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Pearson Algebra 2 Common Core, 2011 View details

6. Absolute Value Equations and Inequalities

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Exercise 3 Page 45

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

-6/5

Practice makes perfect

An absolute value measures an expression's distance from a midpoint on a number line. |x-2|= 4x+8 This equation means that the distance is 4x+8, either in the positive direction or the negative direction. |x-2|= 4x+8 ⇒ lx-2= 4x+8 x-2= - (4x+8) To find the solutions to the absolute value equation, we need to solve both of these cases for x.

|x-2|=4x+8

lc x-2 ≥ 0:x-2 = 4x+8 & (I) x-2 < 0:x-2 = - 4x+8 & (II)

lcx-2=4x+8 & (I) x-2= -(4x+8) & (II)

Distr

(II): Distribute -1

lx-2=4x+8 x-2= -4x-8

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

lx=4x+10 x= -4x-6

SubEqn

(I): LHS-4x=RHS-4x

l- 3x=10 x=- 4x-6

AddEqn

(II): LHS+4x=RHS+4x

l-3x=10 5x= -6

DivEqn

(I): .LHS /(-3).=.RHS /(-3).

lx=- 103 5x=- 6

DivEqn

(II): .LHS /5.=.RHS /5.

lx= - 103 x= - 65

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 - 103.

|x-2|=4x+8

Substitute

x= -10/3

| -10/3-2|? =4( -10/3)+8

▼

Simplify left-hand side

NumberToFrac

a = 3* a/3

|-10/3-6/3|? =4(-10/3)+8

SubFrac

Subtract fractions

|-16/3|? =4(-10/3)+8