Exercise 28 Page 211

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

x=1 or x=-5

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

3|x+2|+4=13

SubEqn

LHS-4=RHS-4

3|x+2|+4-4=13-4

SubTerms

Subtract terms

3|x+2|=9

DivEqn

.LHS /3.=.RHS /3.

|x+2|=3

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

|x+2|= 3 This equation means that the distance is 3, either in the positive direction or the negative direction. |x+2|= 3 ⇒ lx+2= 3 x+2= -3 To find the solutions to the absolute value equation, we need to solve both of these cases for x.

| x+2|=3

lc x+2 ≥ 0:x+2 = 3 & (I) x+2 < 0:x+2 = - 3 & (II)

lcx+2=3 & (I) x+2=-3 & (II)

(I), (II): LHS-2=RHS-2

lx_1=1 x_2=-5

Both x=1 and x=-5 are solutions to the absolute value equation.

Subchapter links

Lesson Check p.210

Practice and Problem-Solving Exercises p.211-213

Standardized Test Prep p.213

Mixed Review p.213

Exercise 28 Page 211

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