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Pearson Algebra 2 Common Core, 2011 View details

6. Absolute Value Equations and Inequalities

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Exercise 48 Page 46

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

5/2

Practice makes perfect

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

7|8-3h|=21h-49

DivEqn

.LHS /7.=.RHS /7.

|8-3h|=21h-49/7

WriteDiffFrac

Write as a difference of fractions

|8-3h|=21h/7-49/7

CalcQuot

Calculate quotient

|8-3h|=3h-7

An absolute value measures an expression's distance from a midpoint on a number line.

|8-3h|= 3h-7 This equation means that the distance is 3h-7, either in the positive direction or the negative direction. |8-3h|= 3h-7 ⇒ l8-3h= 3h-7 8-3h= - (3h-7) To find the solutions to the absolute value equation, we need to solve both of these cases for h.

|8-3h|=3h-7

lc 8-3h ≥ 0:8-3h = (3h-7) & (I) 8-3h < 0:8-3h = - (3h-7) & (II)

lc8-3h=3h-7 & (I) 8-3h= -(3h-7) & (II)

Distr

(II): Distribute - 1

l8-3h=3h-7 8-3h= -3h+7

▼

Solve for h

(I), (II): LHS-8=RHS-8

l-3h=3h-15 -3h= -3h-1

SubEqn

(I): LHS-3h=RHS-3h

l- 6h=- 15 -3h= -3h-1

AddEqn

(II): LHS+3h=RHS+3h

l-6h= -15 0 ≠ -1

DivEqn

(I): .LHS /(- 6).=.RHS /(- 6).

lh= -15- 6 0 ≠ -1

DivNegNeg

(I): - a/- b=a/b

lh= 156 0 ≠ -1

ReduceFrac

(I): a/b=.a /3./.b /3.

lh= 52 0 ≠ -1

Solving the second equation resulted in a contradiction. Thus, there is only one possible solution to the equation. We will now check it.

Checking Our Answers

When solving an absolute value equation, it is important to check for extraneous solutions. We can check our answer by substituting it back into the original equation.

|8-3h|=3h-7

Substitute

h= 5/2

|8-3( 5/2)|? =3( 5/2)-7

MoveLeftFacToNum

a*b/c= a* b/c

|8-15/2|? =15/2-7

NumberToFrac

a = 2* a/2

|16/2-15/2|? =15/2-14/2

SubTerms

Subtract terms