Chapter 8 Systems of Equations and Inequalitiesshubbard/pdfs/HA2 Solutions/chapter 8.pdf · Chapter 8: Systems of Equations and Inequalities ... Systems of Linear Equations: ... Systems
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The solution is 8, 2, 0x y z= = = or using ordered triplets (8, 2, 0).
42. 2 42 4 03 2 11
x yy zx z
+ = −⎧⎪− + =⎨⎪ − = −⎩
Multiply each side of the first equation by 2 and add to the second equation to eliminate y: 4 2 8
2 4 0
4 4 8
x yy z
x z
+ = −− + =
+ = −
Multiply each side of the result by 12
and add to
the original third equation to eliminate z: 2 2 43 2 115 15
3
x zx zx
x
+ = −− = −
= −= −
Substituting and solving for the other variables: 2( 3) 4
6 42
yyy
− + = −− + = −
=
3( 3) 2 119 2 11
2 21
zzzz
− − = −− − = −
− = −=
The solution is 3, 2, 1x y z= − = = or using ordered triplets ( 3, 2, 1)− .
43. 2 3 7
2 43 2 2 10
x y zx y zx y z
− + =⎧⎪ + + =⎨⎪− + − = −⎩
Multiply each side of the first equation by –2 and add to the second equation to eliminate x; and multiply each side of the first equation by 3 and add to the third equation to eliminate x:
2 4 6 142 4
5 5 10
x y zx y z
y z
− + − = −+ + =
− = −
3 6 9 21
3 2 2 10
4 7 11
x y zx y z
y z
− + =− + − = −
− + =
Multiply each side of the first result by 45
and
add to the second result to eliminate y:
4 4 84 7 11
y zy z− = −
− + =
3 3
1zz==
Substituting and solving for the other variables:
1 2
1y
y− = −
= −
2( 1) 3(1) 72 3 7
2
xx
x
− − + =+ + =
=
The solution is 2, 1, 1x y z= = − = or using ordered triplets (2, 1, 1)− .
44. 2 3 02 2 73 4 3 7
x y zx y zx y z
+ − =⎧⎪− + + = −⎨⎪ − − =⎩
Multiply each side of the first equation by –2 and add to the second equation to eliminate y; and multiply each side of the first equation by 4 and add to the third equation to eliminate y:
4 2 6 02 2 7
6 7 7
x y zx y z
x z
− − + =− + + = −
− + = −
8 4 12 03 4 3 7
11 15 7
x y zx y z
x z
+ − =− − =
− =
Multiply each side of the first result by 11 and multiply each side of the second result by 6 to eliminate x:
66 77 7766 90 42
x zx z
− + = −− =
13 353513
z
z
− = −
=
Substituting and solving for the other variables: 356 7 7132456 713
Add the first and second equations to eliminate z: 12 3 2
3 2 3
x y zx y z
x y
− − =+ + =
+ =
Multiply each side of the result by –1 and add to the original third equation to eliminate y:
3 2 33 2 0
0 3
x yx y
− − = −+ =
= −
This equation is false, so the system is inconsistent.
46. 2 3 0
2 53 4 1
x y zx y zx y z
− − =⎧⎪− + + =⎨⎪ − − =⎩
Add the first and second equations to eliminate z; then add the second and third equations to eliminate z: 2 3 0
2 5
5
x y zx y z
x y
− − =− + + =
− =
2 5
3 4 1
2 2 6
x y zx y z
x y
− + + =− − =
− =
Multiply each side of the first result by –2 and add to the second result to eliminate y:
2 2 102 2 6
x yx y
− + = −− =
0 2= − This equation is false, so the system is inconsistent.
47. 1
2 3 43 2 7 0
x y zx y zx y z
− − =⎧⎪− + − = −⎨⎪ − − =⎩
Add the first and second equations to eliminate x; multiply the first equation by –3 and add to the third equation to eliminate x:
12 3 4
4 3
x y zx y z
y z
− − =− + − = −
− = −
3 3 3 33 2 7 0
4 3
x y zx y z
y z
− + + = −− − =
− = −
Multiply each side of the first result by –1 and add to the second result to eliminate y:
4 34 3
0 0
y zy z
− + =− = −
=
The system is dependent. If z is any real number, then 4 3y z= − . Solving for x in terms of z in the first equation:
(4 3) 14 3 1
5 3 15 2
x z zx z z
x zx z
− − − =− + − =
− + == −
The solution is {( , , )x y z 5 2,x z= − 4 3y z= − , z is any real number}.
48. 2 3 03 2 2 2
5 3 2
x y zx y zx y z
− − =⎧⎪ + + =⎨⎪ + + =⎩
Multiply the first equation by 2 and add to the second equation to eliminate z; multiply the first equation by 3 and add to the third equation to eliminate z: 4 6 2 03 2 2 2
7 4 2
x y zx y z
x y
− − =+ + =
− =
6 9 3 0
5 3 2
7 4 2
x y zx y z
x y
− − =+ + =
− =
Multiply each side of the first result by –1 and add to the second result to eliminate y:
Section 8.1: Systems of Linear Equations: Substitution and Elimination
Solving for z in terms of x in the first equation: 2 3
4 22 37
8 4 217
13 47
z x yy y
y y
y
= −
+⎛ ⎞= −⎜ ⎟⎝ ⎠+ −
=
− +=
The solution is 4 2( , , )7 7
x y z x y⎧= +⎨
⎩,
13 4 , is any real number7 7
z y y ⎫= − + ⎬⎭
.
49. 2 2 3 64 3 2 02 3 7 1
x y zx y zx y z
− + =⎧⎪ − + =⎨⎪− + − =⎩
Multiply the first equation by –2 and add to the second equation to eliminate x; add the first and third equations to eliminate x:
4 4 6 124 3 2 0
4 12
x y zx y z
y z
− + − = −− + =
− = −
2 2 3 6
2 3 7 1
4 7
x y zx y z
y z
− + =− + − =
− =
Multiply each side of the first result by –1 and add to the second result to eliminate y:
4 12 4 7
0 19
y zy z
− + =− =
=
This result is false, so the system is inconsistent.
50. 3 2 2 67 3 2 12 3 4 0
x y zx y zx y z
− + =⎧⎪ − + = −⎨⎪ − + =⎩
Multiply the first equation by –1 and add to the second equation to eliminate z; multiply the first equation by –2 and add to the third equation to eliminate z:
3 2 2 67 3 2 1
4 7
x y zx y z
x y
− + − = −− + = −
− = −
6 4 4 122 3 4 0
4 12
x y zx y z
x y
− + − = −− + =
− + = −
Add the first result to the second result to eliminate y:
4 74 12
0 19
x yx y− = −
− + = −
= −
This result is false, so the system is inconsistent.
51. 6
3 2 53 2 14
x y zx y zx y z
+ − =⎧⎪ − + = −⎨⎪ + − =⎩
Add the first and second equations to eliminate z; multiply the second equation by 2 and add to the third equation to eliminate z:
63 2 5
4 1
x y zx y z
x y
+ − =− + = −
− =
6 4 2 10
3 2 14
7 4
x y zx y z
x y
− + = −+ − =
− =
Multiply each side of the first result by –1 and add to the second result to eliminate y:
4 17 4
x yx y
− + = −− =
3 31
xx==
Substituting and solving for the other variables: 4(1) 1
33
yyy
− =− = −
=
3(1) 2(3) 53 6 5
2
zzz
− + = −− + = −
= −
The solution is 1, 3, 2x y z= = = − or using ordered triplets (1, 3, 2)− .
55. Let l be the length of the rectangle and w be the width of the rectangle. Then:
2l w= and 2 2 90l w+ = Solve by substitution:
2(2 ) 2 904 2 90
6 9015 feet2(15) 30 feet
w ww w
wwl
+ =+ =
=== =
The floor is 15 feet by 30 feet.
56. Let l be the length of the rectangle and w be the width of the rectangle. Then:
50l w= + and 2 2 3000l w+ = Solve by substitution:
2( 50) 2 30002 100 2 3000
4 2900725 meters725 50 775 meters
w ww w
wwl
+ + =+ + =
=== + =
The dimensions of the field are 775 meters by 725 meters.
57. Let x = the number of commercial launches and y = the number of noncommercial launches.
Then: 55x y+ = and 2 1y x= + Solve by substitution:
(2 1) 553 54
18
x xxx
+ + ===
2(18) 136 137
yyy
= += +=
In 2005 there were 18 commercial launches and 37 noncommercial launches.
58. Let x = the number of adult tickets sold and y = the number of senior tickets sold. Then:
3259 7 2495
x yx y+ =⎧
⎨ + =⎩
Solve the first equation for y: 325y x= − Solve by substitution: 9 7(325 ) 24959 2275 7 2495
2 220110
x xx x
xx
+ − =+ − =
==
325 110 215y = − = There were 110 adult tickets sold and 215 senior citizen tickets sold.
59. Let x = the number of pounds of cashews. Let y = is the number of pounds in the mixture. The value of the cashews is 5x . The value of the peanuts is 1.50(30) = 45. The value of the mixture is 3y . Then 30x y+ = represents the amount of mixture. 5 45 3x y+ = represents the value of the mixture.
Solve by substitution: 5 45 3( 30)
2 4522.5
x xxx
+ = +==
So, 22.5 pounds of cashews should be used in the mixture.
60. Let x = the amount invested in AA bonds. Let y = the amount invested in the Bank Certificate.
a. Then 150,000x y+ = represents the total investment. 0.10 0.05 12,000x y+ = represents the earnings on the investment.
Solve by substitution: 0.10(150,000 ) 0.05 12,000
15,000 0.10 0.05 12,0000.05 3000
60,000
y yy y
yy
− + =− + =
− = −=
150,000 60,000 90,000x = − = Thus, $90,000 should be invested in AA Bonds and $60,000 in a Bank Certificate.
b. Then 150,000x y+ = represents the total investment. 0.10 0.05 14,000x y+ = represents the earnings on the investment.
Solve by substitution: 0.10(150,000 ) 0.05 14,000
15,000 0.10 0.05 14,0000.05 1000
20,000
y yy y
yy
− + =− + =
− = −=
150,000 20,000 130,000x = − = Thus, $130,000 should be invested in AA Bonds and $20,000 in a Bank Certificate.
61. Let x = the plane’s airspeed and y = the wind speed.
Rate Time DistanceWith Wind 3 600Against 4 600
x yx y+−
( )(3) 600( )(4) 600x yx y+ =⎧
⎨ − =⎩
Multiply each side of the first equation by 13
,
multiply each side of the second equation by 14
,
and add the result to eliminate y 200150
x yx y+ =− =
2 350175
xx==
175 20025
yy
+ ==
The airspeed of the plane is 175 mph, and the wind speed is 25 mph.
62. Let x = the wind speed and y = the distance.
Rate Time Distance
With Wind 150 2Against 150 3
x yx y
+−
(150 )(2)(150 )(3)
x yx y
+ =⎧⎨ − =⎩
Solve by substitution: (150 )(2) (150 )(3)
300 2 450 35 150
30
x xx xxx
+ = −+ = −
==
Thus, the wind speed is 30 mph.
63. Let x = the number of $25-design. Let y = the number of $45-design. Then x y+ = the total number of sets of dishes. 25 45x y+ = the cost of the dishes.
Setting up the equations and solving by substitution:
20025 45 7400
x yx y+ =⎧
⎨ + =⎩
Solve the first equation for y, the solve by substitution: 200y x= −
25 45(200 ) 740025 9000 45 7400
20 160080
x xx x
xx
+ − =+ − =
− = −=
200 80 120y = − = Thus, 80 sets of the $25 dishes and 120 sets of the $45 dishes should be ordered.
64. Let x = the cost of a hot dog. Let y = the cost of a soft drink. Setting up the equations and solving by substitution:
10 5 35.007 4 25.25
x yx y+ =⎧
⎨ + =⎩
10 5 35.00
2 77 2
x yx y
y x
+ =+ =
= −
7 4(7 2 ) 25.257 28 8 25.25
2.752.75
x xx x
xx
+ − =+ − =
− = −=
7 2(2.75) 1.50y = − = A single hot dog costs $2.75 and a single soft drink costs $1.50.
65. Let x = the cost per package of bacon. Let y = the cost of a carton of eggs. Set up a system of equations for the problem:
3 2 13.452 3 11.45x yx y+ =⎧
⎨ + =⎩
Multiply each side of the first equation by 3 and each side of the second equation by –2 and solve by elimination:
9 6 40.354 6 22.90
x yx y+ =
− − = −
5 17.453.49
xx==
Substitute and solve for y: 3(3.49) 2 13.45
10.47 2 13.452 2.98
1.49
yyyy
+ =+ =
==
A package of bacon costs $3.49 and a carton of eggs cost $1.49. The refund for 2 packages of bacon and 2 cartons of eggs will be 2($3.49) + 2($1.49) =$9.96.
Section 8.1: Systems of Linear Equations: Substitution and Elimination
Substitute the expression for 3I into the second equation and simplify:
1 2 2
1 2
1 2
8 4( ) 68 4 104 10 8
I I II I
I I
= + += ++ =
1 2
1 2
8 4 68 6 4
I II I= +− =
Multiply both sides of the first result by –2 and add to the second result to eliminate 1I :
1 2
1 2
8 20 16 8 6 4
I II I
− − = −− =
2
2
26 1212 626 13
I
I
− = −−
= =−
Substituting and solving for the other variables:
1
1
1
1
64 10 813604 813
444131113
I
I
I
I
⎛ ⎞+ =⎜ ⎟⎝ ⎠
+ =
=
=
3 1 211 6 1713 13 13
I I I= + = + =
The solution is 1 2 311 6 17, ,13 13 13
I I I= = = .
75. Let x = the number of orchestra seats. Let y = the number of main seats. Let z = the number of balcony seats. Since the total number of seats is 500,
500x y z+ + = . Since the total revenue is $17,100 if all seats are sold, 50 35 25 17,100x y z+ + = . If only half of the orchestra seats are sold, the revenue is $14,600.
So, 150 35 25 14,6002
x y z⎛ ⎞ + + =⎜ ⎟⎝ ⎠
.
Thus, we have the following system: 500
50 35 25 17,10025 35 25 14,600
x y zx y zx y z
+ + =⎧⎪ + + =⎨⎪ + + =⎩
Multiply each side of the first equation by –25 and add to the second equation to eliminate z; multiply each side of the third equation by –1 and add to the second equation to eliminate z:
25 25 25 12,50050 35 25 17,100
25 10 4600
x y zx y z
x y
− − − = −+ + =
+ =
50 35 25 17,100
25 35 25 14,600x y zx y z+ + =
− − − = −
25 2500
100x
x==
Substituting and solving for the other variables: 25(100) 10 4600
2500 10 460010 2100
210
yyyy
+ =+ =
==
100 210 500310 500
190
zzz
+ + =+ =
=
There are 100 orchestra seats, 210 main seats, and 190 balcony seats.
