __ 3 x + 1 __
3
3x + 2y = 1 → y = - 3 __ 2 x + 1 __
2
This system is consistent and independent. It has one solution.
40. x + 1 __ 2 y = 3 → y = -2x + 6
2x = 6 - y → y = -2x + 6 This system is consistent and dependent. It has
infinitely many solutions.
41. This system is inconsistent. It has no solution.
SOLvINg LINEAr INEqUALITIES
42. ______________ y < 2x - 3 -3 2(0) - 3 -3 < -3 7 (0, -3) is not a solution.
43. ___________ y ≥ x - 3 -1 2 - 3 -1 ≥ -1 3 (2, -1) is a solution.
44. _____________ y > -3x + 4 0 -3(6) + 4 0 -18 + 4 0 > -14 3 (6, 0) is a solution.
45. ___________ y ≤ x - 3 10 10 - 3 10 ≤ 7 7 (10, 10) is not a
solution.
46. Graph the boundary line y = -2x + 5. Use a dashed line for <. The inequality is <, so shade below the line.
2
-2
x
y
0 4 2
47. x - y ≥ 2 ______ -x ___ -x -y ≥ -x + 2 -1(-y) ≤ -1(-x + 2) y ≤ x - 2 Graph the boundary line y = x - 2. Use a solid line
for ≤. The inequality is ≤, so shade below the line.
2
-2
x
y
0 2 -2
48. -x + 2y ≥ 6 _______ +x ___ +x 2y ≥ x + 6
2y
___ 2 ≥ x + 6 _____
2
y ≥ 1 __ 2 x + 3
Graph the boundary line y = 1 __ 2 x + 3. Use a solid
line for ≥. The inequality is ≥, so shade above the line.
4
2
x
y
0 -4 -6 -2
49. Graph the boundary line y = -4x. Use a dashed line for >. The inequality is >, so shade above the line.
-2
x
y
0 2 -2
190 Holt McDougal Algebra 1
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50. x + y + 4 > 0 __________ -x ___ -x y + 4 > -x _____ - 4 ___ -4 y > -x - 4 Graph the boundary line y = -x - 4. Use a dashed
line for >. The inequality is >, so shade above the line.
-4
-2
x y
0 -4 -2
51. 5 - y ≥ 2x ______ -5 ___ -5 -y ≥ 2x - 5 -1(-y) ≤ -(2x - 5) y ≤ -2x + 5 Graph the boundary line y = -2x + 5. Use a
dashed line for ≤. The inequality is ≤, so shade below the line.
4
2
x
y
0 2 -2
52. Let x represent the number of slices of pizza and let y represent the number of bottles of soda.
2x + y ≥ 450 _______ -2x ____ -2x y ≥ -2x + 450 Since the club cannot sell a negative number of
items, the system is graphed only in Quadrant I. Graph the boundary line y = -2x + 450. Use a solid line for ≥. The inequality is ≥, so shade above the line.
Fundraising Needs
Slices of pizza
0
150
300
Bott
les
of le
mon
ade
450
450 300 150
Possible answer: (200, 50), (150, 150)
SOLvINg SySTEmS Of LINEAr INEqUALITIES
53. ______________ y > -2x + 9 3 -2(3) + 9 3 -6 + 9 3 > 3 7
________ y ≥ x 3 ≥ 3 3
(3, 3) is not a solution of the system.
54. ________________ 2x - y > -5 2(-1) - 0 -5 -2 > -5 3
_______________ y ≤ -3x - 3 0 -3(-1) - 3 0 3 - 3 0 ≤ 0 3
(-1, 0) is a solution of the system.
55.
4
-4
x
y
0 4 -4
Possible answer: solutions: (-6, 6), (-10, 0); not solutions: (0, 0), (4, -4)
56.
4
-4
x
y
0 4
Possible answer: solutions: (0, 0), (-5, 0); not solutions: (8, 0), (3, -3)
57. -x + 2y > 6 _______ +x ___ +x 2y > x + 6
2y
___ 2 > x + 6 _____
2
y > 1 __ 2 x + 3
x + y < 4 ______ -x ___ -x y < -x + 4
6
-6
x
y
0 4 2
Possible answer: solutions: (-6, 2), (-8, 1); not solutions: (0, 0), (4, 1)
58. x - y > 7 _______ -x ___ -x -y > -x + 7 -1(-y) < -1(-x + 7) y < x - 7
x + 3y ≤ 15 _______ -x ___ -x 3y ≤ -x + 15
3y
___ 3
≤ -x + 15 _______ 3
y ≤ - 1 __ 3
x + 5
4
-4
y
0 8 4
Possible answer: solutions: (8, -8), (9, 0); not solutions: (0, 0), (0, -4)
59. 4
-6
x
y
4 -6 0
191 Holt McDougal Algebra 1
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60. 4x + 2y ≥ 10 ________ -4x ____ -4x 2y ≥ -4x + 10
2y
___ 2
≥ -4x + 10 ________ 2
y ≥ -2x + 5
6x + 3y < -9 ________ -6x ____ -6x 3y < -6x - 9
3y
___ 3 < -6x - 9 _______
3
y < -2x - 3
4
-6
x
y
6 -6
chAptEr tESt
1. ____________ y = -4x (-4) -4(1) -4 -4 3
_____________ y = 2x - 2 (-4) 2(1) - 2 -4 0 7
(1, -4) is not a solution of the system.
