Exercise 69 Page 258

An absolute value measures an expression's distance from a midpoint on a number line. |3x+2|= 5 This equation means that the distance is 5, either in the positive or the negative direction. |3x+2|= 5 ⇒ l3x+2= 5 3x+2= - 5 To find the solutions to the absolute value equation, we need to solve both of these cases for x.

| 3x+2|=5

lc 3x+2 ≥ 0:3x+2 = 5 & (I) 3x+2 < 0:3x+2 = - 5 & (II)

lc3x+2=5 & (I) 3x+2=- 5 & (II)

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

l3x=3 3x=- 7

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

lx_1=1 x_2=- 73

Both 1 and - 73 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 x_1 = 1.

| 3x + 2 | = 5

Substitute

x= 1

| 3 ( 1) + 2 | = 5

Multiply

Multiply

| 3 + 2 | = 5

AddTerms

Add terms

| 5 | = 5

AbsPos

|5|=5

5 = 5

We will check x_2 = - 73 in the same way.

| 3x +2 | = 5

Substitute

x= - 7/3

| 3 ( - 7/3 ) + 2 | = 5

MoveNegFracToNum

Put minus sign in numerator

| 3 ( - 7/3 ) + 2 | = 5

DenomMultFracToNumber

3 * a/3= a

| - 7 + 2 | = 5

AddTerms

Add terms

| - 5 | = 5

AbsNeg

|-5|=5

5 = 5

We see that x = 1 and x = - 73 satisfy the original equation.