Exercise 52 Page 42

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

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

|3y-12|= 9 This equation means that the distance is 9, either in the positive direction or the negative direction. |3y-12|= 9 ⇒ l3y-12= 9 3y-12= - 9 To find the solutions to the absolute value equation, we need to solve both of these cases for y.

| 3y-12|=9

lc 3y-12 ≥ 0:3y-12 = 9 & (I) 3y-12 < 0:3y-12 = - 9 & (II)

lc3y-12=9 & (I) 3y-12=-9 & (II)

(I), (II): LHS+12=RHS+12

l3y=21 3y=3

(I), (II): .LHS /3.=.RHS /3.

ly=7 y=1

Both y=7 and y=1 are solutions to the absolute value equation. Let's graph these solutions on a number line.