76. Let x = the number of adult tickets. Let y = the number of child tickets. Let z = the number of senior citizen tickets. Since the total number of tickets is 405,
405x y z+ + = . Since the total revenue is $2320, 8 4.50 6 2320x y z+ + = . Twice as many children's tickets as adult tickets are sold. So, 2y x= . Thus, we have the following system:
4058 4.50 6 2320
2
x y zx y z
y x
+ + =⎧⎪ + + =⎨⎪ =⎩
Substitute for y in the first two equations and simplify:
(2 ) 4053 405
x x zx z
+ + =+ =
8 4.50(2 ) 6 232017 6 2320
x x zx z
+ + =+ =
Multiply the first result by –6 and add to the second result to eliminate z:
18 6 243017 6 2320
x zx z
− − = −⎧⎪⎨ + =⎪⎩
110
110x
x− = −
=
22(110)220
y x===
3 4053(110) 405
330 40575
x zzzz
+ =+ =+ =
=
There were 110 adults, 220 children, and 75 senior citizens that bought tickets.
77. Let x = the number of servings of chicken. Let y = the number of servings of corn. Let z = the number of servings of 2% milk.
Protein equation: 30 3 9 66x y z+ + = Carbohydrate equation: 35 16 13 94.5x y z+ + = Calcium equation: 200 10 300 910x y z+ + =
Multiply each side of the first equation by –16 and multiply each side of the second equation by 3 and add them to eliminate y; multiply each side of the second equation by –5 and multiply each side of the third equation by 8 and add to eliminate y:
480 48 144 1056 105 48 39 283.5
375 105 772.5
x y zx y z
x z
− − − = −+ + =
− − = −
175 80 65 472.5
1600 80 2400 7280
1425 2335 6807.5
x y zx y z
x z
− − − = −+ + =
+ =
Multiply each side of the first result by 19 and multiply each side of the second result by 5 to eliminate x:
7125 1995 14,677.57125 11,675 34,037.5
x zx z
− − = −+ =
9680 19,3602
zz==
Substituting and solving for the other variables: 375 105(2) 772.5
375 210 772.5375 562.5
1.5
xx
xx
− − = −− − = −
− = −=
30(1.5) 3 9(2) 6645 3 18 66
3 31
yy
yy
+ + =+ + =
==
The dietitian should serve 1.5 servings of chicken, 1 serving of corn, and 2 servings of 2% milk.
78. Let x = the amount in Treasury bills. Let y = the amount in Treasury bonds. Let z = the amount in corporate bonds.
Since the total investment is $20,000, 20,000x y z+ + =
Since the total income is to be $1390, 0.05 0.07 0.10 1390x y z+ + =
The investment in Treasury bills is to be $3000 more than the investment in corporate bonds. So, 3000x z= +
Substitute for x in the first two equations and simplify: (3000 ) 20,000
2 17,000z y z
y z+ + + =
+ =
5(3000 ) 7 10 139,0007 15 124,000
z y zy z
+ + + =+ =
Section 8.1: Systems of Linear Equations: Substitution and Elimination
Multiply each side of the first result by –7 and add to the second result to eliminate y:
7 14 119,0007 15 124,000
5,000
y zy z
z
− − = −+ =
=
3000 3000 5000 8000x z= + = + = 2 17,000
2(5000) 17,00010,000 17,000
7000
y zyy
y
+ =+ =+ =
=
Kelly should invest $8000 in Treasury bills, $7000 in Treasury bonds, and $5000 in corporate bonds.
79. Let x = the price of 1 hamburger. Let y = the price of 1 order of fries. Let z = the price of 1 drink.
We can construct the system 8 6 6 26.1010 6 8 31.60
x y zx y z+ + =⎧
⎨ + + =⎩
A system involving only 2 equations that contain 3 or more unknowns cannot be solved uniquely.
Multiply the first equation by 12
− and the
second equation by 12
, then add to eliminate y:
4 3 3 13.05 5 3 4 15.80
x y zx y z
− − − = −+ + =
2.75
2.75x z
x z+ =
= −
Substitute and solve for y in terms of z: ( )5 2.75 3 4 15.80
13.75 3 15.803 2.05
1 413 60
z y zy z
y z
y z
− + + =
+ − == +
= +
Solutions of the system are: 2.75x z= − , 1 413 60
y z= + .
Since we are given that 0.60 0.90z≤ ≤ , we choose values of z that give two-decimal-place values of x and y with 1.75 2.25x≤ ≤ and 0.75 1.00y≤ ≤ . The possible values of x, y, and z are shown in the table.
81. Let x = Beth’s time working alone. Let y = Bill’s time working alone. Let z = Edie’s time working alone.
We can use the following tables to organize our work:
Beth Bill EdieHours to do jobPart of job done 1 1 1in 1 hour
x y z
x y z
In 10 hours they complete 1 entire job, so 1 1 110 1
1 1 1 110
x y z
x y z
⎛ ⎞+ + =⎜ ⎟
⎝ ⎠
+ + =
Bill EdieHours to do jobPart of job done 1 1in 1 hour
y z
y z
In 15 hours they complete 1 entire job, so 1 115 1
1 1 115
y z
y z
⎛ ⎞+ =⎜ ⎟
⎝ ⎠
+ =
.
Beth Bill EdieHours to do jobPart of job done 1 1 1in 1 hour
x y z
x y z
With all 3 working for 4 hours and Beth and Bill working for an additional 8 hours, they complete
1 entire job, so 1 1 1 1 14 8 1
12 12 4 1
x y z x y
x y z
⎛ ⎞ ⎛ ⎞+ + + + =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
+ + =
We have the system 1 1 1 1
101 1 1
1512 12 4 1
x y z
y z
x y z
⎧+ + =⎪
⎪⎪
+ =⎨⎪⎪
+ + =⎪⎩
Subtract the second equation from the first equation:
1 1 1 1 10
1 1 1 15
x y z
y z
+ + =
+ =
1 1
3030
xx
=
=
Substitute x = 30 into the third equation: 12 12 4 130
12 4 35
y z
y z
+ + =
+ =.
Now consider the system consisting of the last result and the second original equation. Multiply the second original equation by –12 and add it to the last result to eliminate y:
12 12 1215
12 4 3 5
y z
y z
− − −+ =
+ =
8 315
40zz
− = −
=
Plugging z = 40 to find y: 12 4 3
512 4 3
40 512 1
224
y z
y
yy
+ =
+ =
=
=
Working alone, it would take Beth 30 hours, Bill 24 hours, and Edie 40 hours to complete the job.
82 – 84. Answers will vary.
Section 8.2: Systems of Linear Equations: Matrices
The solution is 1, 2, 0, 6a b c d= = − = = ; so the
equation is 3 2( ) 2 6f x x x= − + .
77. Let x = the number of servings of salmon steak. Let y = the number of servings of baked eggs. Let z = the number of servings of acorn squash. Protein equation: 30 15 3 78x y z+ + = Carbohydrate equation: 20 2 25 59x y z+ + = Vitamin A equation: 2 20 32 75x y z+ + = Set up a matrix and solve:
The dietitian should serve 1.5 servings of salmon steak, 2 servings of baked eggs, and 1 serving of acorn squash.
78. Let x = the number of servings of pork chops. Let y = the number of servings of corn on the cob. Let z = the number of servings of 2% milk. Protein equation: 23 3 9 47x y z+ + = Carbohydrate equation: 16 13 58y z+ = Calcium equation: 10 10 300 630x y z+ + = Set up a matrix and solve:
The dietitian should provide 1 serving of pork chops, 2 servings of corn on the cob, and 2 servings of 2% milk.
79. Let x = the amount invested in Treasury bills. Let y = the amount invested in Treasury bonds. Let z = the amount invested in corporate bonds. Total investment equation: 10,000x y z+ + = Annual income equation: 0.06 0.07 0.08 680x y z+ + = Condition on investment equation: 0.5
2 0z x
x z=
− =
Set up a matrix and solve: 10,0001 1 1
0.06 0.07 0.08 6800 021
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦
( )
2 1 2
3 1 3
2 2
10,0001 1 10.06
0 0.01 0.02 80 10,0000 31
10,0001 1 10 8000 1001 2
10,0000 31
R r rR r r
R r
⎡ ⎤= − +⎢ ⎥ ⎛ ⎞
→ ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥−−−⎣ ⎦⎡ ⎤⎢ ⎥
→ =⎢ ⎥⎢ ⎥−−−⎣ ⎦
( )
1 2 1
3 2 3
3 3
1 3 1
2 3 2
0 20001 10 8000 1 20 0 20001
0 20001 10 8000 1 20 0 20001
0 0 400010 0 4000 1
20 0 20001
R r rR r r
R r
R r rR r r
⎡ ⎤−= − +⎢ ⎥ ⎛ ⎞
→ ⎜ ⎟⎢ ⎥ = +⎝ ⎠⎢ ⎥−−⎣ ⎦⎡ ⎤−⎢ ⎥
→ = −⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤
= +⎢ ⎥→ ⎢ ⎥ = − +⎢ ⎥
⎣ ⎦
⎛ ⎞⎜ ⎟⎝ ⎠
Carletta should invest $4000 in Treasury bills, $4000 in Treasury bonds, and $2000 in corporate bonds.
80. Let x = the fixed delivery charge; let y = the cost of each tree, and let z = the hourly labor charge. 1st subdivision: 250 166 7520x y z+ + = 2nd subdivision: 200 124 5945x y z+ + = 3rd subdivision: 300 200 8985x y z+ + = Set up a matrix and solve:
250 166 75201200 59451 124300 200 89851
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
Section 8.2: Systems of Linear Equations: Matrices
The delivery charge is $250 per job, the cost for each tree is $19.95, and the hourly labor charge is $13.75.
81. Let x = the number of Deltas produced. Let y = the number of Betas produced. Let z = the number of Sigmas produced. Painting equation: 10 16 8 240x y z+ + = Drying equation: 3 5 2 69x y z+ + = Polishing equation: 2 3 41x y z+ + = Set up a matrix and solve:
The company should produce 8 Deltas, 5 Betas, and 10 Sigmas.
82. Let x = the number of cases of orange juice produced; let y = the number of cases of grapefruit juice produced; and let z = the number of cases of tomato juice produced. Sterilizing equation: 9 10 12 398x y z+ + = Filling equation: 6 4 4 164x y z+ + = Labeling equation: 2 58x y z+ + = Set up a matrix and solve:
86. Let x = the amount invested in Treasury bills. Let y = the amount invested in corporate bonds. Let z = the amount invested in junk bonds. Let I = income Total investment equation: 25,000x y z+ + = Annual income equation: 0.07 0.09 0.11x y z I+ + = Set up a matrix and solve:
( )
( )
( )
( )
2 2
1 2 1
12 22
1 1 2
1 1 1 25,0000.07 0.09 0.11
1 1 1 25,000 100
7 9 11 100
1 1 1 25,000 7
0 2 4 100 175,000
1 1 1 25,000
0 1 2 50 87,500
1 0 1112,500 50
0 1 2 50 87,500
I
R rI
R r rI
R rI
IR r r
I
⎡ ⎤⎢ ⎥⎣ ⎦
⎡ ⎤→ =⎢ ⎥
⎣ ⎦⎡ ⎤
→ = −⎢ ⎥−⎣ ⎦⎡ ⎤
→ =⎢ ⎥−⎣ ⎦⎡ − − ⎤
→ = −⎢ ⎥−⎣ ⎦
The matrix in the last step represents the system 112,500 50
2 50 87,500
x z Iy z I
− = −⎧⎨ + = −⎩
Thus, the solution is 112,500 50x I z= − + , 50 87,500 2y I z= − − , z is any real number.
a. 1500I =
( )112,500 50112,500 50 150037,500
x I zz
z
= − +
= − +
= +
( )50 87,500 250 1500 87,500 2
12,500 2
y I zz
z
= − −
= − −
= − −
z is any real number. Since y and z cannot be negative, we must have 0.y z= = Investing all of the money at 7% yields $1750, which is more than the $1500 needed.
b. 2000I =
( )112,500 50112,500 50 200012,500
x T zz
z
= − +
= − +
= +
( )50 87,500 250 2000 87,500 212,500 2
y I zz
z
= − −
= − −
= −
z is any real number.
Possible investment strategies:
Amount Invested At
7% 9% 11%
12,500 12,500 0
15,500 6500 3000
18,750 0 6250 c. 2500I =
( )112,500 50112,500 50 2500
12,500
x T zz
z
= − +
= − +
= − +
( )50 87,500 250 2500 87,500 237,500 2
y I zz
z
= − −
= − −
= −
z is any real number.
Possible investment strategies:
Amount invested at
7% 9% 11%
0 12,500 12,500
1000 10,500 13,500
6250 0 18,750 87. Let x = the amount of supplement 1.
Let y = the amount of supplement 2. Let z = the amount of supplement 3.
58. Any point ( , )x y on the line containing 2 2( , )x y and 3 3( , )x y satisfies:
2 2
3 3
11 01
x yx yx y
=
If the point 1 1( , )x y is on the line containing
2 2( , )x y and 3 3( , )x y [the points are collinear],
then 1 1
2 2
3 3
11 01
x yx yx y
= .
Conversely, if 1 1
2 2
3 3
11 01
x yx yx y
= , then 1 1( , )x y is
on the line containing 2 2( , )x y and 3 3( , )x y , and the points are collinear.
59. Let A = 1 1( , )x y , B = 2 2( , )x y , and C = 3 3( , )x y represent the vertices of a triangle. For simplicity, we position our triangle in the first quadrant. See figure. Let A be the point closest to the y-axis, C be the point farthest from the y-axis, and B be the point “between” points A and C.
y
x1( ,0)x 2( ,0)x 3( ,0)x
1 1( , )x y
2 2( , )x y
3 3( , )x yA
B
C
D E F
Let D = 1( ,0)x , E = 2( ,0)x , and F = 3( ,0)x . We find the area of triangle ABC by subtracting the areas of trapezoids ADEB and BEFC from the area of trapezoid ADFC. Note: 1AD y= ,
2BE y= , 3CF y= , 3 1DF x x= − , 2 1DE x x= − ,
and 3 2EF x x= − . Thus, the areas of the three trapezoids are as follows:
( )( )3 1 1 312ADFCK x x y y= − +
( )( )2 1 1 212ADEBK x x y y= − + , and
( )( )3 2 2 312BEFCK x x y y= − + .
