Systems of Equations and Inequalities
Section 12.1
1. 3 4 8 4 4 1 x x x x + = − = =The solution set is
{ }
1 . 3. inconsistent 5. (3, 2)− 7. b 9. 2 5 5 2 8 x y x y − = + = Substituting the values of the variables: 2(2) ( 1) 4 1 5 5(2) 2( 1) 10 2 8 − − = + = + − = − =
Each equation is satisfied, so x=2,y= −1, or (2, 1)− , is a solution of the system of equations.
11. 3 4 4 1 3 1 2 2 x y x y − = − = −
Substituting the values of the variables: 1 3(2) 4 6 2 4 2 1 1 3 1 (2) 3 1 2 2 2 2 − = − = − = − = −
Each equation is satisfied, so x=2,y= , or 12
( )
12, 2 , is a solution of the system of equations. 13. 3 1 3 2 x y x y − = + =
Substituting the values of the variables, we obtain: 4 1 3 1(4) 1 2 1 3 2 − = + = + =
Each equation is satisfied, so x=4,y=1, or (4, 1), is a solution of the system of equations.
15. 3 3 2 4 0 2 3 8 x y z x y z y z + + = − − = − = −
Substituting the values of the variables: 3(1) 3( 1) 2(2) 3 3 4 4 1 ( 1) 2 1 1 2 0 2( 1) 3(2) 2 6 8 + − + = − + = − − − = + − = − − = − − = −
Each equation is satisfied, so x=1,y= −1, z=2, or (1, 1, 2)− , is a solution of the system of equations. 17. 3 3 2 4 3 10 5 2 3 8 x y z x y z x y z + + = − + = − − =
Substituting the values of the variables:
( ) ( ) ( )
( )
( ) ( ) ( )
3 2 3 2 2 2 6 6 4 4 2 3 2 2 2 6 2 10 5 2 2 2 3 2 10 4 6 8 + − + = − + = − − + = + + = − − − = + − = Each equation is satisfied, so x=2, y= −2, 2
z= , or (2, 2, 2)− is a solution of the system of equations. 19. 8 4 x y x y + = − =
Solve the first equation for y, substitute into the second equation and solve:
8 4 y x x y = − − = (8 ) 4 8 4 2 12 6 x x x x x x − − = − + = = =
Since x=6, y= − =8 6 2. The solution of the system is x=6, y= or using ordered pairs 2
21. 5 21 2 3 12 x y x y − = + = −
Multiply each side of the first equation by 3 and add the equations to eliminate y:
15 3 63 2 3 12 x y x y − = + = − 17 51 3 x x = =
Substitute and solve for y: 5(3) 21 15 21 6 6 y y y y − = − = − = = −
The solution of the system is x=3, y= −6 or using ordered an pair
(
3, 6− .)
23. 3 24 2 0 x x y = + =
Solve the first equation for x and substitute into the second equation:
8 2 0 x x y = + = 8 2 0 2 8 4 y y y + = = − = −
The solution of the system is x=8, y= −4 or using ordered pairs (8, 4)−
25. 3 6 2 5 4 1 x y x y − = + =
Multiply each side of the first equation by 2 and each side of the second equation by 3, then add to eliminate y: 6 12 4 15 12 3 x y x y − = + = 21 7 1 3 x x = =
Substitute and solve for y:
( )
3 1/ 3 6 2 1 6 2 6 1 1 6 y y y y − = − = − = = −The solution of the system is 1, 1
3 6
x= y= − or using ordered pairs 1, 1
3 6 − .
27. 2 1 4 2 3 x y x y + = + =
Solve the first equation for y, substitute into the second equation and solve:
1 2 4 2 3 y x x y = − + = 4 2(1 2 ) 3 4 2 4 3 0 1 x x x x + − = + − = =
This equation is false, so the system is inconsistent.
29. 2 0 4 2 12 x y x y − = + =
Solve the first equation for y, substitute into the second equation and solve:
2 4 2 12 y x x y = + = 4 2(2 ) 12 4 4 12 8 12 3 2 x x x x x x + = + = = = Since 3, 2 3 3 2 2 x= y= =
The solution of the system is 3, 3 2
x= y= or using ordered pairs 3,3 .
2 31. 2 4 2 4 8 x y x y + = + =
Solve the first equation for x, substitute into the second equation and solve:
4 2 2 4 8 x y x y = − + = 2(4 2 ) 4 8 8 4 4 8 0 0 y y y y − + = − + = =
These equations are dependent. The solution of the system is either x= −4 2y, where y is any real number or 4
2 x
y= − , where x is any real number. Using ordered pairs, we write the solution as
{
( , )x y x= −4 2 , is any real numbery y}
or as 4( , ) , is any real number 2 x x y y x − = .
33. 2 3 1 10 11 x y x y − = − + =
Multiply each side of the first equation by –5, and add the equations to eliminate x:
10 15 5 10 11 x y x y − + = + = 16 16 1 y y = =
Substitute and solve for x: 2 3(1) 1 2 3 1 2 2 1 x x x x − = − − = − = =
The solution of the system is x=1, y= or 1 using ordered pairs (1, 1).
35. 2 3 6 1 2 x y x y + = − =
Solve the second equation for x, substitute into the first equation and solve:
2 3 6 1 2 x y x y + = = + 1 2 3 6 2 2 1 3 6 5 5 1 y y y y y y + + = + + = = = Since 1, 1 1 3 2 2
y= x= + = . The solution of the system is 3, 1
2
x= y= or using ordered pairs
37. 1 1 3 2 3 1 2 1 4 3 x y x y + = − = −
Multiply each side of the first equation by –6 and each side of the second equation by 12, then add to eliminate x: 3 2 18 3 8 12 x y x y − − = − − = − 10 30 3 y y − = − =
Substitute and solve for x: 1 1 (3) 3 2 3 1 1 3 2 1 2 2 4 x x x x + = + = = =
The solution of the system is x=4, y= or 3 using ordered pairs (4, 3).
