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Pearson Algebra 1 Common Core, 2011 View details

7. Absolute Value Equations and Inequalities

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Exercise 23 Page 211

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

v=0 or v=6

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

3|v-3|=9

DivEqn

.LHS /3.=.RHS /3.

|v-3|=3

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

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

| v-3|=3

lc v-3 ≥ 0:v-3 = 3 & (I) v-3 < 0:v-3 = - 3 & (II)

lcv-3=3 & (I) v-3=-3 & (II)

(I), (II): LHS+3=RHS+3

lv_1=6 v_2=0

Both v=6 and v=0 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 23 Page 211

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