The area of our triangle ABC is ABC ADFC ADEB BEFCK K K K= − −
( )( ) ( )( )
( )( )
3 1 1 3 2 1 1 2
3 2 2 3
1 12 2
1 2
x x y y x x y y
x x y y
= − + − − +
− − +
( )
3 1 3 3 1 1 1 3 2 1
2 2 1 1 1 2 3 2
3 3 2 2 2 3
3 1 1 3 2 1 1 2 3 2 2 3
3 1 1 3 2 1 1 2 3 2 2 3
1 1 1 1 12 2 2 2 2
1 1 1 1 2 2 2 2
1 1 1 2 2 2
1 1 1 1 1 12 2 2 2 2 212
x y x y x y x y x y
x y x y x y x y
x y x y x y
x y x y x y x y x y x y
x y x y x y x y x y x y
= + − − −
− + + −
− + +
= − − + − +
= − − + − +
Expanding D, we obtain
( )
( )
1 2 3
1 2 3
2 3 1 3 1 21 2 3
1 2 3 2 1 3 3 1 2
1 2 1 3 2 1 2 3 3 1 3 2
12
1 1 1
11 1 1 1 1 12
1 ( ) ( ) ( )212
x x xD y y y
y y y y y yx x x
x y y x y y x y y
x y x y x y x y x y x y
=
⎛ ⎞= − +⎜ ⎟
⎝ ⎠
= − − − + −
= − − + + −
which is the same as the area of triangle ABC.
If the vertices of the triangle are positioned differently than shown in the figure, the signs may be reversed. Thus, the absolute value of D will give the area of the triangle.
If the vertices of a triangle are (2, 3), (5, 2), and (6, 5), then:
2. 2 4y x= − The graph is a parabola. x-intercepts:
2
2
0 4
42, 2
x
xx x
= −
== − =
y-intercept: 20 4 4y = − = − The vertex has x-coordinate:
( )0 0
2 2 1bxa
= − = − = .
The y-coordinate of the vertex is 20 4 4y = − = − .
3. 2 2
2 2
22
2 2
1
1
11 1
y x
x y
yx
= −
− =
− =
The graph is a hyperbola with center (0, 0), transverse axis along the x-axis, and vertices at ( 1,0)− and (1,0) . The asymptotes are y x= − and y x= .
y
x−5(−1, 0)
5
−5
5
(1, 0)
4. 2 2
2 2
22
22
2 2
4 4
4 44 4
14
12 1
x y
x y
x y
yx
+ =
+=
+ =
+ =
The graph is an ellipse with center (0,0) , major axis along the x-axis, vertices at ( 2,0)− and (2,0) . The graph also has y-intercepts at (0, 1)− and (0,1) .
Solve the second equation for a , substitute into the first equation and solve:
10 23
3(10 ) 230 3 2
30 56 4
bb
b bb b
bb a
−=
− =− =
== ⇒ =
10; 2a b b a+ = − =
The ratio of a b+ to b a− is 102 5= .
78. 4314 14
ab
a b a b
⎧ =⎪⎨⎪ + = ⇒ = −⎩
Solve the second equation for a , substitute into the first equation and solve:
( )
14 43
3 14 442 3 4
42 76 8
bb
b bb b
bb a
−=
− =
− === ⇒ =
2; 14a b a b− = + =
The ratio of a b− to a b+ is 2 1714 = .
79. Let x = the width of the rectangle. Let y = the length of the rectangle.
2 2 1615
x yxy
+ =⎧⎨ =⎩
Solve the first equation for y, substitute into the second equation and solve. 2 2 16
2 16 28
x yy xy x
+ == −= −
( )
( )( )
2
2
8 15
8 15
8 15 05 3 0
x x
x x
x xx x
− =
− =
− + =
− − =
5x = or 3x = The dimensions of the rectangle are 3 inches by 5 inches.
80. Let 2x = the side of the first square. Let 3x = the side of the second square.
( ) ( )2 2
2 2
2
2
2 3 52
4 9 52
13 52
42
x x
x x
x
xx
+ =
+ =
=
== ±
Note that we must have 0x > . The sides of the first square are (2)(2) = 4 feet and the sides of the second square are (3)(2) = 6 feet.
81. Let x = the radius of the first circle. Let y = the radius of the second circle.
2 2
2 2 12
20
x y
x y
π + π = π⎧⎪⎨
π + π = π⎪⎩
Solve the first equation for y, substitute into the second equation and solve: 2 2 12
66
x yx y
y x
π + π = π+ =
= −
2 2
2 2
2 2
2 2 2
2
20
20
(6 ) 20
36 12 20 2 12 16 0
6 8 0 ( 4)( 2) 04 or 22 4
x y
x y
x x
x x x x x
x x x xx xy y
π + π = π
+ =
+ − =
+ − + = ⇒ − + =
− + = ⇒ − − == == =
The radii of the circles are 2 centimeters and 4 centimeters.
82. Let x = the length of each of the two equal sides in the isosceles triangle. Let y = the length of the base. The perimeter of the triangle: 18x x y+ + = Since the altitude to the base y is 3, the Pythagorean theorem produces another equation.
Solve the first equation for y, substitute into the second equation and solve.
( )
( )
22
22
2 2
18 29
4324 72 4 9
481 18 9
18 905 18 2 5 8
xx
x x x
x x xxx y
−+ =
− ++ =
− + + =− = −
= ⇒ = − =
The base of the triangle is 8 centimeters.
83. The tortoise takes 9 + 3 = 12 minutes or 0.2 hour longer to complete the race than the hare. Let r = the rate of the hare. Let t = the time for the hare to complete the race. Then t + 0.2 = the time for the tortoise and
0.5r − = the rate for the tortoise. Since the length of the race is 21 meters, the distance equations are:
( )( )
2121
0.5 0.2 21
r t rt
r t
⎧ = ⇒ =⎪⎨⎪ − + =⎩
Solve the first equation for r, substitute into the second equation and solve:
( )
( )
( )( )
2
2
21 0.5 0.2 21
4.221 0.5 0.1 21
4.210 21 0.5 0.1 10 21
210 42 5 210
5 42 05 14 3 0
tt
tt
t t tt
t t t t
t tt t
⎛ ⎞− + =⎜ ⎟⎝ ⎠
+ − − =
⎛ ⎞+ − − = ⋅⎜ ⎟⎝ ⎠
+ − − =
+ − =
− + =
5 14 05 14
14 2.85
tt
t
− ==
= =
or 3 03
tt
+ == −
3t = − makes no sense, since time cannot be negative. Solve for r:
21 7.52.8
r = =
The average speed of the hare is 7.5 meters per hour, and the average speed for the tortoise is 7 meters per hour.
84. Let 1 2 3, ,v v v = the speeds of runners 1, 2, 3. Let 1 2 3, ,t t t = the times of runners 1, 2, 3. Then by the conditions of the problem, we have the following system:
1 1
2 1
3 1
2 2
5280527052605280
v tv tv tv t
=⎧⎪ =⎪⎨ =⎪⎪ =⎩
Distance between the second runner and the third runner after 2t seconds is:
2 23 2 3 1
2 15280 5280
52805280 52605270
10.02
v tv t v t
v t⎛ ⎞
− = − ⎜ ⎟⎝ ⎠⎛ ⎞= − ⎜ ⎟⎝ ⎠
≈
The second place runner beats the third place runner by about 10.02 feet.
85. Let x = the width of the cardboard. Let y = the length of the cardboard. The width of the box will be 4x − , the length of the box will be
4y − , and the height is 2. The volume is ( 4)( 4)(2)V x y= − − .
Solve the system of equations: 216216
2( 4)( 4) 224
xy yx
x y
⎧ = ⇒ =⎪⎨⎪ − − =⎩
Solve the first equation for y, substitute into the second equation and solve.
( )
( )( )
2
2
2
2162 8 4 224
1728432 8 32 224
432 8 1728 32 224
8 240 1728 0
30 216 012 18 0
xx
xx
x x x x
x x
x xx x
⎛ ⎞− − =⎜ ⎟⎝ ⎠
− − + =
− − + =
− + =
− + =
− − =
12 012
xx
− ==
or 18 018
xx
− ==
The cardboard should be 12 centimeters by 18 centimeters.
86. Let x = the width of the cardboard. Let y = the length of the cardboard. The area of the cardboard is: 216xy =
The volume of the tube is: 2 224V r h= π =
where and 2 or 2xh y r x r= π = =π
.
Solve the system of equations:
2 2
216216
224 2242 4
xy yx
x x yy
⎧ = ⇒ =⎪⎪⎨
⎛ ⎞⎪π = ⇒ =⎜ ⎟⎪ π π⎝ ⎠⎩
Solve the first equation for y, substitute into the second equation and solve.
( )2 216224
4216 896
896 13.03216
xx
x
x
πππ
=
=
= ≈
( )2
896216
216216 216 16.57896
yx π π
= = = ≈
The cardboard should be about 13.03 centimeters by 16.57 centimeters.
87. Find equations relating area and perimeter: 2 2 4500
3 3 ( ) 300x yx y x y
⎧ + =⎪⎨
+ + − =⎪⎩
Solve the second equation for y, substitute into the first equation and solve: 4 2 300
2 300 4150 2
x yy xy x
+ == −= −
2 2
2 2
2
2
2
(150 2 ) 4500
22,500 600 4 4500
5 600 18,000 0
120 3600 0
( 60) 060 0
60150 2(60) 30
x x
x x x
x x
x x
xx
xy
+ − =
+ − + =
− + =
− + =
− =− =
== − =
The sides of the squares are 30 feet and 60 feet.
88. Let x = the length of a side of the square. Let r = the radius of the circle. The area of the square is 2x and the area of the circle is 2rπ . The perimeter of the square is 4x and the circumference of the circle is 2 rπ . Find equations relating area and perimeter:
2 2 1004 2 60x r
x r⎧ + π =⎪⎨
+ π =⎪⎩
Solve the second equation for x, substitute into the first equation and solve: 4 2 60
4 60 21152
x rx r
x r
+ π == − π
= − π
22
2 2 2
2 2
115 10021225 15 1004
1 15 125 04
r r
r r r
r r
⎛ ⎞− π + π =⎜ ⎟⎝ ⎠
− π + π + π =
⎛ ⎞π + π − π + =⎜ ⎟⎝ ⎠
2 2 2
2 2
2
14 ( 15 ) 4 (125)41225 5004
100 500 0
b ac ⎛ ⎞− = − π − π + π⎜ ⎟⎝ ⎠⎛ ⎞= π − π + π⎜ ⎟⎝ ⎠
= π − π <
Since the discriminant is less than zero, it is impossible to cut the wire into two pieces whose total area equals 100 square feet.
89. Solve the system for and l w : 2 2l w P
l w A+ =⎧
⎨ =⎩
Solve the first equation for l , substitute into the second equation and solve.
If it is required that length be greater than width, then the solution is:
2 216 16and4 4
P P A P P Aw l− − + −= =
90. Solve the system for and l b :
22 2
2 2
4
P b l b P l
bh l
= + ⇒ = −⎧⎪⎨
+ =⎪⎩
Solve the first equation for b , substitute into the second equation and solve.
( )
2 2 2
22 2
2 2 2 2
2 2
2 2
4 4
4 2 4
4 4 4 4
4 4
44
h b l
h P l l
h P Pl l l
h P Pl
h PlP
+ =
+ − =
+ − + =
+ =
+=
2 2 2 24 42 2
h P P hb PP P+ −
= − =
91. Solve the equation: 2 4(2 4) 0m m− − =
( )
2
2
8 16 0
4 04
m m
mm
− + =
− =
=
Use the point-slope equation with slope 4 and the point (2, 4) to obtain the equation of the tangent line: 4 4( 2) 4 4 8 4 4y x y x y x− = − ⇒ − = − ⇒ = −
92. Solve the system: 2 2 10x y
y mx b⎧ + =⎪⎨
= +⎪⎩
Solve the system by substitution: ( )
( )
22
2 2 2 2
2 2 2
10
2 10 0
1 2 10 0
x mx b
x m x bmx b
m x bmx b
+ + =
+ + + − =
+ + + − =
Note that the tangent line passes through (1, 3). Find the relation between m and b: 3 (1) 3m b b m= + ⇒ = − There is one solution to the quadratic if the discriminant is zero.
( ) ( )( )2 2 2
2 2 2 2 2 2
2 2
2 4 1 10 0
4 4 40 4 40 0
40 4 40 0
bm m b
b m b m m b
m b
− + − =
− + − + =
− + =
Substitute for b and solve: ( )
( )
22
2 2
2
2
2
40 4 3 40 0
40 4 24 36 40 0
36 24 4 0
9 6 1 0
3 1 03 1
13
m m
m m m
m m
m m
mm
m
− − + =
− + − + =
+ + =
+ + =
+ =
= −
= −
1 103 33 3
b m ⎛ ⎞= − = − − =⎜ ⎟⎝ ⎠
The equation of the tangent line is 1 103 3
y x= − + .
93. Solve the system: 2 2y x
y mx b⎧ = +⎪⎨
= +⎪⎩
Solve the system by substitution: 2 22 2 0x mx b x mx b+ = + ⇒ − + − =
Note that the tangent line passes through (1, 3). Find the relation between m and b: 3 (1) 3m b b m= + ⇒ = − Substitute into the quadratic to eliminate b:
2 22 (3 ) 0 ( 1) 0x mx m x mx m− + − − = ⇒ − + − = Find when the discriminant equals 0:
3 3 2 1b m= − = − = The equation of the tangent line is 2 1y x= + .
94. Solve the system: 2 5x y
y mx b⎧ + =⎪⎨
= +⎪⎩
Solve the system by substitution: 2 25 5 0x mx b x mx b+ + = ⇒ + + − =
Note that the tangent line passes through (–2, 1). Find the relation between m and b:
( )1 2 2 1m b b m= − + ⇒ = + Substitute into the quadratic to eliminate b:
( )
2
2
2 1 5 0
2 4 0
x mx m
x mx m
+ + + − =
+ + − =
Find when the discriminant equals 0: ( ) ( )( )
( )
2
2
2
4 1 2 4 0
8 16 0
4 04 0
4
m m
m m
mm
m
− − =
− + =
− =
− ==
( )2 1 2 4 1 9b m= + = + = The equation of the tangent line is 4 9y x= + .
95. Solve the system: 2 22 3 14x y
y mx b⎧ + =⎪⎨
= +⎪⎩
Solve the system by substitution: ( )
( )
22
2 2 2 2
2 2 2
2 3 14
2 3 6 3 14
3 2 6 3 14 0
x mx b
x m x mbx b
m x mbx b
+ + =
+ + + =
+ + + − =
Note that the tangent line passes through (1, 2). Find the relation between m and b: 2 (1) 2m b b m= + ⇒ = − Substitute into the quadratic to eliminate b:
2 2 2
2 2 2 2
(3 2) 6 (2 ) 3(2 ) 14 0
(3 2) (12 6 ) (3 12 2) 0
m x m m x m
m x m m x m m
+ + − + − − =
+ + − + − − = Find when the discriminant equals 0:
2 2 2 2
2
2
2
(12 6 ) 4(3 2)(3 12 2) 0
144 96 16 0
9 6 1 0
(3 1) 03 1 0
13
m m m m m
m m
m m
mm
m
− − + − − =
+ + =
+ + =
+ =+ =
= −
1 72 23 3
b m ⎛ ⎞= − = − − =⎜ ⎟⎝ ⎠
The equation of the tangent line is 1 73 3
y x= − + .