39. 3 5 3 15 5 21 x y x y − = + =
Add the equations to eliminate y:
3 5 3 15 5 21 x y x y − = + = 18 24 4 3 x x = =
Substitute and solve for y:
( )
3 4 / 3 5 3 4 5 3 5 1 1 5 y y y y − = − = − = − =The solution of the system is 4, 1 3 5
x= y= or
using ordered pairs 4 1, 3 5 . 41. 1 1 8 3 5 0 x y x y + = − = Rewrite letting u 1, v 1 x y = = : 8 3 5 0 u v u v + = − =
Solve the first equation for u, substitute into the second equation and solve:
8 3 5 0 u v u v = − − =
3(8 ) 5 0 24 3 5 0 8 24 3 v v v v v v − − = − − = − = − = Since v=3, u= − =8 3 5. Thus, 1 1 5 x u = = , 1 1 3 y v
= = . The solution of the system is 1 1
, 5 3
x= y= or using ordered pairs 1 1, 5 3 . 43. 6 2 3 16 2 4 x y x z y z − = − = + =
Multiply each side of the first equation by –2 and add to the second equation to eliminate x:
2 2 12 2 3 16 2 3 4 x y x z y z − + = − − = − =
Multiply each side of the result by –1 and add to the original third equation to eliminate y:
2 3 4 2 4 4 0 0 y z y z z z − + = − + = = =
Substituting and solving for the other variables: 2 0 4 2 4 2 y y y + = = = 2 3(0) 16 2 16 8 x x x − = = = The solution is x=8, y=2, z=0 or using ordered triples (8, 2, 0). 45. 2 3 7 2 4 3 2 2 10 x y z x y z x y z − + = + + = − + − = −
Multiply each side of the first equation by –2 and add to the second equation to eliminate x; and multiply each side of the first equation by 3 and add to the third equation to eliminate x:
2 4 6 14 2 4 5 5 10 x y z x y z y z − + − = − + + = − = − 3 6 9 21 3 2 2 10 4 7 11 x y z x y z y z − + = − + − = − − + =
Multiply each side of the first result by 4 5 and add to the second result to eliminate y:
4 4 8 4 7 11 y z y z − = − − + = 3 3 1 z z = =
Substituting and solving for the other variables: 1 2 1 y y − = − = − 2( 1) 3(1) 7 2 3 7 2 x x x − − + = + + = =
The solution is x=2, y= −1, z=1 or using ordered triples (2, 1, 1)− . 47. 1 2 3 2 3 2 0 x y z x y z x y − − = + + = + =
Add the first and second equations to eliminate z: 1 2 3 2 3 2 3 x y z x y z x y − − = + + = + =
Multiply each side of the result by –1 and add to the original third equation to eliminate y:
3 2 3 3 2 0 0 3 x y x y − − = − + = = −
This equation is false, so the system is inconsistent.
49. 1 2 3 4 3 2 7 0 x y z x y z x y z − − = − + − = − − − =
Add the first and second equations to eliminate x; multiply the first equation by –3 and add to the third equation to eliminate x:
1 2 3 4 4 3 x y z x y z y z − − = − + − = − − = − 3 3 3 3 3 2 7 0 4 3 x y z x y z y z − + + = − − − = − = −
Multiply each side of the first result by –1 and add to the second result to eliminate y:
4 3 4 3 0 0 y z y z − + = − = − =
The system is dependent. If z is any real number, then y=4z−3.
Solving for x in terms of z in the first equation: (4 3) 1 4 3 1 5 3 1 5 2 x z z x z z x z x z − − − = − + − = − + = = − The solution is {( , , )x y z x=5z−2, y=4z−3, z is any real number}.
51. 2 2 3 6 4 3 2 0 2 3 7 1 x y z x y z x y z − + = − + = − + − =
Multiply the first equation by –2 and add to the second equation to eliminate x; add the first and third equations to eliminate x:
4 4 6 12 4 3 2 0 4 12 x y z x y z y z − + − = − − + = − = − 2 2 3 6 2 3 7 1 4 7 x y z x y z y z − + = − + − = − =
Multiply each side of the first result by –1 and add to the second result to eliminate y:
4 12 4 7 0 19 y z y z − + = − = =
This result is false, so the system is inconsistent.
53. 6 3 2 5 3 2 14 x y z x y z x y z + − = − + = − + − =
Add the first and second equations to eliminate z; multiply the second equation by 2 and add to the third equation to eliminate z:
6 3 2 5 4 1 x y z x y z x y + − = − + = − − = 6 4 2 10 3 2 14 7 4 x y z x y z x y − + = − + − = − =
Multiply each side of the first result by –1 and add to the second result to eliminate y:
4 1 7 4 x y x y − + = − − = 3 3 1 x x = =
Substituting and solving for the other variables: 4(1) 1 3 3 y y y − = − = − = 3(1) 2(3) 5 3 6 5 2 z z z − + = − − + = − = − The solution is x=1, y=3, z= −2 or using ordered triplets (1, 3, 2)− . 55. 2 3 2 4 7 2 2 3 4 x y z x y z x y z + − = − − + = − − + − =
Add the first and second equations to eliminate z; multiply the second equation by 3 and add to the third equation to eliminate z:
2 3 2 4 7 3 2 10 x y z x y z x y + − = − − + = − − = − 6 12 3 21 2 2 3 4 4 10 17 x y z x y z x y − + = − − + − = − = −
Multiply each side of the first result by –5 and add to the second result to eliminate y:
15 10 50 4 10 17 x y x y − + = − = − 11 33 3 x x − = = −
Substituting and solving for the other variables: 3( 3) 2 10 9 2 10 2 1 1 2 y y y y − − = − − − = − − = − =
1 3 2 3 2 3 1 3 1 1 z z z z − + − = − − + − = − − = − = The solution is 3, 1, 1 2 x= − y= z= or using ordered triplets 3, , 11 2 − .
57. Let l be the length of the rectangle and w be the width of the rectangle. Then:
2 l= w and 2l+2w=90 Solve by substitution: 2(2 ) 2 90 4 2 90 6 90 15 feet 2(15) 30 feet w w w w w w l + = + = = = = =
The floor is 15 feet by 30 feet.
59. Let x = the number of commercial launches and y = the number of noncommercial launches. Then: x y+ =81 and y=2x+12 Solve by substitution: (2 12) 81 3 69 23 x x x x + + = = = 2(23) 12 46 12 58 y y y = + = + =
In 2013 there were 23 commercial launches and 58 noncommercial launches.
61. 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 x+30= represents the amount of mixture. y 5x+45 3= y represents the value of the mixture. Solve by substitution: 5 45 3( 30) 2 45 22.5 x x x x + = + = =
So, 22.5 pounds of cashews should be used in the mixture.
