Download Evaluate each expression if x = –4 and y = –9. 2. SOLUTION

1-4 Solving Absolute Value Equations
Evaluate each expression if x = –4 and y = –9.
2. SOLUTION: Substitute –9 for y and solve.
4. SOLUTION: Substitute -4 for x and solve.
Solve each equation. Check your solutions.
6. SOLUTION: There appear to be two solutions, 4 and −20.
Check: Substitute each value in the original equation.
The solution set is
.
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1-4 Solving
Absolute
The solution
set is Value Equations
.
8. SOLUTION: There appear to be two solutions, 10 and 0.
Check: Substitute each value in the original equation.
The solution set is
.
10. SOLUTION: There appear to be two solutions, 3 and 0.
Check: Substitute each value in the original equation.
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1-4 Solving
Absolute Value Equations
The solution set is
.
10. SOLUTION: There appear to be two solutions, 3 and 0.
Check: Substitute each value in the original equation.
The solution set is
.
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1-4 Solving
Absolute Value Equations
The solution set is
.
12. SOLUTION: There appear to be two solutions,
.
Check: Substitute each value in the original equation.
Since 3 ≠ –3, a = 1 is an extraneous solution. The solution is
.
Evaluate each expression if a = –3, b = –5, and c = 4.2.
14. SOLUTION: Substitute 4.2 for c and solve.
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16. Page 4
1-4 Solving
Absolute
Equationssolution. The solution is
Since 3 ≠ –3,
a = 1Value
is an extraneous
.
Evaluate each expression if a = –3, b = –5, and c = 4.2.
14. SOLUTION: Substitute 4.2 for c and solve.
16. SOLUTION: Substitute –3 for a and –5 for b and solve.
18. SOLUTION: Substitute –3 for a and –5 for b and solve.
20. SOLUTION: Substitute –3 for a and 4.2 for c and solve.
22. FOOD To make cocoa powder, cocoa beans are roasted. The ideal temperature for roasting is 300°F, plus or minus 25°. Write and solve an equation describing the maximum and minimum roasting temperatures for cocoa beans.
SOLUTION: Solve the equation
.
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So, the maximum temperature is 325°F and the minimum temperature is 275°F.
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1-4 Solving Absolute Value Equations
22. FOOD To make cocoa powder, cocoa beans are roasted. The ideal temperature for roasting is 300°F, plus or minus 25°. Write and solve an equation describing the maximum and minimum roasting temperatures for cocoa beans.
SOLUTION: Solve the equation
.
So, the maximum temperature is 325°F and the minimum temperature is 275°F.
Solve each equation. Check your solutions.
24. SOLUTION: There appear to be two solutions, 8 and –26.
The solution set is
.
26. SOLUTION: There appear to be two solutions, 41 and –29.
Check: Substitute each value in the original equation.
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1-4 Solving
Absolute
The solution
set is Value Equations
.
26. SOLUTION: There appear to be two solutions, 41 and –29.
Check: Substitute each value in the original equation.
The solution set is
.
28. SOLUTION: There appear to be two solutions, 3 and –11.
Check: Substitute each value in the original equation.
The solution set is
.
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SOLUTION: Page 7
1-4 Solving
Absolute Value Equations
The solution set is
.
30. SOLUTION: Check:
The solution is
.
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1-4 Solving Absolute Value Equations
There appear to be two solutions,
.
Check: Substitute each value in the original equation.
The solution set is
.
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Page 9
The solution set is
.
34. Solving Absolute Value Equations
1-4
SOLUTION: There appear to be two solutions,
.
Check: Substitute the values in the original equation.
z=
z=
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1-4 Solving
Absolute Value Equations
z=
The solution set is
.
Evaluate each expression if q = –8, r = –6, and t = 3.
36. SOLUTION: Substitute –6 for r and 3 for t and solve.
38. SOLUTION: Substitute -8 for q, -6 for r, and 3 for t and solve.
Solve each equation. Check your solutions.
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Solve each equation. Check your solutions.
40. 1-4 Solving Absolute Value Equations
SOLUTION: There appear to be two solutions,
.
Check: Substitute the values in the original equation.
y=
.
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y = 8. Page 12
There appear to be two solutions,
.
Check: Substitute the values in the original equation.
1-4 Solving
Absolute Value Equations
y=
.
y = 8. Since −44 ≠ 60, y = 8 is an extraneous solution. The solution set is .
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1-4 Solving Absolute Value Equations
There appear to be two solutions,
.
Check: Substitute the values in the original equation.
y=
.
y=
.
The solution set is
.
44. MULTIPLE REPRESENTATIONS Draw a number line.
a. GEOMETRIC Label any 5 integers on the number line points A, B, C, D, and F.
b. TABULAR Fill in each blank in the table with either > or < using the points from the number line.
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1-4 Solving
Absolute
The solution
set is Value Equations
.
44. MULTIPLE REPRESENTATIONS Draw a number line.
a. GEOMETRIC Label any 5 integers on the number line points A, B, C, D, and F.
b. TABULAR Fill in each blank in the table with either > or < using the points from the number line.
c. VERBAL Describe the patterns in the table.
d. ALGEBRAIC Describe the patterns algebraically, using the variable x to replace C, D, and F.
SOLUTION: a. Draw a number line and label from -3 to 3. Place points A, B, C, D, and F on the number line. Sample answer:
b. Fill in the table based on the number line drawn in part a. Since A is -2 and B is -1. A < B. Since C is 0, A + C = 2 + 0 or -2. B + C = -1 + 0 or -1. Therefore, A + C < B + C. Similar reasoning is used to complete the table. c. Sample answer: If A is less than B, then any number added to or subtracted from A will be less than the same
number added to or subtracted from B. If B is greater than A, then any number added to or subtracted from B is
greater than the same number added to or subtracted from A.
d. If A < B, then A + x < B + x. If A < B, then A – x < B – x.