96. Solve the system: 2 23 7x y
y mx b⎧ + =⎪⎨
= +⎪⎩
Solve the system by substitution: ( )
( )
22
2 2 2 2
2 2 2
3 7
3 2 7
3 2 7 0
x mx b
x m x mbx b
m x mbx b
+ + =
+ + + =
+ + + − =
Note that the tangent line passes through (–1, 2). Find the relation between m and b: 2 ( 1) 2m b b m= − + ⇒ = + There is one solution to the quadratic if the discriminant equals 0.
Note that the tangent line passes through (2, 1). Find the relation between m and b: 1 (2) 1 2m b b m= + ⇒ = − Substitute into the quadratic to eliminate b:
2 2 2
2 2 2 2
2 2 2 2
(1 ) 2 (1 2 ) (1 2 ) 3 0
(1 ) ( 2 4 ) 1 4 4 3 0
(1 ) ( 2 4 ) ( 4 4 4) 0
m x m m x m
m x m m x m m
m x m m x m m
− − − − − − =
− + − + − + − − =
− + − + + − + − =Find when the discriminant equals 0:
( ) ( )( )
( )
22 2 2
2 3 4 4 3
2
2
2
2 4 4 1 4 4 4 0
4 16 16 16 16 16 16 0
4 16 16 0
4 4 0
2 02
m m m m m
m m m m m m
m m
m m
mm
− + − − − + − =
− + − + − + =
− + =
− + =
− =
= The equation of the tangent line is 2 3y x= − .
98. Solve the system: 2 22 14y x
y mx b⎧ − =⎪⎨
= +⎪⎩
Solve the system by substitution: ( )
( )
2 2
2 2 2 2
2 2 2
2 14
2 4 2 14
2 1 4 2 14 0
mx b x
m x mbx b x
m x mbx b
+ − =
+ + − =
− + + − =
Note that the tangent line passes through (2, 3). Find the relation between m and b: 3 (2) 3 2m b b m= + ⇒ = − There is one solution to the quadratic if the discriminant equals 0.
100. Consider the circle with equation ( ) ( )2 2 2x h y k r− + − = and the third degree
polynomial with equation 3 2y ax bx cx d= + + + . Substituting the second equation into the first equation yields
( ) ( )22 3 2 2x h ax bx cx d k r− + + + + − = .
In order to find the roots for this equation we can expand the terms on the left hand side of the equation. Notice that ( )2x h− yields a 2nd degree
polynomial, and ( )23 2ax bx cx d k+ + + − yields a
6th degree polynomial. Therefore, we need to find the roots of a 6th degree equation, and the Fundamental Theorem of Algebra states that there will be at most six real roots. Thus, the circle and the 3rd degree polynomial will intersect at most six times. Now consider the circle with equation ( ) ( )2 2 2x h y k r− + − = and the polynomial of degree n with equation
2 30 1 2 3 ... n
ny a a x a x a x a x= + + + + + . Substituting the first equation into the first equation yields
( ) ( )22 2 3 20 1 2 3 ... n
nx h a a x a x a x a x k r− + + + + + + − =
In order to find the roots for this equation we can expand the terms on the left hand side of the equation. Notice that ( )2x h− yields a 2nd degree polynomial,
and ( )22 30 1 2 3 ... n
na a x a x a x a x k+ + + + + −
yields a polynomial of degree 2n. Therefore, we need to find the roots of an equation of degree 2n, and the Fundamental Theorem of Algebra states that there will be at most 2n real roots. Thus, the circle and the nth degree polynomial will intersect at most 2n times.
101. Since the area of the square piece of sheet metal is 100 square feet, the sheet’s dimensions are 10 feet by 10 feet. Let x = the length of the cut.
10 – 2x 10 – 2x
x
10
10
xx
The dimensions of the box are: length 10 2 ;x= − width 10 2 ; heightx x= − = . Note that each of these expressions must be positive. So we must have 0 and 10 2 0 5,x x x> − > ⇒ < that is, 0 5x< < . So the volume of the box is given by
( ) ( ) ( )( )( )( )( ) ( )2
length width height
10 2 10 2
10 2
V
x x x
x x
= ⋅ ⋅
= − −
= −
a. In order to get a volume equal to 9 cubic feet, we solve ( ) ( )210 2 9.x x− =
( ) ( )( )
2
2
2 3
10 2 9
100 40 4 9
100 40 4 9
x x
x x x
x x x
− =
− + =
− + =
So we need to solve the equation 3 24 40 100 9 0x x x− + − = .
Graphing 3 21 4 40 100 9y x x x= − + − on a
calculator yields the graph
10−2
−40
80
10−2
−40
80
10−2
−40
80
The graph indicates that there are three real zeros on the interval [0, 6]. Using the ZERO feature of a graphing calculator, we find that the three roots shown occur at 0.093x ≈ ,
4.274x ≈ and 5.632x ≈ . But we’ve already noted that we must have 0< <5x , so the only practical values for the sides of the square base are 0.093x ≈ feet and 4.274x ≈ feet.
2. 3 2 6x y− = The graph is a line. x-intercept: ( )3 2 0 6
3 62
xxx
− =
==
y-intercept: ( )3 0 2 62 6
3
yyy
− =
− == −
3. 2 2 9x y+ = The graph is a circle. Center: (0, 0) ; Radius: 3
4. 2 4y x= + The graph is a parabola. x-intercepts: 2
2
0 4
4, no intercepts
x
x x
= +
= − −
y-intercept: 20 4 4y = + = The vertex has x-coordinate:
( )0 0
2 2 1bxa
= − = − = .
The y-coordinate of the vertex is 20 4 4y = + = .
5. True
6. 2y x= ; right; 2
7. satisfied
8. half-plane
9. False
10. True
11. 0x ≥ Graph the line 0x = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (2, 0). Since 2 ≥ 0 is true, shade the side of the line containing (2, 0).
12. 0y ≥ Graph the line 0y = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 2). Since 2 ≥ 0 is true, shade the side of the line containing (0, 2).
13. 4x ≥ Graph the line 4x = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (5, 0). Since 5 ≥ 0 is true, shade the side of the line containing (5, 0).
14. 2y ≤ Graph the line 2y = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (5, 0). Since 0 ≤ 2 is true, shade the side of the line containing (5, 0).
15. 2 6x y+ ≥ Graph the line 2 6x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 6+ ≥ is false, shade the opposite side of the line from (0, 0).
16. 3 2 6x y+ ≤ Graph the line 3 2 6x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 3(0) 2(0) 6+ ≤ is true, shade the side of the line containing (0, 0).
17. 2 2 1x y+ >
Graph the circle 2 2 1x y+ = . Use a dashed line since the inequality uses >. Choose a test point not on the circle, such as (0, 0). Since 2 20 0 1+ > is false, shade the opposite side of the circle from (0, 0).
18. 2 2 9x y+ ≤
Graph the circle 2 2 9x y+ = . Use a solid line since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since
2 20 0 9+ ≤ is true, shade the same side of the circle as (0, 0).
Graph the parabola 2 1y x= − . Use a solid line since the inequality uses ≤ . Choose a test point not on the parabola, such as (0, 0). Since
20 0 1≤ − is false, shade the opposite side of the parabola from (0, 0).
20. 2 2y x> +
Graph the parabola 2 2y x= + . Use a dashed line since the inequality uses >. Choose a test point not on the parabola, such as (0, 0). Since
20 0 2> + is false, shade the opposite side of the parabola from (0, 0).
21. 4xy ≥ Graph the hyperbola 4xy = . Use a solid line since the inequality uses ≥ . Choose a test point not on the hyperbola, such as (0, 0). Since 0 0 4⋅ ≥ is false, shade the opposite side of the hyperbola from (0, 0).
22. 1xy ≤ Graph the hyperbola 1xy = . Use a solid line since the inequality uses ≤ . Choose a test point not on the hyperbola, such as (0, 0). Since 0 0 1⋅ ≤ is true, shade the same side of the hyperbola as (0, 0).
23. 2
2 4x yx y
+ ≤⎧⎨ + ≥⎩
Graph the line 2x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 4x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 4+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
24. 3 6
2 2x yx y
− ≥⎧⎨ + ≤⎩
Graph the line 3 6x y− = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 3(0) 0 6− ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 2+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution.
Graph the line 2 4x y− = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 4− ≤ is true, shade the side of the line containing (0, 0). Graph the line 3 2 6x y+ = − . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 3(0) 2(0) 6+ ≥ − is true, shade the side of the line containing (0, 0). The overlapping region is the solution.
26. 4 5 02 2
x yx y
− ≤⎧⎨ − ≥⎩
Graph the line 4 5 0x y− = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (2, 0). Since 4(2) 5(0) 0− ≤ is false, shade the opposite side of the line from (2, 0). Graph the line 2 2x y− = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2− ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
27. 2 3 03 2 6
x yx y
− ≤⎧⎨ + ≤⎩
Graph the line 2 3 0x y− = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 3). Since 2(0) 3(3) 0− ≤ is true, shade the side of the line containing (0, 3). Graph the line 3 2 6x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 3(0) 2(0) 6+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution.
28. 4 2
2 2x yx y
− ≥⎧⎨ + ≥⎩
Graph the line 4 2x y− = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 4(0) 0 2− ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 2+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
Graph the line 2 6x y− = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 6− ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 4 0x y− = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 2). Since 2(0) 4(2) 0− ≥ is false, shade the opposite side of the line from (0, 2). The overlapping region is the solution.
30. 4 84 4
x yx y
+ ≤⎧⎨ + ≥⎩
Graph the line 4 8x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 4(0) 8+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 4 4x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 4(0) 4+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
31. 2 22 2
x yx y
+ ≥ −⎧⎨ + ≥⎩
Graph the line 2 2x y+ = − . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2+ ≥ − is true, shade the side of the line containing (0, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
32. 4 44 0
x yx y
− ≤⎧⎨ − ≥⎩
Graph the line 4 4x y− = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 4(0) 4− ≤ is true, shade the side of the line containing (0, 0). Graph the line 4 0x y− = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (1, 0). Since 1 4(0) 0− ≥ is true, shade the side of the line containing (1, 0). The overlapping region is the solution.
Graph the line 2 3 6x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 6+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 3 0x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 2). Since 2(0) 3(2) 0+ ≤ is false, shade the opposite side of the line from (0, 2). Since the regions do not overlap, the solution is an empty set.
34. 2 02 2
x yx y
+ ≥⎧⎨ + ≥⎩
Graph the line 2 0x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (1, 0). Since 2(1) 0 0+ ≥ is true, shade the side of the line containing (1, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
35. 2 2 9
3x y
x y⎧ + ≤⎪⎨
+ ≥⎪⎩
Graph the circle 2 2 9x y+ = . Use a solid line since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since
2 20 0 9+ ≤ is true, shade the same side of the circle as (0, 0). Graph the line 3x y+ = . Use a solid line since the inequality uses ≥ . Choose a test point not on the line, such as (0, 0). Since 0 0 3+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
36. 2 2 9
3x y
x y⎧ + ≥⎪⎨
+ ≤⎪⎩
Graph the circle 2 2 9x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the circle, such as (0, 0). Since
2 20 0 9+ ≥ is false, shade the opposite side of the circle as (0, 0). Graph the line 3x y+ = . Use a solid line since the inequality uses ≤ . Choose a test point not on the line, such as (0, 0). Since 0 0 3+ ≤ is true, shade the same side of the line as (0, 0). The overlapping region is the solution.
Graph the parabola 2 4y x= − . Use a solid line since the inequality uses ≥ . Choose a test point not on the parabola, such as (0, 0). Since
20 0 4≥ − is true, shade the same side of the parabola as (0, 0). Graph the line 2y x= − . Use a solid line since the inequality uses ≤ . Choose a test point not on the line, such as (0, 0). Since 0 0 2≤ − is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
38. 2y xy x
⎧ ≤⎪⎨
≥⎪⎩
Graph the parabola 2y x= . Use a solid line since the inequality uses ≤ . Choose a test point not on the parabola, such as (1, 2). Since 22 1≤ is false, shade the opposite side of the parabola from (1, 2). Graph the line y x= . Use a solid line since the inequality uses ≥ . Choose a test point not on the line, such as (1, 2). Since 2 1≥ is true, shade the same side of the line as (1, 2). The overlapping region is the solution.
39. 2 2
2
16
4
x y
y x
⎧ + ≤⎪⎨
≥ −⎪⎩
Graph the circle 2 2 16x y+ = . Use a sold line since the inequality is not strict. Choose a test point not on the circle, such as (0,0) . Since
2 20 0 16+ ≤ is true, shade the side of the circle containing (0,0) . Graph the parabola
2 4y x= − . Use a solid line since the inequality is not strict. Choose a test point not on the parabola, such as (0,0) . Since 20 0 4≥ − is true, shade the side of the parabola that contains (0,0) . The overlapping region is the solution.
40. 2 2
2
25
5
x y
y x
⎧ + ≤⎪⎨
≤ −⎪⎩
Graph the circle 2 2 25x y+ = . Use a sold line since the inequality is not strict. Choose a test point not on the circle, such as (0,0) . Since
2 20 0 25+ ≤ is true, shade the side of the circle containing (0,0) . Graph the parabola
2 5y x= − . Use a solid line since the inequality is not strict. Choose a test point not on the parabola, such as (0,0) . Since 20 0 5≤ − is false, shade the side of the parabola opposite that which contains the point (0,0) . The overlapping region is the solution.
Graph the hyperbola 4xy = . Use a solid line since the inequality uses ≥ . Choose a test point not on the parabola, such as (0, 0). Since 0 0 4⋅ ≥ is false, shade the opposite side of the hyperbola from (0, 0). Graph the parabola
2 1y x= + . Use a solid line since the inequality uses ≥ . Choose a test point not on the parabola, such as (0, 0). Since 20 0 1≥ + is false, shade the opposite side of the parabola from (0, 0).The overlapping region is the solution.
42. 2
2
1
1
y x
y x
⎧ + ≤⎪⎨
≥ −⎪⎩
Graph the parabola 2 1y x+ = . Use a solid line since the inequality uses ≤ . Choose a test point not on the parabola, such as (0, 0). Since 0 + 02 ≤ 1 is true, shade the same side of the parabola as (0, 0). Graph the parabola
2 1y x= − . Use a solid line since the inequality uses ≥ . Choose a test point not on the parabola, such as (0, 0). Since 20 0 1≥ − is true, shade the same side of the parabola as (0, 0). The overlapping region is the solution.
43.
