Download Graph the system of inequalities in a coordinate plane. . 8.

3-2 Solving Systems of Inequalities by Graphing
.
Solve each system of inequalities by graphing.
10. 7. SOLUTION: Graph the system of inequalities in a coordinate
plane.
SOLUTION: Graph the system of inequalities in a coordinate
plane.
.
.
11. 8. SOLUTION: Graph the system of inequalities in a coordinate
plane.
SOLUTION: Graph the system of inequalities in a coordinate
plane.
.
.
12. 9. SOLUTION: Graph the system of inequalities in a coordinate
plane.
SOLUTION: Graph the system of inequalities in a coordinate
plane.
.
.
10. SOLUTION: Graph the system of inequalities in a coordinate
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plane.
13. SOLUTION: The graph of the system of inequalities is
Page 1
3-2 Solving Systems of Inequalities by Graphing
.
13. SOLUTION: The graph of the system of inequalities is
.
14. SOLUTION: Graph the system of inequalities in a coordinate
plane.
.
15. SOLUTION: Graph the system of inequalities in a coordinate
plane.
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Page 2
3-4 Systems of Equations in Three Variables
Solve each system of equations.
2. SOLUTION: Multiply the third equation by 2 and with the second equation.
Substitute 1 for z in the first equation and solve for y.
Substitute –6 for y and 1 for z in the third equation, and solve for x.
Therefore, the solution is (4, –6, 1).
4. SOLUTION: Eliminate one variable.
Multiply the first equation by 2 and add with the second equation.
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Page 1
3-4 Systems
ofthe
Equations
Three
Variables
Therefore,
solution isin(4,
–6, 1).
4. SOLUTION: Eliminate one variable.
Multiply the first equation by 2 and add with the second equation.
Multiply the second equation by –3, multiply the third equation by 2 and add.
To solve the fourth and fifth equations, add both equations.
Substitute –2 for r in the fifth equation and solve for t.
Substitute –2 for r and –5 for t in the first equation, and solve for s.
Therefore, the solution is (–2, 2, –5).
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SOLUTION: Page 2
3-4 Systems
of Equations in Three Variables
Therefore, the solution is (–2, 2, –5).
6. SOLUTION: Eliminate one variable.
Multiply the second equation by 3 and add with the first equation.
Multiply the second equation by 8 and add with the third equation.
Solve the fourth and fifth equations.
Substitute 3 for a in the fourth equation and solve for c.
Substitute 3 for a and 8 for c in the second equation, and solve for b.
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Therefore, the solution is (3, –4, 8).
Solve each system of equations.
Page 3
3-4 Systems of Equations in Three Variables
Therefore, the solution is (3, –4, 8).
Solve each system of equations.
8. SOLUTION: Eliminate one variable.
Multiply the first equation by –4 and add with the second equation.
Multiply the first equation by 3 and add with the third equation.
Solve the fourth and fifth equations.
Substitute –4 for z in the fourth equation and solve for x.
Substitute –8 and –4 for x and z in the first equation and solve for y.
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Therefore, the solution is (–8, 4, –4).
Page 4
3-4 Systems
of Equations in Three Variables
Substitute –8 and –4 for x and z in the first equation and solve for y.
Therefore, the solution is (–8, 4, –4).
10. SOLUTION: Eliminate one variable.
Multiply the first equation by –2 and add with the second equation.
Multiply the first equation by 3 and add with the third equation.
Solve the fourth and fifth equations.
Substitute –3 for y in the fifth equation and solve for z.
Substitute
and 2 by
forCognero
y and z
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- Powered
in the first equation and solve for x.
Page 5
3-4 Systems of Equations in Three Variables
Substitute –3 and 2 for y and z in the first equation and solve for x.
Therefore, the solution is (8, –3, 2).
12. SOLUTION: Eliminate one variable.
Multiply the first equation by 2 and add with the second equation.
Multiply the second equation by 2 and the third equation by 4 then add.
Solve the fourth and the fifth equation.
This is a false statement. Therefore, there is no solution.
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Eliminate one variable.
