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Big Ideas Math Integrated I, 2016
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Big Ideas Math Integrated I, 2016 View details
4. Solving Absolute Value Equations
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Exercise 22 Page 32

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

Solutions: v=- 3 and v=6
Number Line:

Practice makes perfect

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

- 3|1-2/3v|=- 9
DivEqn

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

|1-2/3v|=3

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

|1-2/3v|= 3 This equation means that the distance from 1 is 3, either in the positive direction or the negative direction. 1-2/3v= 3 and 1-2/3v= - 3 To find the solutions to the absolute value equation, we need to solve both of these cases for v. Let's start with the positive case.

1-2/3v=3
SubEqn

LHS-1=RHS-1

-2/3v=2
MultEqn

LHS * (-3/2)=RHS* (-3/2)

-2/3v(-3/2)=2(-3/2)
MultFracByInverse

a/b* b/a=1

v=2(-3/2)
MoveNegFracToNum

Put minus sign in numerator

v=2(-3/2)
DenomMultFracToNumber

2 * a/2= a

v=- 3

Now let's solve the negative case.

1-2/3v=- 3
SubEqn

LHS-1=RHS-1

-2/3v=- 4
MultEqn

LHS * (-3/2)=RHS* (-3/2)

-2/3v(-3/2)=-4(-3/2)
MultFracByInverse

a/b* b/a=1

v=-4(-3/2)
MoveNegFracToNum

Put minus sign in numerator

v=-4(-3/2)
MoveLeftFacToNum

a*b/c= a* b/c

v=- 4 (-3)/2
MultNegNegOnePar

- a(- b)=a* b

v=12/2
CalcQuot

Calculate quotient

v=6

Both - 3 and 6 are solutions to the absolute value equation. Let's graph these solutions on a number line.

Subchapter links expand_more
arrow_right Communicate Your Answer p.27
4
Communicate Your Answer 5 (Page 27)
arrow_right Monitoring Progress p.28-31
Monitoring Progress 1 (Page 28)
Monitoring Progress 2 (Page 28)
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arrow_right Exercises p.32-34
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