00
2 62 6
xy
x yx y
≥⎧⎪ ≥⎪⎨ + ≤⎪⎪ + ≤⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2 6x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 6+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 6x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 6+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded. Find the vertices:
The x-axis and y-axis intersect at (0, 0). The intersection of 2 6x y+ = and the y-axis is (0, 3). The intersection of 2 6x y+ = and the x-axis is (3, 0). To find the intersection of 2 6x y+ = and 2 6x y+ = , solve the system:
2 62 6
x yx y
+ =⎧⎨ + =⎩
Solve the first equation for x: 6 2x y= − . Substitute and solve: 2(6 2 ) 6
12 4 612 3 6
3 62
y yy y
yyy
− + =− + =
− =− = −
=
6 2(2) 2x = − = The point of intersection is (2, 2). The four corner points are (0, 0), (0, 3), (3, 0), and (2, 2).
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 4x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 4+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 3 6x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 6+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. The graph is unbounded.
Find the vertices: The intersection of 4x y+ = and the y-axis is (0, 4). The intersection of 4x y+ = and the x-axis is (4, 0). The two corner points are (0, 4), and (4, 0).
=
45.
002
2 4
xy
x yx y
≥⎧⎪ ≥⎪⎨ + ≥⎪⎪ + ≥⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 + 0 ≥ 2 is false, shade the opposite side of the line from (0, 0). Graph the line 2 4x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 4+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. The graph is unbounded.
Find the vertices: The intersection of 2x y+ = and the x-axis is (2, 0). The intersection of 2 4x y+ = and the y-axis is (0, 4). The two corner points are (2, 0), and (0, 4).
46.
00
3 62 2
xy
x yx y
≥⎧⎪ ≥⎪⎨ + ≤⎪⎪ + ≤⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 3 6x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 3(0) 0 6+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of 0 and 0x y= = is (0, 0). The intersection of 2 2x y+ = and the x-axis is (1, 0). The intersection of 2 2x y+ = and the y-axis is (0, 2). The three corner points are (0, 0), (1, 0), and (0, 2).
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 3 12x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 12+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 3 12x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 3(0) 0 12+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of 2x y+ = and the y-axis is (0, 2). The intersection of 2x y+ = and the x-axis is (2, 0). The intersection of 2 3 12x y+ = and the y-axis is (0, 4). The intersection of 3 12x y+ = and the x-axis is (4, 0).
To find the intersection of 2 3 12x y+ = and 3 12x y+ = , solve the system:
2 3 123 12
x yx y
+ =⎧⎨ + =⎩
Solve the second equation for y: 12 3y x= − . Substitute and solve: 2 3(12 3 ) 12
2 36 9 127 24
247
x xx x
x
x
+ − =+ − =
− = −
=
24 72 1212 3 127 7 7
y ⎛ ⎞= − = − =⎜ ⎟⎝ ⎠
The point of intersection is 24 12,7 7
⎛ ⎞⎜ ⎟⎝ ⎠
.
The five corner points are (0, 2), (0, 4), (2, 0),
(4, 0), and 24 12,7 7
⎛ ⎞⎜ ⎟⎝ ⎠
.
48.
00210
2 3
xy
x yx yx y
≥⎧⎪ ≥⎪⎪ + ≥⎨⎪ + ≤⎪
+ ≤⎪⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 10x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 10+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 3x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 3+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of 2x y+ = and the y-axis is (0, 2). The intersection of 2 3x y+ = and the y-axis is (0, 3). To find the intersection of 2 3 and 2x y x y+ = + = , solve the system:
2 32
x yx y
+ =⎧⎨ + =⎩
Solve the second equation for y: 2y x= − . Substitute and solve: 2 2 3
1x x
x+ − =
=
2 1 1y = − = The point of intersection is (1, 1). The three corner points are (0, 2), (0, 3), and (1, 1).
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 8x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 8+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 10x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 10+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of 2x y+ = and the y-axis is (0, 2). The intersection of 2x y+ = and the x-axis is (2, 0). The intersection of 8x y+ = and the y-axis is (0, 8). The intersection of 2 10x y+ = and the x-axis is (5, 0). To find the intersection of
8 and 2 10x y x y+ = + = , solve the system: 8
2 10x yx y
+ =⎧⎨ + =⎩
Solve the first equation for y: 8y x= − . Substitute and solve: 2 8 10
2x x
x+ − =
=
8 2 6y = − = The point of intersection is (2, 6). The five corner points are (0, 2), (0, 8), (2, 0), (5, 0), and (2, 6).
50.
0028
2 1
xy
x yx yx y
≥⎧⎪ ≥⎪⎪ + ≥⎨⎪ + ≤⎪
+ ≥⎪⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 8x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 8+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 1x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 1+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of 2x y+ = and the y-axis is (0, 2). The intersection of 2x y+ = and the x-axis is (2, 0). The intersection of 8x y+ = and the y-axis is (0, 8). The intersection of 8x y+ = and the x-axis is (8, 0). The four corner points are (0, 2), (0, 8), (2, 0), and (8, 0).
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2 1x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 1+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 10x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 10+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of 2 1x y+ = and the y-axis is (0, 0.5). The intersection of 2 1x y+ = and the x-axis is (1, 0). The intersection of 2 10x y+ = and the y-axis is (0, 5). The intersection of
2 10x y+ = and the x-axis is (10, 0). The four corner points are (0, 0.5), (0, 5), (1, 0), and (10, 0).
52.
00
2 12 10
28
xy
x yx y
x yx y
≥⎧⎪ ≥⎪⎪ + ≥⎪⎨ + ≤⎪⎪ + ≥⎪
+ ≤⎪⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2 1x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 1+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 2 10x y+ = . Use a solid line since the inequality uses ≤. Choose a
test point not on the line, such as (0, 0). Since 0 2(0) 10+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). Graph the line 8x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 8+ ≤ is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of 2x y+ = and the y-axis is (0, 2). The intersection of 2x y+ = and the x-axis is (2, 0). The intersection of 2 10x y+ = and the y-axis is (0, 5). The intersection of 8x y+ = and the x-axis is (8, 0). To find the intersection of
8x y+ = and 2 10x y+ = , solve the system: 8
2 10x yx y
+ =⎧⎨ + =⎩
Solve the first equation for x: 8x y= − . Substitute and solve: (8 ) 2 10
2y y
y− + =
=
8 2 6x = − = The point of intersection is (6, 2). The five corner points are (0, 2), (0, 5), (2, 0), (8, 0), and (6, 2).
57. a. Let x = the amount invested in Treasury bills, and let y = the amount invested in corporate bonds. The constraints are:
50,000x y+ ≤ because the total investment cannot exceed $50,000.
35,000x ≥ because the amount invested in Treasury bills must be at least $35,000.
10,000y ≤ because the amount invested in corporate bonds must not exceed $10,000.
0, 0x y≥ ≥ because a non-negative amount must be invested. The system is
50,00035,00010,00000
x yxyxy
+ ≤⎧⎪ ≥⎪⎪ ≤⎨⎪ ≥⎪
≥⎪⎩
b. Graph the system.
The corner points are (35,000, 0),
(35,000, 10,000), (40,000, 10,000), (50,000, 0).
58. a. Let x = the # of standard model trucks, and let y = the # of deluxe model trucks. The constraints are:
0, 0x y≥ ≥ because a non-negative number of trucks must be manufactured. 2 3 80x y+ ≤ because the total painting hours worked cannot exceed 80. 3 4 120x y+ ≤ because the total detailing hours worked cannot exceed 120. The system is
59. a. Let x = the # of packages of the economy blend, and let y = the # of packages of the superior blend. The constraints are:
0, 0x y≥ ≥ because a non-negative # of packages must be produced. 4 8 75 16x y+ ≤ ⋅ because the total amount of “A grade” coffee cannot exceed 75 pounds. (Note: 75 pounds = (75)(16) ounces.) 12 8 120 16x y+ ≤ ⋅ because the total amount of “B grade” coffee cannot exceed 120 pounds. (Note: 120 pounds = (120)(16) ounces.) Simplifying the inequalities, we obtain:
4 8 75 162 75 42 300
x yx yx y
+ ≤ ⋅+ ≤ ⋅+ ≤
12 8 120 163 2 120 43 2 480
x yx yx y
+ ≤ ⋅+ ≤ ⋅+ ≤
The system is: 002 300
3 2 480
xyx yx y
≥⎧⎪ ≥⎪⎨ + ≤⎪⎪ + ≤⎩
b. Graph the system.
The corner points are (0, 0), (0, 150), (90, 105), (160, 0).
60. a. Let x = the # of lower-priced packages, and let y = the # of quality packages. The constraints are:
0, 0x y≥ ≥ because a non-negative # of packages must be produced. 8 6 120 16x y+ ≤ ⋅ because the total amount of peanuts cannot exceed 120 pounds. (Note: 120 pounds = (120)(16) ounces.) 4 6 90 16x y+ ≤ ⋅ because the total amount of cashews cannot exceed 90 pounds. (Note: 90 pounds = (90)(16) ounces.) Simplifying the inequalities, we obtain:
8 6 120 164 3 120 84 3 960
x yx yx y
+ ≤ ⋅+ ≤ ⋅+ ≤
4 6 90 162 3 90 82 3 720
x yx yx y
+ ≤ ⋅+ ≤ ⋅+ ≤
The system is 4 3 9602 3 720
0; 0
x yx y
x y
+ ≤⎧⎪ + ≤⎨⎪ ≥ ≥⎩
b. Graph the system.
The corner points are (0, 0), (0, 240), (120, 160), (240, 0).
61. a. Let x = the # of microwaves, and let y = the # of printers. The constraints are:
0, 0x y≥ ≥ because a non-negative # of items must be shipped. 30 20 1600x y+ ≤ because a total cargo weight cannot exceed 1600 pounds. 2 3 150x y+ ≤ because the total cargo volume cannot exceed 150 cubic feet. Note that the inequality 30 20 1600x y+ ≤ can be simplified: 3 2 160x y+ ≤ . The system is:
18. Maximize 2 4z x y= + subject to 0, 0, 2 4, 9x y x y x y≥ ≥ + ≥ + ≤ .
Graph the constraints.
y
x
x + y = 9
2x + y = 4
(0,9)
(9,0)(2,0)
(0,4)
The corner points are (0, 9), (9, 0), (0, 4), (2, 0).
Evaluate the objective function: Vertex Value of 2 4(0, 9) 2(0) 4(9) 36(9, 0) 2(9) 4(0) 18(0, 4) 2(0) 4(4) 16(2, 0) 2(2) 4(0) 4
z x yzzzz
= += + == + == + == + =
The maximum value is 36 at (0, 9).
19. Let x = the number of downhill skis produced, and let y = the number of cross-country skis produced. The total profit is: 70 50P x y= + . Profit is to be maximized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A positive number of skis must be produced.
2 40x y+ ≤ Manufacturing time available. 32x y+ ≤ Finishing time available. Graph the constraints.
y
x
(8,24)
(20,0)(0,0)
(0,32)
2x + y = 40
x + y = 32
To find the intersection of 32x y+ = and
2 40x y+ = , solve the system: 32
2 40x yx y
+ =⎧⎨ + =⎩
Solve the first equation for y: 32y x= − . Substitute and solve: 2 (32 ) 40
8x x
x+ − =
=
32 8 24y = − = The point of intersection is (8, 24). The corner points are (0, 0), (0, 32), (20, 0), (8, 24). Evaluate the objective function:
The maximum profit is $1920, when 16 downhill skis and 16 cross-country skis are produced.
20. Let x = the number of acres of soybeans planted , and let y = the number of acres of wheat planted. The total profit is: 180 100P x y= + . Profit is to be maximized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A non-negative number of acres must be planted.
70x y+ ≤ Acres available to plant. 60 30 1800x y+ ≤ Money available for
Evaluate the objective function: Vertex Value of 180 100(0, 0) 180(0) 100(0) 0(0, 30) 180(0) 100(30) 3000(40, 0) 180(40) 100(0) 7200
P x yP
PP
= += + =
= + == + =
The maximum profit is $7200, when 40 acres of soybeans and 0 acres of wheat are planted.
21. Let x = the number of rectangular tables rented, and let y = the number of round tables rented. The cost for the tables is: 28 52C x y= + . Cost is to be minimized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A non-negative number of tables must be used.
35x y+ ≤ Maximum number of tables. 6 10 250x y+ ≥ Number of guests. 15x ≤ Rectangular tables available.
(15, 16). Evaluate the objective function: Vertex Value of 28 52(0, 25) 28(0) 52(25) 1300(0, 35) 28(0) 52(35) 1820(15, 20) 28(15) 52(20) 1460(15, 16) 28(15) 52(16) 1252
C x yCCCC
= += + == + == + == + =
Kathleen should rent 15 rectangular tables and 16 round tables in order to minimize the cost. The minimum cost is $1252.00.
22. Let x = the number of buses rented, and let y = the number of vans rented. The cost for the vehicles is: 975 350C x y= + . Cost is to be minimized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A non-negative number of buses and vans must be used.
40 8 320x y+ ≥ Number of regular seats. 3 36x y+ ≥ Number of handicapped seats. Graph the constraints.
3(10) 36 30 36 6x = − + = − + = The point of intersection is (6, 10). The corner points are (0, 40), (6, 10), and (36, 0). Evaluate the objective function:
Vertex Value of 975 350(0, 40) 975(0) 350(40) 14,000(6, 10) 975(6) 350(10) 9350 (36, 0) 975(36) 350(0) 35,100
C x yCCC
= += + == + == + =
The college should rent 6 buses and 10 vans for a minimum cost of $9350.00.
23. Let x = the amount invested in junk bonds, and let y = the amount invested in Treasury bills. The total income is: 0.09 0.07I x y= + . Income is to be maximized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A non-negative amount must be invested.
20,000x y+ ≤ Total investment cannot exceed $20,000.
12,000x ≤ Amount invested in junk bonds must not exceed $12,000.
8000y ≥ Amount invested in Treasury bills must be at least $8000.
a. y x≥ Amount invested in Treasury bills must be equal to or greater than the amount invested in junk bonds.
The corner points are (0, 20,000), (0, 8000), (8000, 8000), (10,000, 10,000). Evaluate the objective function:
Vertex Value of 0.09 0.07(0, 20000) 0.09(0) 0.07(20000)
1400(0, 8000) 0.09(0) 0.07(8000)
560(8000, 8000) 0.09(8000) 0.07(8000)
1280(10000, 10000) 0.09(10000) 0.07(10000)
1600
I x yI
I
I
I
= += +== +== +== +=
The maximum income is $1600, when $10,000 is invested in junk bonds and $10,000 is invested in Treasury bills.
b. y x≤ Amount invested in Treasury bills must not exceed the amount invested in junk bonds.