63. Let s = the price of a smartphone and t = the price of a tablet. Then:
965 340 250 270500 s t s t + = + =
Solve the first equation for t: t=965− s
Solve by substitution: 340 250(965 ) 270500 340 241250 250 270500 90 29250 325 s s s s s s + − = + − = = = 965 325 640 t= − =
The price of the smartphone is $325.00 and the price of the tablet is $640.00.
65. Let x = the plane’s average airspeed and y = the average wind speed.
Rate Time Distance With Wind 3 600 Against 4 600 x y x y + − ( )(3) 600 ( )(4) 600 x y x y + = − =
Multiply each side of the first equation by 1 3, multiply each side of the second equation by 1
4, and add the result to eliminate y
200 150 x y x y + = − = 2 350 175 x x = = 175 200 25 y y + = =
The average airspeed of the plane is 175 mph, and the average wind speed is 25 mph. 67. Let x = the number of $25-design.
Let y = the number of $45-design.
Then x y+ = the total number of sets of dishes. 25x+45y = the cost of the dishes.
Setting up the equations and solving by substitution: 200 25 45 7400 x y x y + = + =
Solve the first equation for y, the solve by substitution: y=200−x
25 45(200 ) 7400 25 9000 45 7400 20 1600 80 x x x x x x + − = + − = − = − = 200 80 120 y= − =
Thus, 80 sets of the $25 dishes and 120 sets of the $45 dishes should be ordered.
69. Let x = the cost per package of bacon. Let y = the cost of a carton of eggs.
Set up a system of equations for the problem: 3 2 13.45 2 3 11.45 x y x y + = + =
Multiply each side of the first equation by 3 and each side of the second equation by –2 and solve by elimination: 9 6 40.35 4 6 22.90 x y x y + = − − = − 5 17.45 3.49 x x = =
Substitute and solve for y: 3(3.49) 2 13.45 10.47 2 13.45 2 2.98 1.49 y y y y + = + = = =
A package of bacon costs $3.49 and a carton of eggs cost $1.49. The refund for 2 packages of bacon and 2 cartons of eggs will be
2($3.49) + 2($1.49) = $9.96. 71. Let x = the # of mg of compound 1.
Let y = the # of mg of compound 2. Setting up the equations and solving by substitution: 0.2 0.4 40 vitamin C 0.3 0.2 30 vitamin D x y x y + = + =
Multiplying each equation by 10 yields 2 4 400 6 4 600 x y x y + = + =
Subtracting the bottom equation from the top equation yields
(
)
2 4 6 4 400 600 2 6 200 4 200 50 x y x y x x x x + − + = − − = − − = − =( )
2 50 4 400 100 4 400 4 300 300 75 4 y y y y + = + = = = =So 50 mg of compound 1 should be mixed with 75 mg of compound 2.
73. y ax= 2+ +bx c
At (–1, 4) the equation becomes:
2 4 (–1) ( 1) 4 a b c a b c = + − + = − +
At (2, 3) the equation becomes:
2 3 (2) (2) 3 4 2 a b c a b c = + + = + +
At (0, 1) the equation becomes:
2 1 (0) (0) 1 a b c c = + + =
The system of equations is: 4 4 2 3 1 a b c a b c c − + = + + = =
Substitute c=1 into the first and second equations and simplify:
1 4 3 3 a b a b a b − + = − = = + 4 2 1 3 4 2 2 a b a b + + = + =
Solve the first result for a, substitute into the second result and solve:
4( 3) 2 2 4 12 2 2 6 10 5 3 b b b b b b + + = + + = = − = − 5 4 3 3 3 a= − + = The solution is 4, 5, 1 3 3 a= b= − c= . The equation is 4 2 5 1 3 3 y= x − x+ .
75. 0.06 5000 240 0.06 6000 900 Y r Y r − = + =
Multiply the first equation by −1, the add the result to the second equation to eliminate Y.
0.06 5000 240 0.06 6000 900 Y r Y r − + = − + = 11000 660 0.06 r r = =
Substitute this result into the first equation to find Y. 0.06 5000(0.06) 240 0.06 300 240 0.06 540 9000 Y Y Y Y − = − = = =
The equilibrium level of income and interest rates is $9000 million and 6%.
77. 2 1 3 1 2 2 3 5 3 5 0 10 5 7 0 I I I I I I I = + − − = − − =
Substitute the expression for I2 into the second and third equations and simplify:
1 1 3 1 3 5 3 5( ) 0 8 5 5 I I I I I − − + = − − = − 1 3 3 1 3 10 5( ) 7 0 5 12 10 I I I I I − + − = − − = −
Multiply both sides of the first result by 5 and multiply both sides of the second result by –8 to eliminate I1: 1 3 1 3 40 25 25 40 96 80 I I I I − − = − + = 3 3 71 55 55 71 I I = =
Substituting and solving for the other variables:
1 1 1 1 55 8 5 5 71 275 8 5 71 80 8 71 10 71 I I I I − − = − − − = − − = − = 2 1071 5571 6571 I = + = The solution is 1 10, 2 65, 3 55 71 71 71 I = I = I = .
79. 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,
500
x y z+ + = .
Since the total revenue is $64,250 if all seats are sold, 150x+135y+110z=64, 250.
If only half of the orchestra seats are sold, the revenue is $56,750.
So, 150 1 135 110 56,750 2x + y+ z= . Thus, we have the following system:
500 150 135 110 64, 250 75 135 110 56,750 x y z x y z x y z + + = + + = + + =
Multiply each side of the first equation by –110 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:
110 110 110 55,000 150 135 110 64, 250 40 25 9250 x y z x y z x y − − − = − + + = + = 150 135 110 64, 250 75 135 110 56,750 x y z x y z + + = − − − = − 75 7500 100 x x = =
Substituting and solving for the other variables: 40(100) 25 9250 4000 25 9250 25 5250 210 y y y y + = + = = = 100 210 500 310 500 190 z z z + + = + = = There are 100 orchestra seats, 210 main seats,
81. 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: 30x+3y+9z=66
Carbohydrate equation: 35x+16y+13z=94.5 Calcium equation: 200x+10y+300z=910 Multiply each side of the first equation by –16
and multiply each side of the second equation by 3 and add them to eliminate y; multiply each side of the second equation by –5 and multiply each side of the third equation by 8 and add to eliminate y: 480 48 144 1056 105 48 39 283.5 375 105 772.5 x y z x y z x z − − − = − + + = − − = − 175 80 65 472.5 1600 80 2400 7280 1425 2335 6807.5 x y z x y z x z − − − = − + + = + =
Multiply each side of the first result by 19 and multiply each side of the second result by 5 to eliminate x: 7125 1995 14,677.5 7125 11,675 34,037.5 x z x z − − = − + = 9680 19,360 2 z z = =
Substituting and solving for the other variables: 375 105(2) 772.5 375 210 772.5 375 562.5 1.5 x x x x − − = − − − = − − = − = 30(1.5) 3 9(2) 66 45 3 18 66 3 3 1 y y y y + + = + + = = =
The dietitian should serve 1.5 servings of chicken, 1 serving of corn, and 2 servings of 2% milk.
