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Solution - Absolute value equations

Exact form: x=34,12
x=\frac{3}{4} , \frac{1}{2}
Decimal form: x=0.75,0.5
x=0.75 , 0.5

Other Ways to Solve

Absolute value equations

Step-by-step explanation

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

|x1|+|3x2|=0

Add |3x2| to both sides of the equation:

|x1|+|3x2||3x2|=|3x2|

Simplify the arithmetic

|x1|=|3x2|

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
|x1|=|3x2|
without the absolute value bars:

|x|=|y||x1|=|3x2|
x=+y(x1)=(3x2)
x=y(x1)=(3x2)
+x=y(x1)=(3x2)
x=y(x1)=(3x2)

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||x1|=|3x2|
x=+y , +x=y(x1)=(3x2)
x=y , x=y(x1)=(3x2)

3. Solve the two equations for x

10 additional steps

(x-1)=-(3x-2)

Expand the parentheses:

(x-1)=-3x+2

Add to both sides:

(x-1)+3x=(-3x+2)+3x

Group like terms:

(x+3x)-1=(-3x+2)+3x

Simplify the arithmetic:

4x-1=(-3x+2)+3x

Group like terms:

4x-1=(-3x+3x)+2

Simplify the arithmetic:

4x1=2

Add to both sides:

(4x-1)+1=2+1

Simplify the arithmetic:

4x=2+1

Simplify the arithmetic:

4x=3

Divide both sides by :

(4x)4=34

Simplify the fraction:

x=34

12 additional steps

(x-1)=-(-(3x-2))

NT_MSLUS_MAINSTEP_RESOLVE_DOUBLE_MINUS:

(x-1)=3x-2

Subtract from both sides:

(x-1)-3x=(3x-2)-3x

Group like terms:

(x-3x)-1=(3x-2)-3x

Simplify the arithmetic:

-2x-1=(3x-2)-3x

Group like terms:

-2x-1=(3x-3x)-2

Simplify the arithmetic:

2x1=2

Add to both sides:

(-2x-1)+1=-2+1

Simplify the arithmetic:

2x=2+1

Simplify the arithmetic:

2x=1

Divide both sides by :

(-2x)-2=-1-2

Cancel out the negatives:

2x2=-1-2

Simplify the fraction:

x=-1-2

Cancel out the negatives:

x=12

4. List the solutions

x=34,12
(2 solution(s))

5. Graph

Each line represents the function of one side of the equation:
y=|x1|
y=|3x2|
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.