Holt McDougal Algebra 1 3-7 Solving Absolute-Value Inequalities 3-7 Solving Absolute-Value Inequalities Holt Algebra 1 Warm Up Warm Up Lesson Presentation

Holt McDougal Algebra 1

3-7 Solving Absolute-Value Inequalities3-7 Solving Absolute-Value Inequalities

Holt Algebra 1

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Lesson QuizLesson Quiz



3-7 Solving Absolute-Value Inequalities

Solve compound inequalities in one variable involving absolute-value expressions.

Objectives



Warm UpSolve each inequality and graph the solution. 1. x + 7 < 4

2.

3. 5 + 2x > 1

14x ≥ 28

x < –3 –5 –4 –3 –2 –1 0 1 2 3 4 5

x > –2–5 –4 –3 –2 –1 0 1 2 3 4 5

x ≥ 2 –5 –4 –3 –2 –1 0 1 2 3 4 5



When an inequality contains an absolute-value expression, it can be written as a compound inequality. The inequality |x| < 5 describes all real numbers whose distance from 0 is less than 5 units. The solutions are all numbers between –5 and 5, so |x|< 5 can be rewritten as –5 < x < 5, or as x > –5 AND x < 5.





Additional Example 1A: Solving Absolute-Value Inequalities Involving <

Solve the inequality and graph the solutions.

|x|– 3 < –1

x > –2 AND x < 2

Since 3 is subtracted from |x|, add 3 to both sides to undo the subtraction.

+3 +3|x| < 2

|x|– 3 < –1

Write as a compound inequality.

–2 –1 0 1 2

2 units 2 units



|x – 1| ≤ 2


Additional Example 1B: Solving Absolute-Value Inequalities Involving <


x – 1 ≥ –2 AND x – 1 ≤ 2 Solve each inequality.+1 +1 +1 +1

x ≥ –1 x ≤ 3AND Write as a compound inequality.

–2 –1 0 1 2 3–3



Just as you do when solving absolute-value equations, you first isolate the absolute-value expression when solving absolute-value inequalities.

Helpful Hint



Check It Out! Example 1a


2|x| ≤ 6

x ≥ –3 AND x ≤ 3

Since x is multiplied by 2, divide both sides by 2 to undo the multiplication.|x| ≤ 3


2|x| ≤ 62 2

–2 –1 0 1 2

3 units 3 units

–3 3


3-7 Solving Absolute-Value InequalitiesCheck It Out! Example 1b

|x + 3|– 4.5 ≤ 7.5Since 4.5 is subtracted from |

x + 3|, add 4.5 to both sides to undo the subtraction.

Solve each inequality and graph the solutions.

+ 4.5 +4.5 |x + 3| ≤ 12

|x + 3|– 4.5 ≤ 7.5

x + 3 ≥ –12 AND x + 3 ≤ 12 –3 –3 –3 –3x ≥ –15 AND x ≤ 9


–20 –15 –10 –5 0 5 10 15

Subtract 3 from both sides of each inequality.



The inequality |x| > 5 describes all real numbers whose distance from 0 is greater than 5 units. The solutions are all numbers less than –5 or greater than 5. The inequality |x| > 5 can be rewritten as the compound inequality x < –5 OR x > 5.






Additional Example 2A: Solving Absolute-Value Inequalities Involving >

|x| + 14 ≥ 19

|x| ≥ 5

x ≤ –5 OR x ≥ 5

Since 14 is added to |x|, subtract 14 from both sides to undo the addition.


–10 –8 –6 –4 –2 0 2 4 6 8 10

5 units 5 units

– 14 –14|x| + 14 ≥ 19




3 + |x + 2| > 5Since 3 is added to |x + 2|,

subtract 3 from both sides to undo the addition.

Write as a compound inequality. Solve each inequality.

Additional Example 2B: Solving Absolute-Value Inequalities Involving >

|x + 2| > 2

– 3 – 33 + |x + 2| > 5

x + 2 < –2 OR x + 2 > 2–2 –2 –2 –2

x < –4 OR x > 0 Write as a compound inequality.

–10 –8 –6 –4 –2 0 2 4 6 8 10


3-7 Solving Absolute-Value InequalitiesCheck It Out! Example 2a

|x| + 10 ≥ 12

|x| + 10 ≥ 12– 10 –10

|x| ≥ 2

–5 –4 –3 –2 –1 0 1 2 3 4 5

x ≤ –2 OR x ≥ 2 Write as a compound inequality.

