Exercise 17 Page 46

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. |y-5|= 12This equation means that the distance is 12, either in the positive direction or the negative direction. |y-5|= 12 ⇒ ly-5= 12 y-5= - 12 To find the solutions to the absolute value equation, we need to solve both of these cases for y.

|y-5|=12

lc y-5 ≥ 0:y-5 = 12 & (I) y-5 < 0:y-5 = - 12 & (II)

lcy-5=12 & (I) y-5= -12 & (II)

AddEqn

LHS+5=RHS+5

ly=17 y= -7

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 17.

|y-5|=12

Substitute

y= 17

| 17-5| ? =12

SubTerm

Subtract term

|12| ? =12

AbsPos

|12|=12

12=12

We will check -7 in the same way.

|y-5|=12

Substitute

y= -7

| - 7-5| ? =12

SubTerm

Subtract term

|-12| ? =12

AbsNeg

|-12|=12

12=12

We see that both 17 and -7 satisfy the original equation.