Multiply the second equation by –3 and add with the first equation.
Page 6
14. 3-4 Systems of Equations in Three Variables
SOLUTION: Eliminate one variable.
Multiply the second equation by –3 and add with the first equation.
Multiply the second and third equation by 5 and –2 respectively and add.
Solve the fourth and fifth equations.
Substitute 3 for y in the fourth equation and solve for x.
Substitute –1 and 3 for x and y in the first equation and solve for z.
Therefore, the solution is (–1, 3, 7).
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SOLUTION: Page 7
3-4 Systems
of Equations in Three Variables
Therefore, the solution is (–1, 3, 7).
16. SOLUTION: Eliminate one variable.
Multiply the first and second equation by 5 and 2 respectively then add.
Multiply the first equation by 4 and add with the third equation.
Solve the fourth and fifth equations.
Substitute –8 for a in the fourth equation and solve for b.
Substitute –8 and –7 for a and b in the third equation and solve for c.
Therefore, the solution is (–8, –7, –5).
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18. Page 8
3-4 Systems
of Equations in Three Variables
Therefore, the solution is (–8, –7, –5).
18. SOLUTION: Eliminate one variable.
Subtract the first equation from the third equation.
Multiply the fourth equation by 4 and add with the second equation.
Substitute 3 for y in the fourth equation and solve for z.
Substitute 3 for y in the first equation and solve for x.
Therefore, the solution is (6, 3, –4)
20. CCSS SENSE-MAKING A friend e-mails you the results of a recent high school swim meet. The e-mail states
that 24 individuals placed, earning a combined total of 53 points. First place earned 3 points, second place earned 2
points, and third place earned 1 point. There were as many first-place finishers as second- and third-place finishers
combined.
a. Write a system of three equations that represents how many people finished in each place.
b. How many swimmers finished in first place, in second place, and in third place?
c. Suppose the e-mail had said that the athletes scored a combined total of 47 points. Explain why this statement is
false and the solution is unreasonable.
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SOLUTION: a. Let x, y, and z be the number of swimmers finished in first place, in second place and in third place.
Page 9
3-4 Systems
of Equations in Three Variables
Therefore, the solution is (6, 3, –4)
20. CCSS SENSE-MAKING A friend e-mails you the results of a recent high school swim meet. The e-mail states
that 24 individuals placed, earning a combined total of 53 points. First place earned 3 points, second place earned 2
points, and third place earned 1 point. There were as many first-place finishers as second- and third-place finishers
combined.
a. Write a system of three equations that represents how many people finished in each place.
b. How many swimmers finished in first place, in second place, and in third place?
c. Suppose the e-mail had said that the athletes scored a combined total of 47 points. Explain why this statement is
false and the solution is unreasonable.
SOLUTION: a. Let x, y, and z be the number of swimmers finished in first place, in second place and in third place.
b. Name the equations.
Substitute x for y + z in the first equation and solve for x.
Substitute 12 for x and 5 for y + z in the second equation and solve for y.
Substitute x and y values in the first equation and solve for z.
7 simmers placed third, 5 simmers placed second, and 12 simmers placed first.
c. The statement is false because when you solve for second place, you get a negative as an answer and you cannot
have a negative person.
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Page 10
3-6 Multiplying Matrices
Find each product, if possible.
5. Find each product, if possible.
21. SOLUTION: The inner dimensions of the matrices are equal.
So:
SOLUTION: 7. 22. SOLUTION: The inner dimensions of the matrices are equal.
So:
SOLUTION: 9. SOLUTION: The inner dimensions of the matrices are equal.
So:
23. SOLUTION: 11. 24. SOLUTION: SOLUTION: The inner dimensions of the matrices are equal.
So:
Find each product, if possible.
25. 21. SOLUTION: The inner dimensions of the matrices are not equal.
So, the matrices cannot be multiplied.
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26. 25. SOLUTION: 3-6 Multiplying
Matricesof the matrices are not equal.
The inner dimensions
So, the matrices cannot be multiplied.