2. _______________ 3x - y = 1 3(0) - (-1) 1 0 + 1 1 1 1 3
_______________ x + 5y = -5 0 + 5(-1) -5 -5 -5 3
(0, -1) is a solution of the system.
3. _______________ x - 2y = -1 (3) - 2(2) -1 3 - 4 -1 -1 -1 3
________________ -3x + 2y = 5 -3(3) + 2(2) 5 -9 + 4 5 -5 5 7
(3, 2) is not a solution of the system.
4. x
y
0 6 -2
-6
(0,-3)
y = x - 3
y = -2x - 3
The solution appears to be at (0, -3).
5. 2x + y = -8 _______ -2x ____ -2x y = -2x - 8
x y 0 4 -2
-8
(-3, -2)
y = -2x - 8
y = 1 _ 3 x - 1
The solution appears to be at (-3, -2).
6. x = y + 2 ___ - 2 _____ - 2 x - 2 = y
x
y
0
6
6 -2
-2
(3, 1)
y = -x + 4
y = x - 2
The solution appears to be at (3, 1).
7. y = -6
y = -2x - 2 -6 = -2x - 2 ___ +2 _______ + 2 -4 = -2x
-4 ___ -2
= -2x ____ -2
2 = x (2, -6)
8. -x + y = -4 -x + (2x - 11) = -4 x - 11 = -4 ______ + 11 ____ +11 x = 7
y = 2x - 11 y = 2(7) - 11 y = 14 - 11 y = 3 (7, 3)
9. x - 3y = 3 _____ + 3y ____ +3y x = 3y + 3
2x = 3y 2(3y + 3) = 3y 2(3y) + 2(3) = 3y 6y + 6 = 3y _______ -6y ____ -6y 6 = -3y
6 ___ -3
= -3y
____ -3
-2 = y
x - 3y = 3 x - 3(-2) = 3 x + 6 = 3 _____ - 6 ___ -6 x = -3 (-3, -2)
10. Let x represent the number of days and let y represent the total cost.
Pet Care: y = 30x + 15 Fido’s: y = 28x + 27
y = 28x + 27 30x + 15 = 28x + 27 _________ -28x _________ -28x 2x + 15 = 27 _______ - 15 ____ -15 2x = 12
2x ___ 2 = 12 ___
2
x = 6
y = 30x + 15 y = 30(6) + 15 y = 180 + 15 y = 195 In 6 days, the total cost for each service will be the
same, $195. Fido’s; it will cost less per day after 6 days.
192 Holt McDougal Algebra 1
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11. 3x - y = 7 ___________ + 2x + y = 3 5x + 0 = 10 5x = 10
5x ___ 5 = 10 ___
5
x = 2
2x + y = 3 2(2) + y = 3 4 + y = 3 ______ -4 ___ -4 y = -1 (2, -1)
12. 4x + y = 0 ___________ -(x + y) = 3
4x + y = 0 _____________ + (-x - y = 3) 3x + 0 = 3 3x = 3
3x ___ 3 = 3 __
3
x = 1
x + y = -3 1 + y = -3 ______ -1 ___ -1 y = -4 (1, -4)
13. 2(2x + y = 3) _____________ + x - 2y = -1
4x + 2y = 6 ____________ + x - 2y = -1 5x + 0 = 5 5x = 5
5x ___ 5 = 5 __
5
x = 1
2x + y = 3 2(1) + y = 3 2 + y = 3 ______ -2 ___ -2 y = 1 (1, 1)
14. y = 6x - 1 → y = 6x - 1 6x - y = 1 → y = 6x - 1 This system is consistent and dependent. It has
infinitely many solutions.
15. y = -3x - 3 → y = -3x - 3 3x + y = 3 → y = -3x + 3 This system is inconsistent. It has no solution.
16. 2x - y = 1 → y = 2x - 1 -4x + y = 1 → y = 4x + 1 This system is consistent and independent. It has
one solution.
17. Graph the boundary line y = 2x - 5. Use a dashed line for <. The inequality is <, so shade below the line.
-4
-2
x y
0 3 -3
18. -y ≥ 8 -1(-y) ≤ -1(8) y ≤ -8 Graph the boundary line y = -8. Use a solid line for
≤. The inequality is ≤, so shade below the line.
2
-10
-4
x y
0 2 -2
19. Graph the boundary line y = 1 __ 3 x. Use a dashed line
for >. The inequality is >, so shade above the line.
2
-2
x
y
0 2
20.
-10
x y
0 2 -2
Possible answer: solutions: (2, 0), (4, 0); not solutions: (-2, 0), (-4, 0)
21. 3x - y > 3 _______ -3x ____ -3x -y > -3x + 3 -1(-y) < -1(-3x + 3) y < 3x - 3
2
-2
x
y
0 2 -2
Possible answer: solutions: (6, 4), (9, 0); not solutions: (2, -4), (0, 0)
22. y - 2x < 6 _____ + 2x ____ +2x y < 2x + 6
6
-6
x
y
6 -6
Possible answer: solutions: (0, 1), (0, 2); not solutions: (-6, 0), (4, 1)
193 Holt McDougal Algebra 1
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23. Let x represent the number of coupons Tara sold and let y represent the number of coupons Erza sold.
x + y ≥ 150 y - 2x ≤ 30 The graph should only be in the first quadrant
because sales cannot be negative.
0
40
80
120
60 40 20
Coupon Book Sales
Tara
Ezra
Erza and Tara could have sold any combination represented by an ordered pair of whole numbers in the solution region.
Possible answer: (50, 125), (65, 95)
194 Holt McDougal Algebra 1
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