If B > A, then B + x > A + x. If B > A, then B – x > A – x.
46. CHALLENGE Solve
cases to examine as potential solutions.)
List all cases and resulting equations. (Hint: There are four possible
SOLUTION: The 4 potential solutions are:
1. (2x − 1) ≥ 0 and (5 − x) ≥ 0
2. (2x − 1) ≥ 0 and (5 − x) < 0
3. (2x − 1) < 0 and (5 − x) ≥ 0
4. (2x − 1) < 0 and (5 − x) < 0
The resulting equations corresponding to these cases are:
1. 2x – 1 + 3 = 5 – x : x = 1
2. 2x – 1 + 3 = x – 5 : x = –7
3. 1 – 2x + 3 = 5 – x : x = –1
4. 1 – 2x + 3 = x – 5 : x = 3
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The solutions from case 1 and case 3 work. The others are extraneous. The solution set is {–1, 1}.
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REASONING If a, x, and y are real numbers, determine whether each statement is sometimes, always, or
number added to or subtracted from B. If B is greater than A, then any number added to or subtracted from B is
greater than the same number added to or subtracted from A.
B, then AValue
+ x < Equations
B + x. If A < B, then A – x < B – x.
d. If A <Absolute
1-4 Solving
If B > A, then B + x > A + x. If B > A, then B – x > A – x.
46. CHALLENGE Solve
cases to examine as potential solutions.)
List all cases and resulting equations. (Hint: There are four possible
SOLUTION: The 4 potential solutions are:
1. (2x − 1) ≥ 0 and (5 − x) ≥ 0
2. (2x − 1) ≥ 0 and (5 − x) < 0
3. (2x − 1) < 0 and (5 − x) ≥ 0
4. (2x − 1) < 0 and (5 − x) < 0
The resulting equations corresponding to these cases are:
1. 2x – 1 + 3 = 5 – x : x = 1
2. 2x – 1 + 3 = x – 5 : x = –7
3. 1 – 2x + 3 = 5 – x : x = –1
4. 1 – 2x + 3 = x – 5 : x = 3
The solutions from case 1 and case 3 work. The others are extraneous. The solution set is {–1, 1}.
REASONING If a, x, and y are real numbers, determine whether each statement is sometimes, always, or
never true. Explain your reasoning.
48. If
then SOLUTION: , then x is between –3 or 3. Adding 3 to the absolute value of any of the numbers in this set will
Always; if produce a positive number.
50. OPEN ENDED Write an absolute value equation of the form
that has no solution. Assume that a,
b, c, and
SOLUTION: Sample answer: An absolute value expression cannot be negative. Solve the equation and check the solutions. 52. If 4x – y = 3 and 2x + 3y = 19, what is the value of y?
A2
B3
C4
D5
SOLUTION: Substitute
in the equation 2x + 3y = 19.
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SOLUTION: Sample answer: An absolute value expression cannot be negative. Solve the equation and check the solutions. 1-4 Solving Absolute Value Equations
52. If 4x – y = 3 and 2x + 3y = 19, what is the value of y?
A2
B3
C4
D5
SOLUTION: Substitute
in the equation 2x + 3y = 19.
So, the correct choice is D.
54. Which equation is equivalent to 4(9 – 3x) = 7 – 2(6 – 5x)?
F 8x = 41
G 22x = 41
H 8x = 24
J 22x = 24
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1-4 Solving
Absolute Value Equations
So, the correct choice is D.
54. Which equation is equivalent to 4(9 – 3x) = 7 – 2(6 – 5x)?
F 8x = 41
G 22x = 41
H 8x = 24
J 22x = 24
SOLUTION: Therefore, 22x = 41 is equivalent to 4(9 – 3x) = 7 – 2(6 – 5x).
So, the correct choice is G.
Solve each equation. Check your solution.
56. SOLUTION: Check: Substitute x = 6 in the original equation.
The solution is x = 6.
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1-4 Solving
Absolute
Equations
The solution
is x = Value
6.
58. SOLUTION: Check: Substitute y = 50 in the original equation.
The solution is y = 50.
Name the property illustrated by each equation.
60. (1 + 8) + 11 = 11 + (1 + 8)
SOLUTION: Commutative Property of Addition; the Commutative Property of Addition states that the order in which terms are
added does not affect the sum.
Simplify each expression.
62. 7a + 3b – 4a – 5b
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60. (1 + 8) + 11 = 11 + (1 + 8)
SOLUTION: Commutative Property of Addition; the Commutative Property of Addition states that the order in which terms are
1-4 Solving
Absolute
Value
added does
not affect
the Equations
sum.
Simplify each expression.
62. 7a + 3b – 4a – 5b
SOLUTION: 64. 3(15x – 9y) + 5(4y – x)
SOLUTION: 66. 8(r + 7t) – 4(13t + 5r)
SOLUTION: 68. GEOMETRY The formula for the surface area of a rectangular prism is
where represents the length, w represents the width, and h represents the height. Find the surface area of the rectangular
prism at the right.
SOLUTION: Substitute
in the formula .
The surface area of the rectangular prism is 358 square inches.
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Solve each equation.
70. 2.4y + 4.6 = 20
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1-4 Solving Absolute Value Equations
68. GEOMETRY The formula for the surface area of a rectangular prism is
where represents the length, w represents the width, and h represents the height. Find the surface area of the rectangular
prism at the right.
SOLUTION: Substitute
in the formula .
The surface area of the rectangular prism is 358 square inches.
Solve each equation.
70. 2.4y + 4.6 = 20
SOLUTION: 72. 3(w – 1) = 2w – 6
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1-4 Solving Absolute Value Equations
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