Pearson Algebra 1 Common Core, 2011
PA
Pearson Algebra 1 Common Core, 2011 View details
8. Unions and Intersections of Sets
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Exercise 52 Page 220

Isolate the absolute value on the left-hand side.

y=4/3 or y=-8/3

Practice makes perfect

Before we can solve the given equation, we need to isolate the absolute value expression using the Properties of Equality. 3|3y+2|=18 ⇒ |3y+2|=6 An absolute value measures an expression's distance from a midpoint on a number line. |3y+2|=6 In this case, since 3y+2 can be written as 3y-(-2), it means that the distance between 3y and -2 is 6, either in the positive direction or the negative direction. |3y+2|=6 ⇒ l3y+2= 6 3y+2= - 6 To find the solutions to the absolute value equation, we need to solve both of these cases for y.

| 3y+2|=6

lc 3y+2 ≥ 0:3y+2 = 6 & (I) 3y+2 < 0:3y+2 = - 6 & (II)

lc3y+2=6 & (I) 3y+2=- 6 & (II)

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

l3y=4 3y=- 8

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

ly_1= 43 y_2=- 83