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McGraw Hill Glencoe Algebra 1, 2012

MH

McGraw Hill Glencoe Algebra 1, 2012 View details

5. Solving Equations Involving Absolute Value

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Exercise 42 Page 106

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

Solutions: ${- 2, 2}$
Number Line:

Practice makes perfect

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

\frac{4}{∣ p ∣} + 12 = 14

$LHS - 12 = RHS - 12$

\frac{4}{∣ p ∣} + 12 - 12 = 14 - 12

Subtract terms

\frac{4}{∣ p ∣} = 2

$LHS \cdot ∣ p ∣ = RHS \cdot ∣ p ∣$

\frac{4}{∣ p ∣} \cdot ∣ p ∣ = 2 \cdot ∣ p ∣

FracMultDenomToNumber

$\frac{a}{∣ p ∣} \cdot ∣ p ∣ = a$

4 = 2 \cdot ∣ p ∣

$LHS / 2 = RHS / 2$

2 = ∣ p ∣

Rearrange equation

∣ p ∣ = 2

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

Subchapter links

Check Your Understanding p.105

Practice and Problem Solving p.106-108

H.O.T. Problems p.108

Standardized Test Practice p.109

Spiral Review p.109

Skills Review p.109