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Solution - Absolute value equations

Exact form: x=32,1
x=\frac{3}{2} , 1
Mixed number form: x=112,1
x=1\frac{1}{2} , 1
Decimal form: x=1.5,1
x=1.5 , 1

Step-by-step explanation

1. Rewrite the equation with one absolute value terms on each side

|3x4||x+2|=0

Add |x+2| to both sides of the equation:

|3x4||x+2|+|x+2|=|x+2|

Simplify the arithmetic

|3x4|=|x+2|

2. Rewrite the equation without absolute value bars

Use the rules:
|x|=|y|x=±y and |x|=|y|±x=y
to write all four options of the equation
|3x4|=|x+2|
without the absolute value bars:

|x|=|y||3x4|=|x+2|
x=+y(3x4)=(x+2)
x=y(3x4)=((x+2))
+x=y(3x4)=(x+2)
x=y(3x4)=(x+2)

When simplified, equations x=+y and +x=y are the same and equations x=y and x=y are the same, so we end up with only 2 equations:

|x|=|y||3x4|=|x+2|
x=+y , +x=y(3x4)=(x+2)
x=y , x=y(3x4)=((x+2))

3. Solve the two equations for x

11 additional steps

(3x-4)=(-x+2)

Add to both sides:

(3x-4)+x=(-x+2)+x

Group like terms:

(3x+x)-4=(-x+2)+x

Simplify the arithmetic:

4x-4=(-x+2)+x

Group like terms:

4x-4=(-x+x)+2

Simplify the arithmetic:

4x4=2

Add to both sides:

(4x-4)+4=2+4

Simplify the arithmetic:

4x=2+4

Simplify the arithmetic:

4x=6

Divide both sides by :

(4x)4=64

Simplify the fraction:

x=64

Find the greatest common factor of the numerator and denominator:

x=(3·2)(2·2)

Factor out and cancel the greatest common factor:

x=32

11 additional steps

(3x-4)=-(-x+2)

Expand the parentheses:

(3x-4)=x-2

Subtract from both sides:

(3x-4)-x=(x-2)-x

Group like terms:

(3x-x)-4=(x-2)-x

Simplify the arithmetic:

2x-4=(x-2)-x

Group like terms:

2x-4=(x-x)-2

Simplify the arithmetic:

2x4=2

Add to both sides:

(2x-4)+4=-2+4

Simplify the arithmetic:

2x=2+4

Simplify the arithmetic:

2x=2

Divide both sides by :

(2x)2=22

Simplify the fraction:

x=22

Simplify the fraction:

x=1

4. List the solutions

x=32,1
(2 solution(s))

5. Graph

Each line represents the function of one side of the equation:
y=|3x4|
y=|x+2|
The equation is true where the two lines cross.

Why learn this

We encounter absolute values almost daily. For example: If you walk 3 miles to school, do you also walk minus 3 miles when you go back home? The answer is no because distances use absolute value. The absolute value of the distance between home and school is 3 miles, there or back.
In short, absolute values help us deal with concepts like distance, ranges of possible values, and deviation from a set value.