Exercise 50 Page 211

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

p=-3/5 or p=3/5

Practice makes perfect

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

1.2|5p|=3.6

DivEqn

.LHS /1.2.=.RHS /1.2.

1.2|5p|/1.2=3.6/1.2

CalcQuot

Calculate quotient

|5p|=3

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

|5p|= 3 This equation means that the distance is 3, either in the positive direction or the negative direction. |5p|= 3 ⇒ l5p= 3 5p= -3 To find the solutions to the absolute value equation, we need to solve both of these cases for p.

| 5p|=3

lc 5p ≥ 0:5p = 3 & (I) 5p < 0:5p = - 3 & (II)

lc5p=3 & (I) 5p=-3 & (II)

(I), (II): .LHS /5.=.RHS /5.

lp_1=3/5 p_2=-3/5

Both 35 and - 35 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 50 Page 211

Hints

Check the answer

Practice exercises

Assess your skill