Graph the constraints.
x + y = 20000
y = x
x = 12000
y = 8000(8000,8000)
(10000,10000)
(12000,8000)
The corner points are (12,000, 8000),
(8000, 8000), (10,000, 10,000). Evaluate the objective function:
Vertex Value of 0.09 0.07(12000, 8000) 0.09(12000) 0.07(8000)
1640(8000, 8000) 0.09(8000) 0.07(8000)
1280(10000, 10000) 0.09(10000) 0.07(10000)
1600
I x yI
I
I
= += +== +=
= +=
The maximum income is $1640, when $12,000 is invested in junk bonds and $8000 is invested in Treasury bills.
24. Let x = the number of hours that machine 1 is operated, and let y = the number of hours that machine 2 is operated. The total cost is:
50 30C x y= + . Cost is to be minimized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A positive number of hours must be used.
10x ≤ Time used on machine 1. 10y ≤ Time used on machine 2. 60 40 240x y+ ≥ 8-inch pliers to be produced. 70 20 140x y+ ≥ 6-inch pliers to be produced. Graph the constraints.
(10,10)(0,10)
(10,0)(4,0)
(0,7)
60x + 40y = 24070x + 20y = 140
12 , 21
4( )
To find the intersection of 60 40 240x y+ = and 70 20 140x y+ = , solve the system:
60 40 24070 20 140
x yx y
+ =⎧⎨ + =⎩
Divide the first equation by 2− and add the result to the second equation:
25. Let x = the number of pounds of ground beef, and let y = the number of pounds of ground pork. The total cost is: 0.75 0.45C x y= + . Cost is to be minimized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A positive number of pounds must be used.
200x ≤ Only 200 pounds of ground beef are available.
50y ≥ At least 50 pounds of ground pork must be used.
0.75 0.60 0.70( )x y x y+ ≥ + Leanness condition
(Note that the last equation will simplify to 12
y x≤ .) Graph the constraints.
y
x
(200,100)
(200,50)(100,50)
The corner points are (100, 50), (200, 50),
(200, 100). Evaluate the objective function:
Vertex Value of 0.75 0.45(100, 50) 0.75(100) 0.45(50) 97.50 (200, 50) 0.75(200) 0.45(50) 172.50(200, 100) 0.75(200) 0.45(100) 195
C x yCCC
= += + == + == + =
The minimum cost is $97.50, when 100 pounds of ground beef and 50 pounds of ground pork are used.
26. Let x = the number of gallons of regular, and let y = the number of gallons of premium. The
total profit is: 0.75 0.90P x y= + . Profit is to be maximized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A positive number of gallons must be used.
14
y x≥ At least one gallon of premium
for every 4 gallons of regular. 5 6 3000x y+ ≤ Daily shipping weight limit. 24 20 16(725)x y+ ≤ Available flavoring. 12 20 16(425)x y+ ≤ Available milk-fat (Note: the last two inequalities simplify to
6 5 2900x y+ ≤ and 3 5 1700x y+ ≤ .) Graph the constraints.
y
x
5 + 6 = 3000x y
(0, 340) (400, 100)
300
15014
y x=
24 + 20 = 16(725)x y
12 + 20 = 16(425)x y
The corner points are (0, 0), (400, 100), (0, 340). Evaluate the objective function:
The maximum profit is $280, when 10 racing skates and 15 figure skates are produced.
28. Let x = the amount placed in the AAA bond. Let y = the amount placed in a CD. The total return is: 0.08 0.04R x y= + . Return is to be maximized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A positive amount must be invested in each.
50,000x y+ ≤ Total investment cannot exceed $50,000.
20,000x ≤ Investment in the AAA bond cannot exceed $20,000.
15,000y ≥ Investment in the CD must be at least $15,000.
y x≥ Investment in the CD must exceed or equal the investment in the bond.
Graph the constraints.
y
x
y = x
x + y = 50000
x = 20000
y = 15000
(0,50000)
(0,15000)
(15000,15000)
(20000,30000)
(20000,20000)
The corner points are (0, 50,000), (0, 15,000),
(15,000, 15,000), (20,000, 20,000), (20,000, 30,000). Evaluate the objective function:
Vertex Value of 0.08 0.04(0, 50000) 0.08(0) 0.04(50000)
2000(0, 15000) 0.08(0) 0.04(15000)
600(15000, 15000) 0.08(15000) 0.04(15000)
1800(20000, 20000) 0.08(20000) 0.04(20000)
2400(20000, 30000
R x yR
R
R
R
= += +== +== +== +=
) 0.08(20000) 0.04(30000)2800
R = +=
The maximum return is $2800, when $20,000 is invested in a AAA bond and $30,000 is invested in a CD.
29. Let x = the number of metal fasteners, and let y = the number of plastic fasteners. The total cost is: 9 4C x y= + . Cost is to be minimized, so this is the objective function. The constraints are:
2, 2x y≥ ≥ At least 2 of each fastener must be made.
6x y+ ≥ At least 6 fasteners are needed. 4 2 24x y+ ≤ Only 24 hours are available. Graph the constraints.
y
x
(2,8)
(5,2)
(4,2)
(2,4)
The corner points are (2, 4), (2, 8), (4, 2), (5, 2).
Evaluate the objective function: Vertex Value of 9 4(2, 4) 9(2) 4(4) 34(2, 8) 9(2) 4(8) 50(4, 2) 9(4) 4(2) 44(5, 2) 9(5) 4(2) 53
C x yCCCC
= += + == + == + == + =
The minimum cost is $34, when 2 metal fasteners and 4 plastic fasteners are ordered.
30. Let x = the amount of “Gourmet Dog,” and let y = the amount of “Chow Hound.” The total
cost is: 0.40 0.32C x y= + . Cost is to be minimized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A non-negative number of cans must be purchased.
20 35 1175x y+ ≥ At least 1175 units of vitamins per month.
75 50 2375x y+ ≥ At least 2375 calories per month.
60x y+ ≤ Storage space for 60 cans. Graph the constraints.
y
x
x + y = 60
75x+50y=2375
20x+35y=1175
(0,47.5)
(0,60)
(60,0)
(58.75,0)
(15,25)
The corner points are (0, 47.5), (0, 60), (60, 0),
(58.75, 0), (15, 25). Evaluate the objective function:
Vertex Value of 0.40 0.32(0, 47.5) 0.40(0) 0.32(47.5) 15.20(0, 60) 0.40(0) 0.32(60) 19.20(60, 0) 0.40(60) 0.32(0) 24.00
The minimum cost is $14, when 15 cans of "Gourmet Dog" and 25 cans of “Chow Hound” are purchased.
31. Let x = the number of first class seats, and let y = the number of coach seats. Using the hint,
the revenue from x first class seats and y coach seats is ,Fx Cy+ where 0.F C> > Thus, R Fx Cy= + is the objective function to be maximized. The constraints are: 8 16x≤ ≤ Restriction on first class seats. 80 120y≤ ≤ Restriction on coach seats.
a. 112
xy
≤ Ratio of seats.
The constraints are: 8 16x≤ ≤ 80 120y≤ ≤ 12x y≤ Graph the constraints.
The corner points are (8, 96), (8, 120), and (10, 120). Evaluate the objective function:
Vertex Value of (8, 96) 8 96(8, 120) 8 120(10, 120) 10 120
R Fx CyR F CR F CR F C
= += += += +
Since 0,C > 120 96 ,C C> so 8 120 8 96 .F C F C+ > + Since 0,F > 10 8 ,F F> so 10 120 8 120 .F C F C+ > + Thus, the maximum revenue occurs when the aircraft is configured with 10 first class seats and 120 coach seats.
b. 18
xy
≤
The constraints are: 8 16x≤ ≤ 80 120y≤ ≤ 8x y≤ Graph the constraints.
The corner points are (8, 80), (8, 120),
(15, 120), and (10, 80). Evaluate the objective function:
Vertex Value of (8, 80) 8 80(8, 120) 8 120(15, 120) 15 120(10, 80) 10 80
R Fx CyR F CR F CR F CR F C
= += += += += +
Since 0F > and 0,C > 120 96 ,C C> the maximum value of R occurs at (15, 120). The maximum revenue occurs when the aircraft is configured with 15 first class seats and 120 coach seats.
c. Answers will vary.
32. Answers will vary.
Chapter 8 Review Exercises
1. 2 55 2 8
x yx y
− =⎧⎨ + =⎩
Solve the first equation for y: 2 5y x= − . Substitute and solve: 5 2(2 5) 8
5 4 10 89 18
2
x xx x
xx
+ − =+ − =
==
2(2) 5 4 5 1y = − = − = − The solution is 2, 1x y= = − or (2, 1)− .
2. 2 3 27 3x yx y
+ =⎧⎨ − =⎩
Solve the second equation for y: 7 3y x= − Substitute into the first equation and solve: 2 3(7 3) 2
2 21 9 223 11
1123
x xx x
x
x
+ − =+ − =
=
=
11 77 69 87 323 23 23 23
y ⎛ ⎞= − = − =⎜ ⎟⎝ ⎠
The solution is 11 8,23 23
x y= = or 11 8,23 23
⎛ ⎞⎜ ⎟⎝ ⎠
.
3. 3 4 4
132
x y
x y
− =⎧⎪⎨
− =⎪⎩
Solve the second equation for x: 132
x y= +
Substitute into the first equation and solve: 13 3 4 4239 4 42
Multiply each side of the second equation by –y and add the equations to eliminate y:
2 2
2 2
3 2
0
2 2 1
x x y y
x x y y
xx
− + + = −
− + − − =
− = −=
If 1:x = 2 2
2
1 3(1) 2
0( 1) 0
0 or 1
y y
y yy y
y y
− + + = −
+ =+ =
= = − Note that 0y ≠ because that would cause division by zero in the original system. Solution: (1, –1)
78. Multiply each side of the second equation by –x and add the equations to eliminate x:
2 2 2 2
2
2
2
2 221 2
2
2 0 ( 2) 0
0 or 2
x x y y x x y yyx x x y
xy y
y yy yy y
⎧ + + = + → + + = +⎪⎨ −
+ = → − − = −⎪⎩
=
− =− =
= =
2 2 2
2 2 2
If 0 :
0 0 2 2 0( 1)( 2) 0 1 or 2
If 2 :
2 2 2 0( 1) 0 0 or 1
y
x x x xx x x x
y
x x x xx x x x
=
+ + = + → + − =→ − + = → = = −
=
+ + = + → + =→ + = → = = −
Note that 0x ≠ because that would cause division by zero in the original system. Solutions: (1, 0), (–2, 0), (–1, 2)
79. 3 4 12x y+ ≤ Graph the line 3 4 12x y+ = . Use a solid line since the inequality uses ≤ . Choose a test point not on the line, such as (0, 0). Since
( ) ( )3 0 4 0 12+ ≤ is true, shade the side of the line containing (0, 0).
y
x−5
3 + 4 x y 12≤
5
5
−5
80. 2 3 6x y− ≥ Graph the line 2 3 6x y− = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since ( ) ( )2 0 3 0 6− ≥ is false, shade the side of the line opposite (0, 0).
Graph the parabola 2y x= . Use a solid curve since the inequality uses ≤ . Choose a test point not on the parabola, such as (0, 1). Since 20 1≤ is false, shade the opposite side of the parabola from (0, 1).
y
x−5 5
5
−5
2y x≤
82. 2x y≥
Graph the circle 2x y= . Use a solid curve since the inequality uses ≥ . Choose a test point not on the parabola, such as (1, 0). Since 21 0≥ is true, shade the same side of the parabola as (1, 0).
y
x−5 5
5
−5
2x y≥
83. 2 2
2x yx y
− + ≤⎧⎨ + ≥⎩
Graph the line 2 2x y− + = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since
2(0) 0 2− + ≤ is true, shade the side of the line containing (0, 0). Graph the line 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
y
–5
5
x–5 5
–2x + y = 2
x + y = 2
The graph is unbounded. Find the vertices: To find the intersection of 2x y+ = and
2 2x y− + = , solve the system: 2
2 2x yx y
+ =⎧⎨− + =⎩
Solve the first equation for x: 2x y= − . Substitute and solve:
2(2 ) 24 2 2
3 62
y yy y
yy
− − + =− + + =
==
2 2 0x = − = The point of intersection is (0, 2). The corner point is (0, 2).
84. 2 6
2 2x yx y
− ≤⎧⎨ + ≥⎩
Graph the line 2 6x y− = . Use a solid line since the inequality uses ≤ . Choose a test point not on the line, such as (0, 0). Since 0 2(0) 6− ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
y
–5
5
x–2 8
x – 2y = 6
(2, –2)
2x + y = 2 The graph is unbounded. Find the vertices: To find the intersection of 2 6x y− = and 2 2x y+ = , solve the system:
2 62 2x y
x y− =⎧
⎨ + =⎩
Solve the first equation for x: 2 6x y= + . Substitute and solve: 2(2 6) 2
4 12 25 10
2
y yy y
yy
+ + =+ + =
= −= −
2( 2) 6 2x = − + = The point of intersection is (2, 2)− . The corner point is (2, 2)− .
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 4x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 4+ ≤ is true, shade the side of the line containing (0, 0). Graph the line 2 3 6x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 6+ ≤ is true, shade the side of the line containing (0, 0).
y
–2
8
x8
(0, 2)
(3, 0)(0, 0)
x + y = 4
2x + 3y = 6
The overlapping region is the solution. The graph is bounded. Find the vertices: The x-axis and y-axis intersect at (0, 0). The intersection of 2 3 6x y+ = and the y-axis is (0, 2). The intersection of 2 3 6x y+ = and the x-axis is (3, 0). The three corner points are (0, 0), (0, 2), and (3, 0).
86.
00
3 62 2
xy
x yx y
≥⎧⎪ ≥⎪⎨ + ≥⎪⎪ + ≥⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 3 6x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 3(0) + 0 ≥ 6 is false, shade the opposite side of the line from (0, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) + 0 ≥ 2 is false, shade the opposite side of the line from (0, 0).
y
–2
8
x–2 8
(0, 6)
(2, 0)
3x + y = 6
2x + y = 2
The overlapping region is the solution. The graph
is unbounded. Find the vertices: The intersection of 3 6x y+ = and the y-axis is (0, 6). The intersection of 3 6x y+ = and the x-axis is (2, 0). The two corner points are (0, 6), and (2, 0).
87.
00
2 82 2
xy
x yx y
≥⎧⎪ ≥⎪⎨ + ≤⎪⎪ + ≥⎩
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 2 8x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) + 0 ≤ 8 is true, shade the side of the line containing (0, 0). Graph the line 2 2x y+ = . Use a solid line since the inequality uses ≥ . Choose a test point not on the line, such as (0, 0). Since 0 + 2(0) ≥ 2 is false, shade the opposite side of the line from (0, 0).
y
–1
9
x–1 9
2x + y = 8
x + 2y = 2
(2, 0)
(4, 0)(0, 1)
(0, 8)
The overlapping region is the solution. The graph
is bounded. Find the vertices: The intersection of 2 2x y+ = and the y-axis is (0, 1). The
intersection of 2 2x y+ = and the x-axis is (2, 0). The intersection of 2 8x y+ = and the y-axis is (0, 8). The intersection of 2 8x y+ = and the x-axis is (4, 0). The four corner points are (0, 1), (0, 8), (2, 0), and (4, 0).
