of 5/5
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

• View
1

0

Embed Size (px)

### Text of Absolute Value Equations and Inequalities...3-7 Absolute Value Equations and Inequalities Write T...

• 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

Cop

yrig

by

Pear

son

Educ

atio

n, In

c. o

r its

aff

iliat

es. A

ll Ri

ghts

Res

erve

d.

Vocabulary

Review

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

expression (noun) ek SPRESH un

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

Main Idea: An expression is a word or phrase that communicates an idea. A mathematical expression is a mathematical phrase. A mathematical expression may be numerical or algebraic.

Use Your VocabularyWrite an expression for each word phrase.

5. m increased by 8 6. y divided by 9 7. u more than 7

m y u

8. Cross out the expression that is NOT algebraic.

3y 2 12 4 1 18 2 3 12 1 x

9. Cross out the expression that is NOT numeric.

3 2 12 4 1 18q 2 3 12 1 5

numerical expression18 (6 3)

algebraic expression4 k 7

HSM11A1MC_0307.indd 102HSM11A1MC_0307.indd 102 2/14/09 12:52:19 PM2/14/09 12:52:19 PM

absolutevaluee thedistanceanumberisfromzeroonthenumberlinealwayspositive

121 2151 5

Ei Ei E

Tts12 tests 3

Int 7 Im1 3x or

n n m rm

3 3 g f C 714 10 1,1 4 Igt

21

t 0ortr yorr Y QoF

• 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

X Tz 3 2 2r 8 5 or r 8 5 law1 4 14111 7sets 8 8 4ft or 4ft1 7r orr B z or 2w 4I a a a f EEEw zorw f 2orf

6 to 41 1 1 ti tim 2 3 orMI 72 1 21 8 t t 42231122 2

or m s 3 2 755 2 8 44 12 4 42 2 4 4 422311o A 3 243 3 2233

ZEE 3762y oory

T

is 12 Ty I3 2 91 23 3 Incl 31291 23

NosolutionLNosolutionC

S TS 7 To13 61 2 12 31 5

Nosolution 2 3 13 or 2 1332x 822 22 a 2X iorx

• 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

2 N 2 n 3 or n3

go.to gjyE E7ggT

IE 234 or8Th p 7 3 or p 7 3 Sm 9 24 orSm9324ne 3 vn tfcf.pt

t ta ta ta 9

Esme Eg2333FTd t mE 3ormz6

Fi6 O 6

2 4 E S or 2 425 yt8c 7 or Lts 3d 72 28 or 3d 772y 4 Y Y g g 8 8 t 7 7 t221 21 22212 3dL 21 3d 35xe 4 or zt Y borY z g z g

d or d 11FREE Tiz4 2 O iz y td8 O 8

• 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?

II E Issa If I 9E2 92 2 ezag e Is 21 ELI2EcE MERESTei

it b 13121773 z fe222 E E

1 31 5

1017 221824 O E FT

1100 snot1 40

er

Actual target allowabledifference 1W 2131E 5

5 EW 21325

hetwactualweighqqyqqgIosewtI3

s

• o

Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Education
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Education
Documents
Documents
Documents