83. Let x = the price of 1 hamburger. Let y = the price of 1 order of fries. Let z = the price of 1 drink. We can construct the system
8 6 6 26.10 10 6 8 31.60 x y z x y z + + = + + =
A system involving only 2 equations that contain 3 or more unknowns cannot be solved uniquely. Multiply the first equation by 1
2
− and the second equation by 1
2, then add to eliminate y: 4 3 3 13.05 5 3 4 15.80 x y z x y z − − − = − + + = 2.75 2.75 x z x z + = = −
Substitute and solve for y in terms of z:
(
)
5 2.75 3 4 15.80 13.75 3 15.80 3 2.05 1 41 3 60 z y z y z y z y z − + + = + − = = + = +Solutions of the system are: x=2.75−z, 1 41
3 60
y= z+ .
Since we are given that 0.60≤ ≤z 0.90, we choose values of z that give two-decimal-place values of x and y with 1.75≤ ≤x 2.25 and
0.75≤ ≤y 1.00.
The possible values of x, y, and z are shown in the table. x y z 2.13 0.89 0.62 2.10 0.90 0.65 2.07 0.91 0.68 2.04 0.92 0.71 2.01 0.93 0.74 1.98 0.94 0.77 1.95 0.95 0.80 1.92 0.96 0.83 1.89 0.97 0.86 1.86 0.98 0.89
85. Let x = Beth’s time working alone. Let y = Bill’s time working alone. Let z = Edie’s time working alone.
We can use the following tables to organize our work:
Beth Bill Edie Hours to do job
Part of job done 1 1 1 in 1 hour
x y z
x y z
In 10 hours they complete 1 entire job, so 1 1 1 10 1 1 1 1 1 10 x y z x y z + + = + + = Bill Edie Hours to do job
Part of job done 1 1 in 1 hour
y z
y z
In 15 hours they complete 1 entire job, so 1 1 15 1 1 1 1 15 y z y z + = + = .
Beth Bill Edie Hours to do job
Part of job done 1 1 1 in 1 hour
x y z
x y z
With all 3 working for 4 hours and Beth and Bill working for an additional 8 hours, they complete 1 entire job, so 4 1 1 1 8 1 1 1 12 12 4 1 x y z x y x y z + + + + = + + = We have the system
1 1 1 1 10 1 1 1 15 12 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 30 30 x x = =
Substitute x = 30 into the third equation: 12 12 4 1 30 12 4 3 5 y z y z + + = + = .
Now consider the system consisting of the last result and the second original equation. Multiply the second original equation by –12 and add it to the last result to eliminate y:
12 12 12 15 12 4 3 5 y z y z − +− =− + = 8 3 15 40 z z − = − = Plugging z = 40 to find y: 12 4 3 5 12 4 3 40 5 12 1 2 24 y z y y y + = + = = =
Working alone, it would take Beth 30 hours, Bill 24 hours, and Edie 40 hours to complete the job. 87. Answers will vary.
89.
91. sin 1 sin 10
9 π
− − follows the form of the
equation f−1
( )
f x( )
=sin−1(
sin( )
x)
= , but x we cannot use the formula directly since 109 π −
is not in the interval , 2 2 π π − . We need to find an angle θ in the interval ,
2 2 π π − for which 10 sin sin 9 π θ − = . The angle 10 9 π − is in quadrant II so sine is positive. The reference angle of 10 9 π − is 9 π and we want θ to be in quadrant I so sine will still be positive. Thus, we have sin 10 sin
9 9 π π − = . Since 9 π is in the interval , 2 2 π π −
, we can apply the equation above and get
1 10 1
sin sin sin sin
9 9 9 π π π − − = − = .
Section 12.2
1. matrix 3. third; fifth 5. b7. Writing the augmented matrix for the system of equations: 5 5 1 5 5 4 3 6 4 3 6 x y x y − = − → + = 9. 2 3 6 0 4 6 2 0 x y x y + − = − + =
Write the system in standard form and then write the augmented matrix for the system of equations:
2 3 6 2 3 6 4 6 2 4 6 2 x y x y + = → − = − − −
11. Writing the augmented matrix for the system of equations: 0.01 0.03 0.06 0.01 0.03 0.06 0.13 0.10 0.20 0.13 0.10 0.20 x y x y − = − → + =
13. Writing the augmented matrix for the system of equations: 10 1 1 1 10 3 3 5 3 3 0 5 2 2 1 1 2 2 x y z x y x y z − + = − + = → + + =
15. Writing the augmented matrix for the system of equations: 2 1 1 1 2 3 2 2 3 2 0 2 5 3 1 5 3 1 1 x y z x y x y z + − = − − = → − + − = −
17. Writing the augmented matrix for the system of equations: 10 1 1 1 10 2 2 1 2 1 2 1 3 4 5 3 4 0 5 4 5 0 4 5 1 0 x y z x y z x y x y z − − = − − + + = − − → − + = − − + = − 19. 1 3 2 3 2 2 5 5 2 5 5 x y x y − − − = − → − − = 2 21 2 R = − r +r 1 3 2 1 3 2 2 5 5 2(1) 2 2( 3) 5 2( 2) 5 1 3 2 0 1 9 − − − − → − − + − − − − + − − →
