Exercise 27 Page 211

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

y=0 or y=3

Practice makes perfect

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

4|2y-3|-1=11

AddEqn

LHS+1=RHS+1

4|2y-3|-1+1=11+1

AddTerms

Add terms

4|2y-3|=12

DivEqn

.LHS /4.=.RHS /4.

|2y-3|=3

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

|2y-3|= 3 This equation means that the distance is 3, either in the positive direction or the negative direction. |2y-3|= 3 ⇒ l2y-3= 3 2y-3= -3 To find the solutions to the absolute value equation, we need to solve both of these cases for y.

| 2y-3|=3

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

lc2y-3=3 & (I) 2y-3=-3 & (II)

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

lc2y=6 & (I) 2y=0 & (II)

(I), (II): .LHS /2.=.RHS /2.

ly_1=3 y_2=0

Both y=3 and y=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 27 Page 211

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