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)