21. 1 3 4 3 3 4 3 3 5 6 6 3 5 6 6 5 3 4 6 5 3 4 6 x y z x y z x y z − − + = − → − + = − − + + = R2 = −3r1+ r2 1 3 4 3 3 5 6 6 5 3 4 6 1 3 4 3 3(1) 3 3( 3) 5 3(4) 6 3(3) 6 5 3 4 6 1 3 4 3 0 4 6 3 5 3 4 6 − − − − → − + − − − − + − + − − → − − − 3 51 3 R = r + r 1 3 4 3 0 4 6 3 5 3 4 6 1 3 4 3 0 4 6 3 5(1) 5 5( 3) 3 5(4) 4 5(3) 6 1 3 4 3 0 4 6 3 0 12 24 21 − − − − − → − − − − + + + − → − − − 23. 1 3 2 6 3 2 6 2 5 3 4 2 5 3 4 3 6 4 6 3 6 4 6 x y z x y z x y z − − − + = − − − → − + = − − − − − + = R2 = −2r1+ r2 1 3 2 6 2 5 3 4 3 6 4 6 1 3 2 6 2(1) 2 2( 3) 5 2(2) 3 2( 6) 4 3 6 4 6 1 3 2 6 0 1 1 8 3 6 4 6 − − − − − − − − → − + − − − − + − − − − − − − → − − − 3 31 3 R = r + r 1 3 2 6 0 1 1 8 3 6 4 6 1 3 2 6 0 1 1 8 3(1) 3 3( 3) 6 3(2) 4 3( 6) 6 1 3 2 6 0 1 1 8 0 15 10 12 − − − − − − − → − − − − + − + − − → − − − 25. 5 3 1 2 5 3 2 2 5 6 2 2 5 6 2 4 1 4 6 4 4 6 x y z x y z x y z − − − + = − − − → − + = − − − + + = R1= −2r2+ r1 5 3 1 2 2 5 6 2 4 1 4 6 2(2) 5 2( 5) 3 2(6) 1 2( 2) 2 2 5 6 2 4 1 4 6 1 7 11 2 2 5 6 2 4 1 4 6 − − − − − − + − − − − + − − − → − − − − → − − − 3 22 3 R = r +r 1 7 11 2 2 5 6 2 4 1 4 6 1 7 11 2 2 5 6 2 2(2) ( 4) 2( 5) 1 2(6) 4 2( 2) 6 1 7 11 2 2 5 6 2 0 9 16 2 − − − − − → − − + − − + + − + − → − − − 27. 5 1 x y = = −
Consistent; x=5,y= − or using ordered pairs 1, (5, 1)− . 29. 1 2 0 3 x y = = = Inconsistent
31. 2 1 4 2 0 0 x z y z + = − − = − = Consistent; 1 2 2 4
is any real number
x z y z z = − − = − + or {( , , ) |x y z x= − −1 2 ,z y= − +2 4 ,z z is any real number} 33. 1 2 4 3 4 1 2 2 3 x x x x x = + = + = Consistent; 1 2 4 3 4 4 1 2 3 2
is any real number x x x x x x = = − = − or {( , , , ) |x x x x1 2 3 4 x1=1, x2 = −2 x4,
3 3 2 ,4 4 is any real number}
x = − x x 35. 1 4 2 3 4 4 2 3 3 0 0 x x x x x + = + + = = Consistent; 1 4 2 3 4 3 4 2 4 3 3
, are any real numbers
x x x x x x x = − = − − or {( , , , ) |x x x x1 2 3 4 x1= −2 4 , x x4 2 = − −3 x3 3 ,x4
3 and are any real numbers}4
x x 37. 1 4 2 4 3 4 2 2 2 0 0 0 x x x x x x + = − + = − = = Consistent; 1 4 2 4 3 4 4 2 2 2
is any real number
x x x x x x x = − − = − = or {( , , , ) |x x x x1 2 3 4 x1= − −2 x x4, 2 = −2 2 ,x4
3 4, is any real number}4
x =x x 39. 8 4 x y x y + = − =
Write the augmented matrix:
(
)
(
)
(
)
2 1 2 1 2 2 2 1 2 1 8 8 1 1 1 1 0 1 1 4 2 4 8 1 1 0 1 2 0 6 1 0 1 2 R r r R r R r r → = − + − − − → = − → = − + The solution is x=6,y=2 or using ordered pairs (6, 2). 41. 2 4 2 3 2 3 x y x y − = − + =
Write the augmented matrix:
(
)
(
)
(
)
(
)
1 1 2 1 2 1 2 1 2 8 2 3 4 1 2 1 2 1 3 4 2 4 2 1 2 1 3 2 3 3 2 3 1 2 1 3 0 8 6 1 2 1 0 1 0 1 2 0 1 R r R r r R r R r r − − − − = → − − = − + → − − = → = + → The solution is 1, 3 2 4x= y= or using ordered pairs 1 3 , 2 4 . 43. 2 4 2 4 8 x y x y + = + =
Write the augmented matrix:
(
2 1 2)
1 2 4 1 2 4 2 8 0 0 0 2 4 R r r = − + → This is a dependent system. 2 4 4 2 x y x y + = = −
The solution is x= −4 2 , is any real numbery y
45. 2 3 6 1 2 x y x y + = − =
Write the augmented matrix:
(
)
(
)
(
)
(
)
3 1 2 1 1 2 1 1 2 2 3 2 2 1 2 5 5 2 2 3 2 2 2 2 5 3 3 2 1 2 1 2 3 6 2 1 3 1 1 1 1 3 1 0 3 1 0 1 1 0 1 0 1 1 R r R r r R r R r r → = − − → = − + − − → = − → = − + The solution is 3, 1 2 x= y= or 3, 1 2 . 47. 3 5 3 15 5 21 x y x y − = + = Write the augmented matrix:
(
)
(
)
(
)
(
)
5 1 3 1 1 3 5 3 2 1 2 5 3 1 2 30 2 1 5 4 5 3 1 3 2 1 1 5 3 5 3 1 1 15 5 21 15 5 21 1 1 15 0 30 6 1 1 0 1 0 1 0 1 R r R r r R r R r r − − → = − → = − + − → = → = + The solution is 4, 1 3 5 x= y= or 4 1, 3 5 . 49. 6 2 3 16 2 4 x y x z y z − = − = + = Write the augmented matrix:
(
)
(
)
2 1 2 3 1 2 2 2 2 0 6 1 1 0 3 16 2 0 2 1 4 0 6 1 1 0 2 3 4 2 0 2 1 4 0 6 1 1 0 1 2 0 2 1 4 R r r R r − − − → − = − + − − → = (
)
3 2 1 2 1 3 2 3 2 3 3 2 3 1 3 3 2 4 3 1 2 3 1 3 2 2 3 2 0 8 1 0 1 2 2 0 0 4 0 0 8 1 0 1 2 0 0 1 0 0 0 8 1 0 1 0 2 0 0 1 0 R r r R r r R r R r r R r r − = + − → = − + − − → = = + → = + The solution is x=8,y=2,z=0 or (8, 2, 0). 51. 2 3 7 2 4 3 2 2 10 x y z x y z x y z − + = + + = − + − = − Write the augmented matrix:
(
)
2 1 2 3 1 3 1 2 5 2 1 2 1 3 2 3 3 7 1 2 2 1 1 4 3 2 2 10 3 7 1 2 2 0 5 5 10 3 0 4 7 11 3 7 1 2 0 1 1 2 0 4 7 11 0 3 1 1 2 0 1 1 2 4 0 0 3 3 R r r R r r R r R r r R r r − − − − − = − + → − − = + − − → − − = − = + → − − = + (
1)
3 3 3 0 3 1 1 0 1 1 2 0 0 1 1 R r → − − = 1 3 1 2 3 2 0 0 1 2 0 1 0 1 0 0 1 1 R r r R r r = − + → − = + The solution is x=2,y= −1,z= or (2, 1,1)1 − . 53. 2 2 2 2 2 3 2 3 2 0 x y z x y z x y − − = + + = + = Write the augmented matrix:
(
)
(
)
1 1 2 1 2 1 2 3 1 3 3 2 3 2 2 2 2 3 2 1 2 3 2 0 0 1 1 1 1 3 2 1 2 3 2 0 0 1 1 1 1 2 0 5 3 0 3 0 5 3 3 1 1 1 1 0 5 3 0 0 0 0 3 R r R r r R r r R r r − − − − → = − − = − + → = − + − − − → = − + − There is no solution. The system is inconsistent.