Graph 0; 0x y≥ ≥ . Shaded region is the first quadrant. Graph the line 3 9x y+ = . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 3(0) 0 9+ ≤ is true, shade the side of the line containing (0, 0).
y
–1
9
x–1 9
(0, 2)
(0, 9)
(3, 0)
2x + 3y = 6
3x + y = 9
Graph the line 2 3 6x y+ = . Use a solid line since
the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 6+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. The graph is bounded. Find the vertices: The intersection of 2 3 6x y+ = and the y-axis is (0, 2). The intersection of 2 3 6x y+ = and the x-axis is (3, 0). The intersection of 3 9x y+ = and the y-axis is (0, 9). The intersection of 3 9x y+ = and the x-axis is (3, 0). The three corner points are (0, 2), (0, 9), and (3, 0).
89. Graph the system of inequalities: 2 2 16
2x y
x y⎧ + ≤⎪⎨
+ ≥⎪⎩
Graph the circle 2 2 16x y+ = .Use a solid line since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since
2 20 0 16+ ≤ is true, shade the side of the circle containing (0, 0). Graph the line 2x y+ = . Use a solid line since the inequality uses ≥ . Choose a test point not on the line, such as (0, 0). Since 0 0 2+ ≥ is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
y
–5
5
x–5 5
x + y = 2
x2+ y2 = 16
90. Graph the system of inequalities: 2 1
3y x
x y⎧ ≤ −⎪⎨
− ≤⎪⎩
Graph the parabola 2 1y x= − . Use a solid line since the inequality uses ≤ . Choose a test point not on the parabola, such as (0, 0). Since
20 0 1≤ − is false, shade the opposite side of the parabola from (0, 0). Graph the line 3x y− = . Use a solid line since the inequality uses ≤ . Choose a test point not on the line, such as (0, 0). Since 0 0 3− ≤ is true, shade the same side of the line as (0, 0). The overlapping region is the solution.
y
–5
5
x–2 8
x – y = 3
y2 = x – 1
91. Graph the system of inequalities: 2
4y x
xy⎧ ≤⎪⎨
≤⎪⎩
Graph the parabola 2y x= . Use a solid line since the inequality uses ≤ . Choose a test point not on the parabola, such as (1, 2). Since 22 1≤ is false, shade the opposite side of the parabola from (1, 2). Graph the hyperbola 4xy = . Use a solid line since the inequality uses ≤ . Choose a test point not on the hyperbola, such as (1, 2). Since 1 2 4⋅ ≤ is true, shade the same side of the hyperbola as (1, 2). The overlapping region is the solution.
Graph the circle 2 2 1x y+ = . Use a solid line since the inequality uses ≥ . Choose a test point not on the circle, such as (0, 0). Since
2 20 0 1+ ≥ is false, shade the opposite side of the circle from (0, 0). Graph the circle 2 2 4x y+ = . Use a solid line since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since
2 20 0 4+ ≤ is true, shade the same side of the circle as (0, 0). The overlapping region is the solution.
y
–5
5
x–5 5
x2 + y2 = 4
x2 + y2 = 1
93. Maximize 3 4z x y= + subject to 0x ≥ , 0y ≥ , 3 2 6x y+ ≥ , 8x y+ ≤ . Graph the constraints.
y
x
(0,8)
(8,0)(2,0)
(0,3)
The corner points are (0, 3), (2, 0), (0, 8), (8, 0).
Solve the first equation for a, substitute into the second equation and solve:
4( 1) 2 04 4 2 0
6 423
b bb b
b
b
− − − =− − − =
− =
= −
2 113 3
a = − = −
The quadratic function is 21 2 13 3
y x x= − − + .
100. 2 2 0x y Dx Ey F+ + + + = At (0, 1) the equation becomes:
2 20 1 (0) (1) 01
D E FE F
+ + + + =+ = −
At (1, 0) the equation becomes: 2 21 0 (1) (0) 0
1D E F
D F+ + + + =
+ = −
At (–2, 1) the equation becomes: 2 2( 2) 1 ( 2) (1) 0
2 5D E F
D E F− + + − + + =
− + + = −
The system of equations is: 11
2 5
E FD F
D E F
+ = −⎧⎪ + = −⎨⎪− + + = −⎩
Substitute 1E F+ = − into the third equation and solve for D:
2 ( 1) 52 4
2
DDD
− + − = −− = −
=
Substitute and solve: 2 1
3FF
+ = −= −
( 3) 12
EE
+ − = −=
The equation of the circle is 2 2 2 2 3 0x y x y+ + + − = .
101. Let x = the number of pounds of coffee that costs $6.00 per pound, and let y = the number of pounds of coffee that costs $9.00 per pound. Then 100x y+ = represents the total amount of coffee in the blend. The value of the blend will be represented by the equation: 6 9 6.90(100)x y+ = . Solve the system of equations:
1006 9 690
x yx y
+ =⎧⎨ + =⎩
Solve the first equation for y: 100y x= − . Solve by substitution: 6 9(100 ) 690
6 900 9 6903 210
70
x xx x
xx
+ − =+ − =
− = −=
100 70 30y = − = The blend is made up of 70 pounds of the $6.00-per-pound coffee and 30 pounds of the $9.00-per-pound coffee.
102. Let x = the number of acres of corn, and let y = the number of acres of soybeans. Then
1000x y+ = represents the total acreage on the farm. The total cost will be represented by the equation: 65 45 54,325x y+ = . Solve the system of equations:
100065 45 54,325
x yx y
+ =⎧⎨ + =⎩
Solve the first equation for y: 1000y x= − Solve by substitution: 65 45(1000 ) 54,325
65 45,000 45 54,32520 9325
466.25
x xx x
xx
+ − =+ − =
==
1000 466.25 533.75y = − = Corn should be planted on 466.25 acres and soybeans should be planted on 533.75 acres.
103. Let x = the number of small boxes, let y = the number of medium boxes, and let z = the number of large boxes. Oatmeal raisin equation: 2 2 15x y z+ + = Chocolate chip equation: 2 10x y z+ + = Shortbread equation: 3 11y z+ =
2 2 152 103 11
x y zx y z
y z
+ + =⎧⎪ + + =⎨⎪ + =⎩
Multiply each side of the second equation by –1 and add to the first equation to eliminate x:
2 2 152 10
3 11 5
x y zx y z
y zy
+ + =⎧⎪⎨− − − = −⎪⎩
+ ==
Substituting and solving for the other variables: 5 3 11
3 62
zzz
+ ===
5 2(2) 10
9 101
xx
x
+ + =+ =
=
Thus, 1 small box, 5 medium boxes, and 2 large boxes of cookies should be purchased.
104. a. Let x = the number of lower-priced packages, and let y = the number of quality packages. Peanut inequality: 8 6 120(16)
4 3 960x yx y
+ ≤+ ≤
Cashew inequality: 4 6 72(16)2 3 576
x yx y
+ ≤+ ≤
The system of inequalities is:
00
4 3 9602 3 576
xy
x yx y
≥⎧⎪ ≥⎪⎨ + ≤⎪⎪ + ≤⎩
b. Graphing:
To find the intersection of 2 3 576x y+ = and 4 3 960x y+ = , solve the system:
4 3 9602 3 576
x yx y
+ =⎧⎨ + =⎩
Subtract the second equation from the first:
4 3 9602 3 576
2 384 192
x yx y
xx
+ =− − =
==
Substitute and solve: 2(192) 3 576
3 19264
yyy
+ ===
The corner points are (0, 0), (0, 192), (240, 0), and (192, 64).
105. Let x = the speed of the boat in still water, and let y = the speed of the river current. The distance from Chiritza to the Flotel Orellana is 100 kilometers.
106. Let x = the speed of the jet stream, and let d = the distance from Chicago to Ft. Lauderdale. The jet stream flows from Chicago to Ft. Lauderdale because the time is shorter in that direction.
Rate Time DistanceChicago to
475 5 / 2Ft. LauderdaleFt. Lauderdale
475 17 / 6to Chicago
x d
x d
+
−
The system of equations is: ( )( )
( )( )475 5 / 2
475 17 / 6
y d
y d
+ =⎧⎪⎨
− =⎪⎩
Simplifying the system, we obtain: 2 5 2375
6 17 8075d x
d x− =⎧
⎨ + =⎩
Multiply the first equation by 3− and add the result to the second equation:
6 15 71256 17 8075
d xd x
− + = −+ =
32 950950 47532 16
x
x
=
= =
The speed of the jet stream is approximately 475 /16 29.69≈ miles per hour.
107. Let x = the number of hours for Bruce to do the job alone, let y = the number of hours for Bryce to do the job alone, and let z = the number of hours for Marty to do the job alone. Then 1/x represents the fraction of the job that Bruce does in one hour. 1/y represents the fraction of the job that Bryce does in one hour. 1/z represents the fraction of the job that Marty does in one hour. The equation representing Bruce and Bryce working together is:
( )1 1 1 3 0.75
4 / 3 4x y+ = = =
The equation representing Bryce and Marty working together is:
( )1 1 1 5 0.625
8 / 5 6y z+ = = =
The equation representing Bruce and Marty working together is:
( )1 1 1 3 0.375
8 / 3 8x z+ = = =
Solve the system of equations: 1 1
1 1
1 1
0.75
0.625
0.375
x y
y z
x z
− −
− −
− −
⎧ + =⎪⎪ + =⎨⎪ + =⎪⎩
Let 1 1 1, ,u x v y w z− − −= = = 0.750.6250.375
u vv wu w
+ =⎧⎪ + =⎨⎪ + =⎩
Solve the first equation for u: 0.75u v= − . Solve the second equation for w: 0.625w v= − . Substitute into the third equation and solve: (0.75 ) (0.625 ) 0.375
2 10.5
v vvv
− + − =− = −
=
0.75 0.5 0.25u = − = 0.625 0.5 0.125w = − =
Solve for , , andx y z : 4, 2, 8 (reciprocals)x y z= = =
Bruce can do the job in 4 hours, Bryce in 2 hours, and Marty in 8 hours.
108. Let x = the number of dancing girls produced, and let y = the number of mermaids produced. The total profit is: 25 30P x y= + . Profit is to be maximized, so this is the objective function. The constraints are:
0, 0x y≥ ≥ A non-negative number of figurines must be produced.
3 3 90x y+ ≤ 90 hours are available for molding.
6 4 120x y+ ≤ 120 hours are available for painting.
2 3 60x y+ ≤ 60 hours are available for glazing. Graph the constraints.
The maximum profit is $660, when 12 dancing girl and 12 mermaid figurines are produced each day. To determine the excess, evaluate each constraint at 12x = and 12y = :
The minimum cost is $24,000, when 35 gasoline engines and 15 diesel engines are produced. The excess capacity is 15 gasoline engines, since only 20 gasoline engines had to be delivered.
110. Answers will vary.
Chapter 8 Test
1. 2 74 3 9
x yx y
− + = −⎧⎨ + =⎩
Substitution: We solve the first equation for y, obtaining
2 7y x= − Next we substitute this result for y in the second equation and solve for x.
( )4 3 9
4 3 2 7 94 6 21 9
10 3030 310
x yx x
x xx
x
+ =
+ − =
+ − ==
= =
We can now obtain the value for y by letting 3x = in our substitution for y.
The solution of the system is 3x = , 1y = − or (3, 1)− .
Elimination: Multiply each side of the first equation by 2 so that the coefficients of x in the two equations are negatives of each other. The result is the equivalent system
4 2 144 3 9
x yx y
− + = −⎧⎨ + =⎩
We can replace the second equation of this system by the sum of the two equations. The result is the equivalent system
4 2 145 5
x yy
− + = −⎧⎨ = −⎩
Now we solve the second equation for y. 5 5
5 15
y
y
= −−
= = −
We back-substitute this value for y into the original first equation and solve for x.
( )2 7
2 1 72 6
6 32
x yx
x
x
− + = −
− + − = −
− = −−
= =−
The solution of the system is 3x = , 1y = − or (3, 1)− .
2. 1 2 13
5 30 18
x y
x y
⎧ − =⎪⎨⎪ − =⎩
We choose to use the method of elimination and multiply the first equation by 15− to obtain the equivalent system
5 30 155 30 18
x yx y
− + = −⎧⎨ − =⎩
We replace the second equation by the sum of the two equations to obtain the equivalent system
5 30 150 3
x y− + = −⎧⎨ =⎩
The second equation is a contradiction and has no solution. This means that the system itself has no solution and is therefore inconsistent.
3. 2 5 (1)
3 4 2 (2)5 2 3 8 (3)
x y zx y z
x y z
− + =⎧⎪ + − = −⎨⎪ + + =⎩
We use the method of elimination and begin by eliminating the variable y from equation (2). Multiply each side of equation (1) by 4 and add the result to equation (2). This result becomes our new equation (2). 2 5 4 4 8 20
3 4 2 3 4 2
7 7 18 (2)
x y z x y zx y z x y z
x z
− + = − + =+ − = − + − = −
+ =
We now eliminate the variable y from equation (3) by multiplying each side of equation (1) by 2 and adding the result to equation (3). The result becomes our new equation (3). 2 5 2 2 4 105 2 3 8 5 2 3 8
7 7 18 (3)
x y z x y zx y z x y z
x z
− + = − + =+ + = + + =
+ =
Our (equivalent) system now looks like 2 5 (1)
7 7 18 (2)7 7 18 (3)
x y zx zx z
− + =⎧⎪ + =⎨⎪ + =⎩
Treat equations (2) and (3) as a system of two equations containing two variables, and eliminate the x variable by multiplying each side of equation (2) by 1− and adding the result to equation (3). The result becomes our new equation (3). 7 7 18 7 7 187 7 18 7 7 18 0 0 (3)
x z x zx z x z
+ = − − = −+ = + =
=
We now have the equivalent system 2 5 (1)
7 7 18 (2)0 0 (3)
x y zx z
− + =⎧⎪ + =⎨⎪ =⎩
This is equivalent to a system of two equations with three variables. Since one of the equations contains three variables and one contains only two variables, the system will be dependent. There are infinitely many solutions. We solve equation (2) for x and determine that
We start by clearing the fraction in equation (2) by multiplying both sides of the equation by 3.
