math 140 quiz 5 - summer 2006 solution review (small white numbers next to problem number represent...
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Math 140Quiz 5 - Summer 2006
Solution Review
(Small white numbers next to problem number represent its difficulty as per cent getting it wrong.)
Problem 1 (05)
Solve the system. (7/3)x + (5/4)y = 4 (a)
(5/6)x - 2y = 21 (b) Using the method of substitution, select an equation, say (b), to solve for a selected variable, say, y:
y = (5/12)x - (21/2). (c)
Then, substitute this for y in (a) & solve for x:
(7/3)x + (5/4) [(5/12)x - (21/2)] = 4
Hence, x = (4+105/8)/[(7/3)+(5/4) (5/12)] = 6.
Put this in (c) to get: y = (5/12)(6) - (21/2) = -8. (6, -8)
Problem 2 (33)
Determine the number of solutions for the given system without solving the system.
4x - 3y = 5 (a)
16x - 12y = 20 (b)
Replace (b) by (b) minus 4 times (a): 0 = 0.
This is indicates a consistent system with an infinite number of solutions.
Check by noting (b) is a line of same slope (4/3) and y-intercept (–5/3) as (a). Thus, it is the same line.
Problem 3 (33)
Determine the number of solutions for the given system without solving the system.
3x + 3y = -2 (a)
12x + 12y = 3 (b)
Replace (b) by: (b) minus 4 times (a) => 0 = 11 !!!
This is indicates an inconsistent system with no solution.
Note: (a) & (b) are lines of same slope (-4/3) but differing y-intercepts (2/3) & (-1/4). Thus, they are parallel lines and do not intersect.
Problem 4 (19)
Use a calculator to solve the system of equations.
y = 2.12x - 31 (a)
y = -0.7x + 24 (b)
Substitute for y in (a) the y given in (b) & solve for x.
Then, from (b): y = -0.7(33.09153) + 24 = 0.83593
Problem 5 (43) Solve: A twin-engine aircraft can fly 1190 miles
from city A to city B in 5 hours with the wind and make the return trip in 7 hours against the wind. What is the speed of the wind?
Let speeds be p for plane & w for wind. Then,
5(p + w) = 1190 (a)
7(p - w) = 1190 (b)
After dividing (a) by 5 & (b) by 7, subtract them.
2w = 238 – 170 = 68
w = 34 miles per hour
Problem 6 (24) Solve the system of equations.
Use augmented matrix for system & manipulate to row echelon form by row operations.
752
2945
842
zyx
zyx
zyx
7512
29145
8142
1430
49140
421
2
72
1
2
19
4
192
7
4
12
1
00
10
421
R3 = -r1 + r3, R1 = r1/2,
R2 = -5r1 + r2
R2 = -r2/14,
R3 = 3r2 + r3
Problem 6 cont’d (24) Solve the system of equations.
Use augmented matrix for system & manipulate to row echelon form by row operations.
x = -4 - z/2 - 2y = -4 - (-2)/2 -2(-3) = 3
752
2945
842
zyx
zyx
zyx
2100
10
421
2
7
4
12
1
z = -2
y = -7/2 - z/4 = -7/2 - (-2)/4 = -3
R3 = r3/(14/19)
Problem 7 (0!) Write the augmented matrix for the system.
The augmented matrix for the system is obtained by just copying the coefficients in the standard equations.
13324
9773
28228
zyx
zyx
zyx
13324
9773
28228
Problem 8 (0!) Write a system of equations associated with the
augmented matrix. Do not try to solve.
The standard equations are obtained from the augmented matrix for the system by just copying the coefficients into their places.
983
575
yx
yx
983
575
Problem 9 (29) Perform in order (a), (b), and (c) on the augmented
matrix. (a) R2 = -2r1 + r2
(b) R3 = 4r1 + r3 (c) R3 = 3r2 + r3
6454
5252
2531
6454
11210
2531
1416170
11210
2531
1720140
11210
2531
Problem 10 (19) Find the value of the determinant.
ab
73
babaab
73)7(373
Problem 11 (38) Find the value of the determinant.
143
515
442
19023
515
501
143
515
501
143
515
442
134)23)(5()19)(1(1923
51
143
515
501
143
515
442
Use, e.g., (a) R1 = r1 + r3, (b) R3 = -4r2 + r3, & expand down column 2.
Problem 12 (81) Use the properties of determinants to find the value of
the second determinant, given the value of the first.
Note the matrix in D2 differs from that in D1 only by (a) a row swap R1 = r3, R3 = r1; & (b) by the factor of 3 in row 1 of D2. In the R1 row expansion of D2 these yield a (-1) overall & an overall factor of 3 compared to the R3 row expansion of D1. Details are on next slide. The result is D2 = -3D1 = 12.
4
111
1
wvu
zyx
D ?