55. 1 2 3 4 3 2 7 0 x y z x y z x y z − + + = − − + − = − − − =
Write the augmented matrix:
(
1 1)
2 1 2 3 1 3 1 2 1 3 2 3 1 1 1 1 4 3 1 2 2 3 7 0 1 1 1 1 4 3 1 2 2 3 7 0 1 1 1 1 4 0 1 3 3 4 0 1 3 2 0 5 1 4 0 1 3 0 0 0 0 R r R r r R r r R r r R r r − − − − − − − − − − − → − = − − − − − = + − → − = − + − − − − = + − → − = − + The matrix in the last step represents the system
5 2 4 3 0 0 x z y z − = − − = − = or, equivalently, 5 2 4 3 0 0 x z y z = − = − =
The solution is x=5z− , 2 y=4z− , z is any 3 real number or {( , , ) |x y z x=5z−2,y=4z− z 3, is any real number}.
57. 2 2 3 6 4 3 2 0 2 3 7 1 x y z x y z x y z − + = − + = − + − =
Write the augmented matrix: 2 3 6 2 3 0 4 2 2 3 7 1 − − − −
(
)
3 2 1 1 2 1 3 2 2 1 2 3 1 3 5 2 1 2 1 3 2 3 3 1 1 3 0 4 2 2 3 7 1 3 1 1 4 4 0 1 12 2 4 0 1 7 0 9 1 4 0 1 12 0 0 0 19 R r R r r R r r R r r R r r − − → = − − − = − + − → − = + − − − = + − → − = − + 59. 6 3 2 5 3 2 14 x y z x y z x y z + − = − + = − + − =
Write the augmented matrix: 6 1 1 1 2 3 1 5 2 3 1 14 − − − −
(
)
2 1 2 3 1 3 23 4 1 5 5 2 5 2 7 1 5 5 23 4 1 2 1 5 5 3 2 3 6 3 5 5 6 1 1 1 3 23 0 5 4 0 2 1 8 6 1 1 1 0 1 0 2 1 8 0 1 0 1 2 0 0 R r r R r r R r R r r R r r − = − + − → − = − + − − − → = − − − = − + − → = − + − (
)
7 1 5 5 23 4 5 5 5 3 3 3 1 1 5 3 1 4 2 5 3 2 0 1 0 1 2 0 0 1 0 0 1 1 0 1 0 3 2 0 0 1 R r R r r R r r − − → = − = + → = + − The solution is x=1,y=3,z= − , or (1, 3, 2)2 − . 61. 2 3 2 4 7 2 2 3 4 x y z x y z x y z + − = − − + = − − + − = Write the augmented matrix: 3 1 2 1 4 7 2 1 2 2 3 4 − − − − − −
(
)
2 1 2 3 1 3 3 1 1 8 8 2 8 2 3 1 2 1 2 8 0 3 1 2 2 0 6 5 3 1 2 1 0 1 2 0 6 5 R r r R r r R r − − = − + − → − = + − − − − − → = − − − (
)
13 1 4 4 3 1 1 2 1 8 8 3 2 3 11 11 4 4 13 1 4 4 3 1 4 8 8 3 11 3 1 1 3 1 1 4 2 3 2 8 3 2 0 1 2 0 1 6 0 0 0 1 0 1 0 0 1 1 0 0 3 1 0 1 0 0 0 1 1 R r r R r r R r R r r R r r − − = − + − → = − + − − − − − → = − − = + → = + The solution is 3, 1, 1 2 x= − y= z= or 3, ,11 2 − . 63. 2 3 3 2 1 8 4 2 3 x y z x y z x y + − = − + = + = Write the augmented matrix:
(
)
2 3 8 3 1 1 2 3 3 9 1 1 3 1 8 3 3 1 1 2 1 1 1 0 4 2 1 2 1 1 1 0 4 2 R r − − − → − = (
)
1 1 2 3 3 9 5 5 5 2 1 2 3 3 9 3 1 3 16 2 4 3 3 9 1 1 2 3 3 9 1 3 3 2 5 2 16 2 4 3 3 9 1 3 1 1 2 1 1 3 3 2 3 3 2 3 1 2 0 4 0 1 0 1 1 0 0 0 1 0 1 1 0 0 2 2 R r r R r r R r R r r R r r − = − + − → = − + − − → − = − = − + − → − = − + (
)
(
)
1 3 1 1 3 3 2 3 1 3 2 3 2 3 2 0 0 1 0 1 1 0 0 1 1 0 0 1 0 1 0 0 0 1 1 R r R r r − → − = → = + The solution is 1, 2, 1 3 3 x= y= z= or 1 2, , 1 3 3 . 65. 4 2 0 3 2 6 2 2 2 1 x y z w x y z x y z w x y z w + + + = − + = + + − = − − + = − Write the augmented matrix:
1 1 1 1 4 0 0 2 1 1 3 2 1 1 6 2 2 1 2 1 − − − − − 2 1 2 3 1 3 4 1 4 1 1 1 1 4 2 2 8 0 3 1 3 2 4 6 0 1 0 3 3 1 5 R r r R r r R r r = − + − − − − → = − + − − − − = − + − − −
(
)
2 3 2 2 1 2 1 3 2 3 4 2 4 1 1 1 1 4 Interchange 2 4 6 0 1 and 2 8 0 3 1 0 3 3 1 5 1 1 1 1 4 0 1 2 4 6 2 8 0 3 1 0 3 3 1 5 2 0 3 1 1 0 1 2 4 6 3 0 0 5 10 10 3 0 0 3 13 13 r r R r R r r R r r R r r − − − − → − − − − − − − → = − − − − − − − − − − − = − + → = + = + (
1)
3 5 3 1 3 1 2 3 2 4 3 4 2 0 3 1 1 0 1 2 4 6 0 0 1 2 2 0 0 3 13 13 0 0 0 1 1 0 1 0 0 2 2 0 0 1 2 2 3 0 0 0 7 7 R r R r r R r r R r r − − − → = − = + → = − + = − + (
