Exercise 54 Page 181

An absolute value measures an expression's distance from a midpoint on a number line. |2y-3|= 12 This equation means that the distance is 12, either in the positive direction or the negative direction. |2y-3|= 12 ⇒ l2y-3= 12 2y-3= -12 To find the solutions to the absolute value equation, we need to solve both of these cases for y.

| 2y-3|=12

lc 2y-3 ≥ 0:2y-3 = 12 & (I) 2y-3 < 0:2y-3 = - 12 & (II)

lc2y-3=12 & (I) 2y-3=- 12 & (II)

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

lc2y=15 & (I) 2y=-9 & (II)

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

ly_1= 152 y_2=- 92

Both 152 and - 92 are solutions to the absolute value equation.

Checking Our Answers

When solving an absolute value equation, it is important to check for extraneous solutions.We can check our answers by substituting them back into the original equation. Let's start with 152.

|2y-3|=12

Substitute

y= 15/2

|2( 15/2)-3|? =12

▼

Simplify left-hand side

DenomMultFracToNumber

2 * a/2= a

|15-3|? =12

SubTerm

Subtract term

|12|? =12

AbsPos

|12|=12

12=12 ✓

We will check - 92 in the same way.

|2y-3|=12

Substitute

y= -9/2

|2( -9/2)-3|? =12

▼

Simplify left-hand side

DenomMultFracToNumber

2 * a/2= a

|-9-3|? =12

SubTerm

Subtract term

|-12|? =12

AbsNeg

|-12|=12

12=12 ✓

We see that both solutions satisfy the original equation.