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Big Ideas Math Integrated I, 2016
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Big Ideas Math Integrated I, 2016 View details
4. Solving Absolute Value Equations
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Exercise 23 Page 32

How many cases do we have after we remove the absolute value?

Solutions: s=20
Number Line:

Practice makes perfect

Before we can solve this equation, we need to isolate the absolute value expression using the Properties of Equality.

3=- 2|1/4s-5|+3
SubEqn

LHS-3=RHS-3

0=- 2|1/4s-5|
DivEqn

.LHS /(- 2).=.RHS /(- 2).

0=|1/4s-5|
RearrangeEqn

Rearrange equation

|1/4s-5|=0

An absolute value measures an expression's distance from a midpoint on a number line.

|1/4s-5|= 0 This equation means that the distance between 14s and 5 is 0. Therefore the expression inside the absolute value needs to equal 0. |1/4s-5|= 0 ⇕ 1/4s-5=0 Let's solve this equation for s.

1/4s-5=0
AddEqn

LHS+5=RHS+5

1/4s=5
MultEqn

LHS * 4=RHS* 4

s=20

The solution to absolute value equation is 20. Let's graph this solution on a number line.

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arrow_right Communicate Your Answer p.27
4
Communicate Your Answer 5 (Page 27)
arrow_right Monitoring Progress p.28-31
Monitoring Progress 1 (Page 28)
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arrow_right Exercises p.32-34
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