1)
4 7 4 1 4 1 3 4 3 0 0 0 1 1 0 1 0 0 2 0 0 1 2 2 0 0 0 1 1 0 0 0 1 1 0 1 0 0 2 2 0 0 1 0 0 0 0 0 1 1 R r R r r R r r − → = = + → = − + The solution is x=1,y=2,z=0,w=1 or (1, 2, 0, 1). 67. 2 1 2 2 2 3 3 3 x y z x y z x y z + + = − + = + + = Write the augmented matrix: 1 2 1 1 2 1 2 2 3 1 3 3 −
(
)
2 1 2 3 1 3 3 2 3 1 2 1 1 2 0 5 0 0 3 0 5 0 0 1 2 1 1 0 5 0 0 0 0 0 0 R r r R r r R r r = − + → − = − + − → − = − + The matrix in the last step represents the system 2 1 5 0 0 0 x y z y + + = − = =
Substitute and solve: 5 0 0 y y − = = 2(0) 1 1 x z z x + + = = −
number or {( , , ) |x y z y=0,z= −1 x, x is any real number}. 69. 5 3 2 2 0 x y z x y z − + = + − =
Write the augmented matrix: 1 1 1 5 3 2 2 0 − −
(
)
(
)
(
)
2 1 2 1 2 5 2 1 2 1 1 1 1 5 3 0 5 5 15 1 1 1 5 0 1 1 3 1 0 0 2 0 1 1 3 R r r R r R r r − → − − = − + − → − − = → − − = + The matrix in the last step represents the system 2 3 x y z = − = − or, equivalently, 2 3 x y z = = −
Thus, the solution is x=2, y= −z 3, z is any real number or {( , , ) |x y z x=2,y= −z 3, z is any real number}.
71. 2 3 3 0 0 3 5 x y z x y z x y z x y z + − = − − = − + + = + + =
Write the augmented matrix: 2 3 1 3 1 1 1 0 1 1 1 0 1 1 3 5 − − − − 1 2 2 1 2 3 1 3 4 1 4 1 1 1 0 interchange 2 3 1 3 and 1 1 1 0 1 1 3 5 1 1 1 0 2 0 5 1 3 0 0 0 0 0 2 4 5 r r R r r R r r R r r − − − → − − − = − + → = + = − + 3 4 1 1 1 0 interchange 0 5 1 3 and 0 2 4 5 0 0 0 0 r r − − →
(
)
(
)
2 3 2 1 2 1 3 2 3 1 3 3 19 18 18 1 1 1 0 0 1 7 7 2 0 2 4 5 0 0 0 0 1 0 8 7 0 1 7 7 2 0 0 18 19 0 0 0 0 7 1 0 8 7 0 1 7 = 0 0 1 0 0 0 0 R r r R r r R r r R r − − − − → = − + − − − − = + → = − + − − − − → The matrix in the last step represents the system
8 7 7 7 19 18 x z y z z − = − − = − =
Substitute and solve: 19 7 7 18 7 18 y y − = = 19 8 7 18 13 9 x x − = − = Thus, the solution is 13
9 x= , 7 18 y= , 19 18 z= or 13 7 19 , , 9 18 18 . 73. 4 4 2 3 3 x y z w x y z w + + − = − + + =
Write the augmented matrix:
1 2 4 1 1 1 4 1 1 2 3 3 interchange 1 1 2 3 3 and 4 1 1 1 4 r r − − − → −
(
2 1 2)
1 1 2 3 3 4 0 5 7 13 8 R r r − → = − + − − − The matrix in the last step represents the system
2 3 3 5 7 13 8 x y z w y z w − + + = − − = −
5 7 13 8 5 7 13 8 7 13 8 5 5 5 y z w y z w y z w − − = − = + − = + − The first equation yields
2 3 3 3 2 3 x y z w x y z w − + + = = + − − Substituting for y: 8 7 13 3 2 3 5 5 5 3 2 7 5 5 5 x z w z w x z w = + − + + − − = − − +
Thus, the solution is 3 2 7
5 5 5
x= − z− w+ ,
7 13 8
5 5 5
y= z+ w− , z and w are any real numbers or 3 2 7 7 13 8 ( , , , ) , , 5 5 5 5 5 5 x y z w x z w y z w = − − + = + −
and are any real numbers
z w
.
75. Each of the points must satisfy the equation
2 y ax= + +bx c. (1, 2) : 2 ( 2, 7) : 7 4 2 (2, 3) : 3 4 2 a b c a b c a b c = + + − − − = − + − − = + + Set up a matrix and solve:
(
)
2 1 2 3 1 3 1 5 1 2 6 2 2 2 1 1 2 2 1 2 1 5 1 2 2 3 2 3 1 1 1 2 2 7 4 1 3 4 2 1 1 1 1 2 4 0 6 3 15 4 2 0 3 11 1 1 1 2 0 1 2 0 3 11 0 1 0 1 2 2 6 0 0 R r r R r r R r R r r R r r − − − = − + → − − − = − + − − − = − → − − − − = − + → = + − − (
)
1 1 2 2 1 1 5 3 2 3 2 2 1 1 2 3 1 1 2 2 3 2 1 0 0 1 0 0 1 3 1 0 0 2 0 1 0 1 0 0 1 3 R r R r r R r r − → → = − − = − + → = − + The solution is a= −2,b=1,c=3; so the equation is y= −2x2+ +x 3.