Since 10 is added to |x|, subtract 10 from both sides to undo the addition.


2 units 2 units



Check It Out! Example 2b

Since is added to |x + 2 |, subtract from both sides to undo the addition.

OR


x ≤ –6 x ≥ 1


Write as a compound inequality. Solve each inequality.



Check It Out! Example 2b Continued


–7 –6 –5 –4 –3 0 1 2 3–2 –1



Additional Example 3: ApplicationA pediatrician recommends that a baby’s bath water be 95°F, but it is acceptable for the temperature to vary from this amount by as much as 3°F. Write and solve an absolute-value inequality to find the range of acceptable temperatures. Graph the solutions.

Let t represent the actual water temperature.

The difference between t and the ideal temperature is at most 3°F.

t – 95 ≤ 3



Additional Example 3 Continued

t – 95 ≤ 3

|t – 95| ≤ 3

t – 95 ≥ –3 AND t – 95 ≤ 3 Solve the two inequalities.+95 +95 +95 +95

t ≥ 92 AND t ≤ 98

The range of acceptable temperature is 92 ≤ t ≤ 98.

98 10096949290



Check It Out! Example 3

A dry-chemical fire extinguisher should be pressurized to 125 psi, but it is acceptable for the pressure to differ from this value by at most 75 psi. Write and solve an absolute-value inequality to find the range of acceptable pressures. Graph the solution.

Let p represent the desired pressure.

The difference between p and the ideal pressure is at most 75 psi.

p – 125 ≤ 75



Check It Out! Example 3 Continued

p – 125 ≤ 75

|p – 125| ≤ 75

p – 125 ≥ –75 AND p – 125 ≤ 75 Solve the two inequalities.+125 +125 +125 +125

p ≥ 50 AND p ≤ 200

The range of pressure is 50 ≤ p ≤ 200.

200 225175150125100755025



When solving an absolute-value inequality, you may get a statement that is true for all values of the variable. In this case, all real numbers are solutions of the original inequality. If you get a false statement when solving an absolute-value inequality, the original inequality has no solutions.



Additional Example 4A: Special Cases of Absolute-Value Inequalities

Solve the inequality.

|x + 4|– 5 > – 8

|x + 4|– 5 > – 8 + 5 + 5

|x + 4| > –3

Add 5 to both sides.

Absolute-value expressions are always nonnegative. Therefore, the statement is true for all real numbers.

All real numbers are solutions.


3-7 Solving Absolute-Value InequalitiesAdditional Example 4B: Special Cases of Absolute-Value

Inequalities


|x – 2| + 9 < 7

|x – 2| + 9 < 7 – 9 – 9

|x – 2| < –2

Subtract 9 from both sides.

Absolute-value expressions are always nonnegative. Therefore, the statement is false for all values of x.

The inequality has no solutions.



An absolute value represents a distance, and distance cannot be less than 0.

Remember!



Check It Out! Example 4a


|x| – 9 ≥ –11

|x| – 9 ≥ –11 +9 ≥ +9

|x| ≥ –2Add 9 to both sides.

Absolute-value expressions are always nonnegative. Therefore, the statement is true for all real numbers.

All real numbers are solutions.



Check It Out! Example 4b


4|x – 3.5| ≤ –8

4|x – 3.5| ≤ –8

4 4|x – 3.5| ≤ –2 Absolute-value expressions are

always nonnegative. Therefore, the statement is false for all values of x.

Divide both sides by 4.

The inequality has no solutions.


3-7 Solving Absolute-Value InequalitiesLesson Quiz: Part I


1. 3|x| > 15 x < –5 or x > 5

0–5–10 5 10

2. |x + 3| + 1 < 3 –5 < x < –1

–2 0–1–3–4–5–6

3. A number, n, is no more than 7 units away from 5. Write and solve an inequality to show the range of possible values for n.|n– 5| ≤ 7; –2 ≤ n ≤ 12



Solve each inequality.

4. |3x| + 1 < 1

5. |x + 2| – 3 ≥ – 6 all real numbers

no solutions

Lesson Quiz: Part II

Documents

Holt McDougal Algebra 1 3-7 Solving Absolute-Value Inequalities 3-7 Solving Absolute-Value Inequalities Holt Algebra 1 Warm Up Warm Up Lesson Presentation