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Pearson Algebra 2 Common Core, 2011 View details

6. Absolute Value Equations and Inequalities

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Exercise 51 Page 46

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

-71/36

Practice makes perfect

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

1/4|4x+7|=8x+16

MultEqn

LHS * 4=RHS* 4

|4x+7|=4(8x+16)

Distr

Distribute 4

|4x+7|=32x+64

An absolute value measures an expression's distance from a midpoint on a number line. |4x+7|= 32x+64

This equation means that the distance is 32x+64, either in the positive direction or the negative direction. |4x+7|= 32x+64 ⇓ l4x+7= 32x+64 4x+7= - (32x+64) To find the solutions to the absolute value equation, we need to solve both of these cases for x.

|4x+7|=32x+64

lc 4x+7 ≥ 0:4x+7 = (32x+64) & (I) 4x+7 < 0:4x+7 = - (32x+64) & (II)

lc4x+7=32x+64 & (I) 4x+7= -(32x+64) & (II)

Distr

(II): Distribute - 1

l4x+7=32x+64 4x+7= -32x-64

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

l4x=32x+57 4x= -32x-71

SubEqn

(I): LHS-32x=RHS-32x

l- 28x=57 4x= -32x-71

AddEqn

(II): LHS+32x=RHS+32x

l-28x=57 36x= -71

DivEqn

(I): .LHS /(-28).=.RHS /(-28).

lx=- 5728 36x= -71

DivEqn

(II): .LHS /36.=.RHS /36.

lx = - 5728 x=- 7136

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 - 5728.

|4x+7|=32x+64

Substitute

x= -57/28

|4( -57/28)+7|? =32( -57/28)+64

▼

Simplify

MultPosNeg

a(- b)=- a * b

|- 4*57/28+7|? =- 32*57/28+64

MoveLeftFacToNum

a*b/c= a* b/c

|- 4* 57/28+7|? =- 32* 57/28+64