77. Each of the points must satisfy the equation
3 2 ( ) f x =ax +bx + +cx d. ( 3) 112 : 27 9 3 112 ( 1) 2 : 2 (1) 4 : 4 (2) 13 : 8 4 2 13 f a b c d f a b c d f a b c d f a b c d − = − − + − + = − − = − − + − + = − = + + + = = + + + =
Set up a matrix and solve: 27 9 3 1 112 2 1 1 1 1 1 1 1 1 4 8 4 2 1 13 − − − − − − 3 1 1 1 1 1 4 Interchange 2 1 1 1 1 and 27 9 3 1 112 8 4 2 1 13 r r − − − → − − − 2 1 2 3 1 3 4 1 4 1 1 1 1 4 0 2 0 2 2 27 4 0 36 24 28 8 4 6 0 7 19 R r r R r r R r r = + → = + − = − + − − − −
(
1)
2 2 2 1 2 1 3 2 3 4 2 4 1 1 1 1 4 0 1 0 1 1 4 0 36 24 28 4 6 0 7 19 0 0 3 1 1 0 1 0 1 1 36 8 40 0 0 24 4 6 0 0 3 15 R r R r r R r r R r r → = − − − − − = − + → = − + − − = + − − − (
1)
3 24 3 5 5 3 3 0 0 3 1 1 0 1 0 1 1 0 0 1 6 0 0 3 15 R r → = − − − − − (
)
1 14 3 3 1 3 1 5 1 4 3 4 3 3 1 14 3 3 1 4 5 4 5 1 3 3 1 1 3 4 1 2 4 2 1 3 3 4 3 0 0 1 0 1 0 1 1 6 0 0 1 25 0 0 0 5 0 0 1 0 1 0 1 1 0 0 1 0 0 0 1 5 0 0 0 3 1 4 0 1 0 0 0 0 1 0 0 0 0 0 1 5 R r r R r r R r R r r R r r R r r = − + → = + − − − − → = − − − = − + − → = − + = + The solution is a=3,b= −4,c=0,d =5; so the equation is f x( ) 3= x3−4x2+5.
79. 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: 30x+15y+3z=78
Carbohydrate equation: 20x+2y+25z=59 Vitamin A equation: 2x+20y+32z=75 Set up a matrix and solve:
(
)
3 1 1 1 2 1 30 15 3 78 20 2 25 59 20 32 75 2 20 32 75 2 Interchange 20 2 25 59 and r 30 15 3 78 10 16 37.5 1 20 2 25 59 30 15 3 78 r R r → → = 2 1 2 3 1 3 10 16 37.5 1 20 0 198 295 691 30 0 285 477 1047 R r r R r r = − + → − − − = − + − − − (
)
(
)
95 3 66 2 3 3457 3457 66 66 66 3 3457 3 10 16 37.5 1 0 198 295 691 0 0 10 16 37.5 1 0 198 295 691 0 0 1 1 R r r R r → − − − = − + − − → − − − = − Substitute z=1 and solve: 198 295(1) 691 198 396 2 y y y − − = − − = − = 10(2) 16(1) 37.5 36 37.5 1.5 x x x + + = + = =
The dietitian should serve 1.5 servings of salmon steak, 2 servings of baked eggs, and 1 serving of acorn squash.
81. 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:
10,000
x y z+ + =
Annual income equation: 0.06x+0.07y+0.08z=680 Condition on investment equation:
0.5 2 0 z x x z = − =
Set up a matrix and solve:
10,000 1 1 1 0.06 0.07 0.08 680 0 2 0 1 −
(
)
2 1 2 3 1 3 2 2 10,000 1 1 1 0.06 0 0.01 0.02 80 10,000 0 1 3 10,000 1 1 1 0 1 2 8000 100 10,000 0 1 3 R r r R r r R r = − + → = − + − − − → = − − − (
)
1 2 1 3 2 3 3 3 1 3 1 2 3 2 0 2000 1 1 0 1 2 8000 0 0 1 2000 0 2000 1 1 0 1 2 8000 0 0 1 2000 0 0 4000 1 0 1 0 4000 2 0 0 1 2000 R r r R r r R r R r r R r r − = − + → = + − − − → = − = + → = − + Carletta should invest $4000 in Treasury bills, $4000 in Treasury bonds, and $2000 in corporate bonds.
83. Let x = the number of Deltas produced. Let y = the number of Betas produced. Let z = the number of Sigmas produced. Painting equation: 10x+16y+8z=240 Drying equation: 3x+5y+2z=69 Polishing equation: 2x+3y z+ =41 Set up a matrix and solve:
(
)
(
)
1 2 1 2 1 2 3 1 3 1 2 2 2 10 16 8 240 3 5 2 69 2 3 1 41 1 1 2 33 3 5 2 69 3 2 3 1 41 1 1 2 33 3 0 2 4 30 2 0 1 3 25 1 1 2 33 0 1 2 15 0 1 3 25 1 0 4 48 0 1 2 15 0 0 1 10 R r r R r r R r r R r → = − + = − + → − − = − + − − → − − = − − → − − − − 1 1 2 3 3 2 R r r R r r = − = − (
3 3)
1 3 1 2 3 2 1 0 4 48 0 1 2 15 0 0 1 10 1 0 0 8 4 0 1 0 5 2 0 0 1 10 R r R r r R r r → − − = − = − + → = + The company should produce 8 Deltas, 5 Betas, and 10 Sigmas.
85. Rewrite the system to set up the matrix and solve: 2 2 4 1 1 4 3 1 1 3 3 4 1 1 3 4 4 8 2 0 2 4 8 5 5 8 4 3 3 4 0 I I I I I I I I I I I I I I I I − + − = = = + + = → = + + = + = − − = 2 1 1 2 2 2 3 1 3 4 1 4 0 2 0 0 4 0 0 5 8 1 0 3 0 1 4 0 0 1 1 1 0 0 5 8 1 Interchange 0 2 0 0 4 and 0 3 0 1 4 0 0 1 1 1 0 0 5 8 1 0 1 0 0 2 4 0 0 3 5 6 8 0 0 1 r r R r R r r R r r − − → − − = → = − + − − = − + − − −