3 2 8 3 (1)3 2 3 3 (2)
6 3 15 8 (3)
x y zx y z
x y z
+ − = −⎧⎪− − + =⎨⎪ − + =⎩
We use the method of elimination and begin by eliminating the variable x from equation (2). The coefficients on x in equations (1) and (2) are negatives of each other so we simply add the two equations together. This result becomes our new equation (2). 3 2 8 3
3 2 3 3
5 0 (2)
x y zx y z
z
+ − = −− − + =
− =
We now eliminate the variable x from equation (3) by multiplying each side of equation (1) by
2− and adding the result to equation (3). The result becomes our new equation (3). 3 2 8 3 6 4 16 66 3 15 8 6 3 15 8
7 31 14 (3)
x y z x y zx y z x y z
y z
+ − = − − − + =− + = − + =
− + =
Our (equivalent) system now looks like 3 2 8 3 (1)
5 0 (2)7 31 14 (3)
x y zz
y z
+ − = −⎧⎪ − =⎨⎪ − + =⎩
We solve equation (2) for z by dividing both
sides of the equation by 5− . 5 0
0zz
− ==
Back-substitute 0z = into equation (3) and solve for y.
7 31 147 31(0) 14
7 142
y zy
yy
− + =− + =
− == −
Finally, back-substitute 2y = − and 0z = into equation (1) and solve for x.
3 2 8 33 2( 2) 8(0) 3
3 4 33 1
13
x y zx
xx
x
+ − = −+ − − = −
− = −=
=
The solution of the original system is 13
x = , 2y = − , 0z = or 1 , 2, 03
⎛ ⎞−⎜ ⎟⎝ ⎠
.
5. 4 5 02 6 19
5 5 10
x y zx y
x y z
− + =⎧⎪− − + = −⎨⎪ + − =⎩
We first check the equations to make sure that all variable terms are on the left side of the equation and the constants are on the right side. If a variable is missing, we put it in with a coefficient of 0. Our system can be rewritten as
4 5 02 0 25
5 5 10
x y zx y z
x y z
− + =⎧⎪− − + = −⎨⎪ + − =⎩
The augmented matrix is 4 5 1 02 1 0 25
1 5 5 10
−⎡ ⎤⎢ ⎥− − −⎢ ⎥
−⎢ ⎥⎣ ⎦
6. The matrix has three rows and represents a system with three equations. The three columns to the left of the vertical bar indicate that the system has three variables. We can let x, y, and z denote these variables. The column to the right of the vertical bar represents the constants on the right side of the equations. The system is
9. AC cannot be computed because the dimensions are mismatched. To multiply two matrices, we need the number of columns in the first matrix to be the same as the number of rows in the second matrix. Matrix A has 2 columns, but matrix C has 3 rows. Therefore, the operation cannot be performed.
10. Here we are taking the product of a 2 3× matrix and a 3 2× matrix. Since the number of columns in the first matrix is the same as the number of rows in the second matrix (3 in both cases), the operation can be performed and will result in a 2 2× matrix.
( ) ( ) ( ) ( )( ) ( )
1 1 2 0 5 3 1 1 2 4 5 2
0 1 3 0 1 3 0 1 3 4 1 2
1 11 2 5
0 40 3 1
3 2
16 173 10
BA
⋅ + − ⋅ + ⋅ ⋅ − + − ⋅ − + ⋅
⋅ + ⋅ + ⋅ ⋅ − + − + ⋅
−⎡ ⎤−⎡ ⎤ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦
= ⎡ ⎤⎢ ⎥⎣ ⎦⎡ ⎤
= ⎢ ⎥−⎣ ⎦
11. We first form the matrix
[ ]2
3 2 1 0|
5 4 0 1A I
⎡ ⎤= ⎢ ⎥
⎣ ⎦
Next we use row operations to transform [ ]2|A I into reduced row echelon form.
( )1 1
2 113 33
3 2 1 05 4 0 1
1 0
5 4 0 1R r
⎡ ⎤⎢ ⎥⎣ ⎦
⎡ ⎤→ =⎢ ⎥
⎢ ⎥⎣ ⎦
( )
( )
( )
2 1 2
2 2
1 2 1
2 13 3
523 32 13 3 3
25 32 2
25 3 32 2
1 0 5
0 1
1 0
0 1
1 0 2 1
0 1
R r r
R r
R rr
⎡ ⎤→ = − +⎢ ⎥
−⎢ ⎥⎣ ⎦⎡ ⎤
→ =⎢ ⎥−⎢ ⎥⎣ ⎦
−⎡ ⎤→ = − +⎢ ⎥−⎣ ⎦
Therefore, 15 32 2
2 1A−
−⎡ ⎤= ⎢ ⎥−⎣ ⎦
.
12. We first form the matrix
[ ]3
1 1 1 1 0 0| 2 5 1 0 1 0
2 3 0 0 0 1B I
−⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦
Next we use row operations to transform [ ]3|B I into reduced row echelon form.
We start by writing the augmented matrix for the system.
6 3 122 1 2
⎡ ⎤⎢ ⎥− −⎣ ⎦
Next we use row operations to transform the augmented matrix into row echelon form.
( )
( )
1 2
2 1
11 12
2 1 2
12
12
6 3 12 2 1 2
2 1 2 6 3 12
1 1
6 3 12
1 1 6
0 6 18
R rR r
R r
R r r
=− − ⎛ ⎞⎡ ⎤ ⎡ ⎤→ ⎜ ⎟⎢ ⎥ ⎢ ⎥− − =⎣ ⎦ ⎣ ⎦ ⎝ ⎠
⎡ ⎤− −→ =⎢ ⎥
⎢ ⎥⎣ ⎦⎡ ⎤− −
→ = − +⎢ ⎥⎢ ⎥⎣ ⎦
( )
( )
12 26
12 2 12
12
12
1 1
0 1 3
1 0
0 1 3
R r
R r r
⎡ ⎤− −→ =⎢ ⎥
⎢ ⎥⎣ ⎦⎡ ⎤
→ = +⎢ ⎥⎢ ⎥⎣ ⎦
The solution of the system is 12x = , 3y = or
( )12 , 3
14. 1 74
8 2 56
x y
x y
⎧ + =⎪⎨⎪ + =⎩
We start by writing the augmented matrix for the system.
141 7
8 2 56⎡ ⎤⎢ ⎥⎣ ⎦
Next we use row operations to transform the augmented matrix into row echelon form.
14
2 1 2
14
1 788 2 56
1 70 0 0
R R r⎡ ⎤⎢ ⎥ = − +⎣ ⎦⎡ ⎤⎢ ⎥⎣ ⎦
The augmented matrix is now in row echelon form. Because the bottom row consists entirely of 0’s, the system actually consists of one equation in two variables. The system is dependent and therefore has an infinite number of solutions. Any ordered pair satisfying the
equation 1 74
x y+ = , or 4 28y x= − + , is a
solution to the system.
15. 2 4 32 7 15 124 7 13 10
x y zx y zx y z
+ + = −⎧⎪ + + = −⎨⎪ + + = −⎩
We start by writing the augmented matrix for the system.
1 2 4 32 7 15 124 7 13 10
−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−⎣ ⎦
Next we use row operations to transform the augmented matrix into row echelon form.
( )
2 1 2
3 1 3
2 3
3 2
3 2 3
1 2 4 32 7 15 124 7 13 10
1 2 4 32
0 3 7 6 4
0 1 3 2
1 2 4 30 1 3 2 0 3 7 6
1 2 4 30 1 3 2 30 0 2 0
1 2 4 30 1 3 20 0 1 0
R r rR r r
R rR r
R r r
−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−⎣ ⎦
−⎡ ⎤= − +⎛ ⎞⎢ ⎥→ − ⎜ ⎟⎢ ⎥ = − +⎝ ⎠⎢ ⎥− −⎣ ⎦
−⎡ ⎤= −⎛ ⎞⎢ ⎥→ − ⎜ ⎟⎢ ⎥ =⎝ ⎠⎢ ⎥−⎣ ⎦
−⎡ ⎤⎢ ⎥→ − = − +⎢ ⎥⎢ ⎥−⎣ ⎦
−⎡ ⎤⎢→ −⎢⎢⎣
( )13 32 R r⎥ = −⎥
⎥⎦
The matrix is now in row echelon form. The last row represents the equation 0z = . Using 0z = we back-substitute into the equation 3 2y z+ = − (from the second row) and obtain
( )3 2
3 0 22
y zy
y
+ = −
+ = −
= −
Using 2y = − and 0z = , we back-substitute into the equation 2 4 3x y z+ + = − (from the first row) and obtain
( ) ( )2 4 3
2 2 4 0 31
x y zx
x
+ + = −
+ − + = −
=
The solution is 1x = , 2y = − , 0z = or (1, 2, 0)− .
The solution of the system is 1x = , 1y = − , 4z = or (1, 1, 4)− .
21. 2 2
2
3 12
9
x y
y x
⎧ + =⎪⎨
=⎪⎩
Substitute 9x for 2y into the first equation and solve for x:
( )2
2
2
3 9 12
3 9 12 0
3 4 0( 1)( 4) 0
x x
x x
x xx x
+ =
+ − =
+ − =− + =
1 or 4x x= = − Back substitute these values into the second equation to determine y:
1x = : 2 9(1) 93
yy
= == ±
4x = − : 2 9( 4) 36
36 (not real)
y
y
= − = −
= ± −
The solutions of the system are (1, 3)− and (1, 3) .
22. 2 22 3 5
1 1y x
y x y x⎧ − =⎨
− = ⇒ = +⎩
Substitute 1x + for y into the first equation and solve for x:
( )( )
2 2
2 2
2 2
2
2
2 1 3 5
2 2 1 3 5
2 4 2 3 5
4 3 0
4 3 0( 1)( 3) 0
x x
x x x
x x x
x x
x xx x
+ − =
+ + − =
+ + − =
− + − =
− + =− − =
1 or 3x x= = Back substitute these values into the second equation to determine y:
1x = : 1 1 2y = + = 3x = : 3 1 4y = + =
The solutions of the system are (1, 2) and (3, 4) .
23. 2 2 100
4 3 0x y
x y⎧ + ≤⎨
− ≥⎩
Graph the circle 2 2 100x y+ = . Use a solid curve since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since
2 20 0 100+ ≤ is true, shade the same side of the circle as (0, 0); that is, inside the circle.
Graph the line 4 3 0x y− = . Use a solid line since the inequality uses ≥ . Choose a test point not on the line, such as (0, 1). Since 4(0) 3(1) 0− ≥ is false, shade the opposite side of the line from (0, 1). The overlapping region is the solution.
The denominator contains the repeated linear factor 3x + . Thus, the partial fraction decomposition takes on the form
( ) ( )2 2
3 733 3
x A Bxx x
+= +
++ +
Clear the fractions by multiplying both sides by ( )23x + . The result is the identity
( )3 7 3x A x B+ = + + or
( )3 7 3x Ax A B+ = + + We equate coefficients of like powers of x to obtain the system
37 3
AA B
=⎧⎨ = +⎩
Therefore, we have 3A = . Substituting this result into the second equation gives
( )7 37 3 32
A BB
B
= +
= +
− =
Thus, the partial fraction decomposition is
( ) ( )2 2
3 7 3 233 3
xxx x
+ −= +
++ +.
25. ( )
2
22
4 3
3
x
x x
−
+
The denominator contains the linear factor x and the repeated irreducible quadratic factor 2 3x + . The partial fraction decomposition takes on the form
( ) ( )2
2 2 22 2
4 333 3
x A Bx C Dx Ex xx x x
− + += + +
++ +
We clear the fractions by multiplying both sides
by ( )22 3x x + to obtain the identity
( ) ( )( ) ( )2 2224 3 3 3x A x x x Bx C x Dx E− = + + + + + +
Collecting like terms yields ( ) ( )( ) ( )
2 4 3 24 3 6 3
3 9
x A B x Cx A B D x
C E x A
− = + + + + +
+ + +
Equating coefficients, we obtain the system
00
6 3 43 0
9 3
A BC
A B DC E
A
+ =⎧⎪ =⎪⎪ + + =⎨⎪ + =⎪
= −⎪⎩
From the last equation we get 13
A = − .
Substituting this value into the first equation
gives 13
B = . From the second equation, we
know 0C = . Substituting this value into the fourth equation yields 0E = .
Substituting 13
A = − and 13
B = into the third
equation gives us ( ) ( )1 1
3 36 3 4
2 1 45
D
DD
− + + =
− + + ==
Therefore, the partial fraction decomposition is
( ) ( ) ( )
2
2 222 2
1 13 34 3 5
33 3
xx xx xx x x
−− = + +++ +
26. 002 8
2 3 2
xyx y
x y
≥⎧⎪ ≥⎪⎨ + ≥⎪⎪ − ≥⎩
The inequalities 0x ≥ and 0y ≥ require that the graph be in quadrant I.
2 81 42
x y
y x
+ ≥
≥ − +
Test the point ( )0,0 .
( )2 8
0 2 0 8 ?0 8 false
x y+ ≥
+ ≥
≥
The point ( )0,0 is not a solution. Thus, the graph of the inequality 2 8x y+ ≥ includes the
half-plane above the line 1 42
y x= − + . Because
the inequality is non-strict, the line is also part of the graph of the solution.
The point ( )0,0 is not a solution. Thus, the graph of the inequality 2 3 2x y− ≥ includes the
half-plane below the line 2 23 3
y x= − .
Because the inequality is non-strict, the line is also part of the graph of the solution. The overlapping shaded region (that is, the shaded region in the graph below) is the solution to the system of linear inequalities.
The graph is unbounded. The corner points are ( )4,2 and ( )8,0 .
27. The objective function is 5 8z x y= + . We seek the largest value of z that can occur if x and y are solutions of the system of linear inequalities
02 8
3 3
xx y
x y
≥⎧⎪ + ≤⎨⎪ − ≤ −⎩
2 82 8
x yy x
+ == − +
3 33 3
1 13
x yy x
y x
− = −− = − −
= +
The graph of this system (the feasible points) is shown as the shaded region in the figure below. The corner points of the feasible region are ( )0,1 , ( )3,2 , and ( )0,8 .
4
8
y
x4 8
2 8x y+ =
3 3x y− = −
(0, 1)(3, 2)
(0, 8)
( )
( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )
Corner point, , Value of obj. function, 0,1 5 0 8 1 83,2 5 3 8 2 310,8 5 0 8 8 64
x y zz
zz
= + == + == + =
From the table, we can see that the maximum value of z is 64, and it occurs at the point ( )0,8 .
28. Let j = unit price for flare jeans, c = unit price for camisoles, and t = unit price for t-shirts. The given information yields a system of equations with each of the three women yielding an equation.
The sum of each row is 1 (or 100%). These represent the three possibilities of educational achievement for a parent of a child, unless someone does not attend school at all. Since these are rounded percents, chances are the other possibilities are negligible.
4.
2
2
0.8 0.18 0.020.4 0.5 0.10.2 0.6 0.2
0.716 0.246 0.0380.54 0.382 0.0780.44 0.456 0.104
P⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
Grandchild of a college graduate is a college graduate: entry (1, 1): 0.716. The probability is 71.6%
5. Grandchild of a high school graduate finishes college: entry (2,1): 0.54. The probability is 54%.