26. SOLUTION: 27. SOLUTION: 28. SOLUTION: eSolutions Manual - Powered by Cognero
Page 2
3-7 Solving Systems of Equations Using Cramer's Rule
Evaluate each determinant.
29. 26. SOLUTION: SOLUTION: Rewrite the first two columns in the right of the
determinant.
27. Find the product of the element of the diagonal.
SOLUTION: 28. Find the sum of each group.
SOLUTION: Subtract the value of the second group from the first
group.
29. Therefore, the value of the determinant is –135.
SOLUTION: Rewrite the first two columns in the right of the
determinant.
30. SOLUTION: Rewrite the first two columns in the right of the
determinant.
Find the product of the element of the diagonal.
Find the product of the element of the diagonal.
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3-7 Solving
Systems of Equations Using Cramer's Rule
Therefore, the value of the determinant is –135.
Therefore, the value of the determinant is 124.
31. 30. SOLUTION: Rewrite the first two columns in the right of the
determinant.
SOLUTION: Rewrite the first two columns in the right of the
determinant.
Find the product of the element of the diagonal.
Find the product of the element of the diagonal.
Find the sum of each group.
Find the sum of each group.
Subtract the value of the second group from the first
group.
Subtract the value of the second group from the first
group.
Therefore, the value of the determinant is –459.
Therefore, the value of the determinant is 124.
32. 31. SOLUTION: Rewrite the first two columns in the right of the
determinant.
SOLUTION: Rewrite the first two columns in the right of the
determinant.
Find the product of the element of the diagonal.
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Find the product of the element of the diagonal.
Page 2
group.
group.
3-7 Solving
Systems of Equations Using Cramer's Rule
Therefore, the value of the determinant is –459.
32. Therefore, the value of the determinant is 63.
33. SOLUTION: Rewrite the first two columns in the right of the
determinant.
SOLUTION: Rewrite the first two columns in the right of the
determinant.
Find the product of the element of the diagonal.
Find the product of the element of the diagonal.
Therefore, the value of the determinant is 0.
Find the sum of each group.
Subtract the value of the second group from the first
group.
Therefore, the value of the determinant is 63.
34. SOLUTION: Rewrite the first two columns in the right of the
determinant.
33. SOLUTION: Rewrite the first two columns in the right of the
determinant.
Find the product of the element of the diagonal.
Find the product of the element of the diagonal.
Find the sum of each group.
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Page 3
Subtract the value of the second group from the first
3-7 Solving
Systems of Equations Using Cramer's Rule
Therefore, the value of the determinant is 0.
34. Therefore, the value of the determinant is –13.
35. SOLUTION: Rewrite the first two columns in the right of the
determinant.
SOLUTION: Rewrite the first two columns in the right of the
determinant.
Find the product of the element of the diagonal.
Find the product of the element of the diagonal.
Find the sum of each group.
Find the sum of each group.
Subtract the value of the second group from the first
group.
Subtract the value of the second group from the first
group.
Therefore, the value of the determinant is –13.
Therefore, the value of the determinant is 728.
35. 36. SOLUTION: Rewrite the first two columns in the right of the
determinant.
SOLUTION: Rewrite the first two columns in the right of the
determinant.
Find the product of the element of the diagonal.
Find the product of the element of the diagonal.
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Page 4
group.
3-7 Solving
Systems of Equations Using Cramer's Rule
Therefore, the value of the determinant is 728.
Therefore, the value of the determinant is –120.
37. 36. SOLUTION: Rewrite the first two columns in the right of the
determinant.
SOLUTION: Rewrite the first two columns in the right of the
determinant.
Find the product of the element of the diagonal.
Find the product of the element of the diagonal.
Find the sum of each group.
Find the sum of each group.
Subtract the value of the second group from the first
group.
Subtract the value of the second group from the first
group.
Therefore, the value of the determinant is –120.
Therefore, the value of the determinant is –952.
37. SOLUTION: Rewrite the first two columns in the right of the
determinant.
Find the product of the element of the diagonal.
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Page 5
3-8 Solving Systems of Equations Using Inverse Matrices
Since
Determine whether each pair of matrices are
inverses of each other.