333
2
zyx
wvuD
Problem 12 cont’d (81) Use the properties of determinants to find the value of
the second determinant, given the value of the first.
yx
vu
zx
wu
zy
wv
yx
vu
zx
wu
zy
wvD 33332
The R1 row expansion of D2 is:
4
111
1
wvu
zyx
D ?
333
2
zyx
wvuD
yx
vu
zx
wu
zy
wvD 3332
.1233 1
D
vu
yx
wx
zx
wv
zy
Since & above is D1 ’s R3 row expansion.ba
dc
dc
ba
Problem 13 (05) Use Cramer's rule to solve the linear system.
Construct & evaluate the determinants:
18.649 0.267 - 2.498
15.855 0.205 - 2.110
yx
yx
-0.051280.267 - 2.498
0.205 - 2.110D
-0.410240.267 - 18.649
0.205 - 15.855xD 2564.0
18.649 2.498
15.855 2.110yD
Then, x = Dx /D = -0.41024/(-.05128) = 8
y = Dy /D = - 0.2564/(-.05128) = 5
Problem 14 (10) Use Cramer's rule to solve the linear system.
Construct & evaluate the determinants:
10 4
6- 3 2 -
11 - 2 2
zyx
zyx
zyx
3
114
321
122
D 3
1110
326-
1211
xD
15
1104
361
1112
yD
Then, x = Dx /D = 3/3= 1, y = Dy /D = 15/3 = 5,
______________ z = Dz /D = 3/3= 1.
3
1014
621
1122
zD
Problem 15 (10)
Solve the system. x2 + y2 = 100 (a)
x + y = 2 (b) Using the method of substitution, select an equation, preferably (b), to solve for a selected variable, say, y:
y = -x + 2. (c)
Then, substitute this for y in (a) & solve for x:
x2 + (-x + 2)2 = 100 => 2x2 –4x – 96 = 0.
Divide by 2 & factor: (x-8)(x+6)=0 => x = 8 or -6.
Put this in (c): y = -6 or 8. => {(8, -6), (-6, 8)}
Problem 16 (10)
Solve the system. xy = 20 (a)
x + y = 9 (b) Using the method of substitution, select an equation, preferably (b), to solve for a selected variable, say, y:
y = - x + 9. (c)
Then, substitute this for y in (a) & solve for x:
x(- x + 9) = 20 => x2 - 9x + 20 = 0.
Factor: (x – 4)(x – 5) = 0 => x = 4 or 5.
Put this in (c): y = 5 or 4. => {(4, 5), (5, 4)} See graph.
Problem 16 cont’d (10)
Solve the system. xy = 20 (a)
x + y = 9 (b) Solution is: {(4, 5), (5, 4)}.
Zoomed inGraph of (a) & (b)
Problem 17 (10)
Solve the system. x2 + y2 = 25 (a)
x2 – y2 = 25 (b) Using the method of elimination, add & subtract the equations:
2x2 = 50 and 2y2 = 0.
Thus, solving this for x & y:
x = +5 and y = 0.
{(-5, 0), (5, 0)}
Graph of (a) & (b)
Problem 18 (33)
Solve the system. 2x2 + y2 = 66 (a)
x2 + y2 = 41 (b) Using the method of elimination, subtract the equations & back substitute result:
x2 = 25 and 25 + y2 = 41.
Thus, solving this for x & y:
x = +5 and y = + 4.
{(-5, 4), (5, 4), (-5, -4), (5, -4)}
Graph of (a) & (b)
Problem 19 (52)
Solve the system. x2 - xy + y2 = 3 (a)
2x2 + xy + 2y2 = 12 (b) Using the method of elimination, add the equations, simplify, & back substitute result:
x2 + y2 = 5 and xy = 2.
Then, solving this for x & y:
y = 2/x => x2 + (2/x)2 = 5,
x4 - 5x2 + 4 = 0 => x2 = 1 or 4.
{(-1, -2), (1, 2), (-2, -1), (2, 1)} Graph of (a) & (b)
Problem 20 (33) Solve: A rectangular piece of tin has area 736 in.2. A
square of 3 in. is cut from each corner, and an open box is made by turning up the ends and sides. If the volume of the box is 1200 in.3, what were the original dimensions of the piece of tin?
Tin area: hw = 736 (a)
Box volume: 3(h - 6)(w - 6) = 1200 (b)
w h3
3Let sides of tin be h & w. Then,
Problem 20 cont’d (29)
hw = 736 (a)
3(h - 6)(w - 6) = 1200 (b)
Solve (c) for w = 62 - h & substitute in (a),
h (62 -h) = 736 => h2 - 62 h +736 = (h -16) (h -46) =0
So h = 16 or 46 & w = 46 or 16. Tin is 16 in. by 46 in.
After dividing (b) by 3 & expanding (b)’s product
hw - 6w - 6h +36 = 400.
Simplifying this with use of (a) we replace (b) with:
h + w = 62 (c)