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Big Ideas Math Integrated I, 2016

BI

Big Ideas Math Integrated I, 2016 View details

5. Rewriting Equations and Formulas

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Exercise 54 Page 42

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

Solutions: s=-3, s=- 15
Number Line:

Practice makes perfect

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

-4|s+9|=-24

.LHS /-4.=.RHS /-4.

-4|s+9|/-4=-24/-4

Calculate quotient

|s+9|=6

An absolute value measures an expression's distance from a midpoint on a number line. |s+9|= 6 This equation means that the distance is 6, either in the positive direction or the negative direction. |s+9|= 6 ⇒ ls+9= 6 s+9= - 6 To find the solutions to the absolute value equation, we need to solve both of these cases for s.

| s+9|=6

lc s+9 ≥ 0:s+9 = 6 & (I) s+9 < 0:s+9 = - 6 & (II)

lcs+9=6 & (I) s+9=-6 & (II)

(I), (II): LHS-9=RHS-9

ls_1=-3 s_2=- 15

Both s=-3 and s=-15 are solutions to the given equation. We can show our solutions on a number line.

Subchapter links

Communicate Your Answer p.35

Monitoring Progress p.36-39

Exercises p.40-42