, they are not inverses.
Find the inverse of each matrix, if it exists.
17. 13. SOLUTION: SOLUTION: If K and L are inverses, then
.
Since the determinant does not equal 0, the inverse
exists.
Since
Substitute
, they are not inverses.
and .
15. SOLUTION: If P and Q are inverses, then
.
19. SOLUTION: Since
, they are not inverses.
Find the inverse of each matrix, if it exists.
Since the determinant does not equal 0, the inverse
exists.
17. SOLUTION: Substitute
and .
Since the determinant does not equal 0, the inverse
exists.
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3-8 Solving Systems of Equations Using Inverse Matrices
19. 21. SOLUTION: SOLUTION: Since the determinant does not equal 0, the inverse
exists.
Since the determinant does not equal 0, the inverse
exists.
Substitute
and .
Substitute
and .
23. 21. SOLUTION: SOLUTION: Since the determinant does not equal 0, the inverse
exists.
Since the determinant does not equal 0, the inverse
exists.
Substitute
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and .
Substitute
and .
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3-8 Solving Systems of Equations Using Inverse Matrices
23. 25. SOLUTION: SOLUTION: Since the determinant does not equal 0, the inverse
exists.
Since the determinant does not equal 0, the inverse
exists.
Substitute
and .
and Substitute
.
CCSS PERSEVERANCE Use a matrix equation to solve each system of equations.
27. −x + y = 4
−x + y = −4
25. SOLUTION: SOLUTION: Since the determinant does not equal 0, the inverse
exists.
Since the determinant is equal to 0, the inverse does
not exist.
Therefore, the system has no solution.
29. Substitute
and .
SOLUTION: The matrix equation is
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.
Find the inverse of the coefficient matrix.
Page 3
Since the determinant is equal to 0, the inverse does
not exist.
3-8 Solving
Systems
of Equations
Using Inverse Matrices
Therefore,
the system
has no solution.
The solution is (–1, 5).
31. y − x = 5
2y − 2x = 8
29. SOLUTION: Rewrite the given system as below.
SOLUTION: The matrix equation is
−x + y = 5
−2x + 2y = 8
.
Find the inverse of the coefficient matrix.
Substitute
and .
Since the determinant is equal to 0, the inverse does
not exist.
Therefore, the system has no solution.
33. 1.6y − 0.2x = 1
0.4y −0.1x = 0.5
SOLUTION: Rewrite the given system as below.
− 0.2x + 1.6y = 1
−0.1x + 0.4y = 0.5
Multiply each side of the matrix equation by the
inverse matrix.
The matrix equation is
.
Find the inverse of the coefficient matrix.
Substitute
and .
The solution is (–1, 5).
31. y − x = 5
2y − 2x = 8
SOLUTION: Rewrite the given system as below.
−x + y = 5
−2x + 2y = 8
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Multiply each side of the matrix equation by the
inverse matrix.
Page 4
Since the determinant is equal to 0, the inverse does
not exist.Systems of Equations Using Inverse Matrices
3-8 Solving
Therefore, the system has no solution.
The solution is (–5, 0).
35. 2y − 4x = 3
4x − 3y = −6
33. 1.6y − 0.2x = 1
0.4y −0.1x = 0.5
SOLUTION: Rewrite the given system as below.
SOLUTION: Rewrite the given system as below.
− 0.2x + 1.6y = 1
−0.1x + 0.4y = 0.5
− 4x + 2y = 3
4x − 3y = −6
The matrix equation is
The matrix equation is
.
.
Find the inverse of the coefficient matrix.
Find the inverse of the coefficient matrix.
Substitute
and .
and Substitute
.
Multiply each side of the matrix equation by the
inverse matrix.
Multiply each side of the matrix equation by the
inverse matrix.
The solution is (–5, 0).
35. 2y − 4x = 3
4x − 3y = −6
SOLUTION: Rewrite the given system as below.
The solution is
.
− 4x + 2y = 3
4x − 3y = −6
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The matrix equation is
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