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In the Solve It, Serena’s distance from Darius decreases and then increases. You can use absolute value to model such changes. You can solve absolute value equations and inequalities by first isolating the absolute value expression, if necessary. Then write an equivalent pair of linear equations or inequalities. PROBLEM 1: SOLVING AN ABSOLUTE VALUE EQUATION Solve each equation. Check your solutions. a) |"| +2=9 b) || −5=2 c) −3|+| = −9 d) |, | −3=7 e) 5|.| = 20 f) |0| 1 =3 3-7 Absolute Value Equations and Inequalities absolutevaluee the distance a number isfrom zero on the number line alwayspositive 1 21 2 151 5 Ei Ei E Tts 1 2 tests 3 Int 7 Im 1 3 x or n n m rm 3 3 g f C 7 14 10 1,1 4 Igt 21 t 0ort r yorr Y QoF

Absolute Value Equations and Inequalities...3-7 Absolute Value Equations and Inequalities Write T for true or F for false. 1. To indicate the absolute value of 28, you write u28 u

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  • In the Solve It, Serena’s distance from Darius decreases and then increases. You can use absolute value to model such changes. You can solve absolute value equations and inequalities by first isolating the absolute value expression, if necessary. Then write an equivalent pair of linear equations or inequalities. PROBLEM 1: SOLVING AN ABSOLUTE VALUE EQUATION Solve each equation. Check your solutions. a) |"| + 2 = 9 b) |'| − 5 = 2 c) −3|+| = −9 d) |,| − 3 = 7 e) 5|.| = 20 f) |0|1 = 3

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    Chapter 3 102

    3-7 Absolute Value Equations and Inequalities

    Write T for true or F for false.

    1. To indicate the absolute value of 28, you write u28 u .

    2. The absolute value of 28 is 28, since 28 is 8 units to the left of 0 on the number line.

    3. The absolute value of 28 is 8, since 28 is 8 units away from 0 on the number line.

    4. According to the definition of absolute value, if u r u 5 3, then r 5 3 or r 5 23.

    Vocabulary Builder

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    Related Words: express (verb), phrase (noun)

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  • Some equations, such as |2" − 5| = 13, have variable expressions within absolute value symbols. The equation |2" − 5| = 13 means that the distance on a number line from 2" − 5 to 0 is 13 units. There are two points that are 13 units from 0: 13 and -13. So to find the values of x, solve 2" − 5 = −13 and 2" − 5 = 13. You can generalize this process as follows: KEY CONCEPT: SOLVING ABSOLUTE VALUE EQUATIONS To solve an equation in the form |4| = 5, where 4 represents a variable expression and 5 > 7, solve: 4 = −5 and 4 = 5 PROBLEM 2: SOLVING AN ABSOLUTE VALUE EQUATION Solve each equation. Check your solutions. a) |. − 8| = 5 b) −3|29| = −12 c) |4; + 1| − 2 = 5 d) |+ − 2| = 3 e) |3, − 2| − 6 = 2 f) 4|2= − 3| − 1 = 11 Recall that an absolute value represents distance from 0 on a number line. Distance is always nonnegative. So any equation that states that the absolute value of an expression is negative has no solution. PROBLEM 3: SOLVING AN ABSOLUTE VALUE EQUATION WITH NO SOLUTION Solve each equation. Check your solutions. If there is no solution, write no solution. a) 3|2" + 9| + 12 = 10 b) −4|>| = 12 c) |3" − 6| − 5 = −7 d) −6|2" − 3| = −30

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  • You can write absolute value inequalities as compound inequalities. Think about some numbers that would make each one of these inequalities true. |'| < 2 |'| > 3 KEY CONCEPT: SOLVING ABSOLUTE VALUE INEQUALITIES To solve an inequality in the form |4| < 5, where 4 is a variable expression and 5 > 7, solve the compound inequality −5 < 4 < 5. To solve an inequality in the form |4| > 5, where 4 is a variable expression and 5 > 7, solve the compound inequality 4 < −5AB4 > 5. Similar rules are true for ≤ DEF ≥. PROBLEM 4: SOLVING AN ABSOLUTE VALUE INEQUALITY INVOLVING ≥ Solve each inequality. Graph the solution. a) |8n| ≥ 24 b) |p − 7| > 3 c) |5+ − 9| ≥ 24 d) |2" + 4| ≥ 5 e) |= + 8| > 7 f) |3J − 7| > 28

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  • PROBLEM 5: SOLVING AN ABSOLUTE VALUE INEQUALITY INVOLVING ≤ Solve each inequality. Graph your solution. a). |" − 2| < 5 b) |2K − 5| ≤ 9 c) |2L − 1| ≤ 9 d) |" + 3| < 1 e) |2; + 9| ≤ 13 f) 2|" − 3| + 5 ≤ 15 PROBLEM 6: REAL-WORLD PROBLEM SOLVING a) Starting at 100 feet away, your friend skates toward you and then passes by you. She skates at a constant speed of 20 ft/s. Her distance d from you in feet after t seconds is given by J = |100 − 20,|. At what times is she 40 feet away from you? b) A company makes boxes of crackers that should weigh 213 g. A quality-control inspector randomly selcts boxes to weigh. Any box that varies from the weight by more than 5 g is sent back. What is the range of allowable weights for a box of crackers?

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