Exercise 5 Page 29

How many cases do we have after we remove the absolute value?

x=- 1.5 and x=- 5.5

Practice makes perfect

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

4|2x+7|=16

DivEqn

.LHS /4.=.RHS /4.

|2x+7|=4

Solving for x

An absolute value measures an expression's distance from a midpoint on a number line. |2x+7|= 4This equation means that the distance is 4, either in the positive direction or the negative direction. |2x+7|= 4 ⇒ l2x+7= 4 2x+7= - 4 To find the solutions to the absolute value equation, we need to solve both of these cases for x.

| 2x+7|=4

lc 2x+7 ≥ 0:2x+7 = 4 & (I) 2x+7 < 0:2x+7 = - 4 & (II)

lc2x+7=4 & (I) 2x+7=- 4 & (II)

(I), (II): LHS-7=RHS-7

l2x=- 3 2x=- 11

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

lx_1=- 32 x_2=- 112

(I), (II): Write as a decimal

lx_1=-1.5 x_2=-5.5

Both -1.5 and -5.5 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= - 32.

4|2x+7|=16

Substitute

x= -1.5

4|2( -1.5)+7|? =16

MultPosNeg

a(- b)=- a * b

4|- 3+7|? =16

AddTerms

Add terms

4|4|? =16

AbsPos

|4|=4