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Solution - Absolute value equations

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

Other Ways to Solve

Absolute value equations

Step-by-step explanation

1. 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
|x5|=|3x1|
without the absolute value bars:

|x|=|y||x5|=|3x1|
x=+y(x5)=(3x1)
x=y(x5)=(3x1)
+x=y(x5)=(3x1)
x=y(x5)=(3x1)

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||x5|=|3x1|
x=+y , +x=y(x5)=(3x1)
x=y , x=y(x5)=(3x1)

2. Solve the two equations for x

13 additional steps

(x-5)=(3x-1)

Subtract from both sides:

(x-5)-3x=(3x-1)-3x

Group like terms:

(x-3x)-5=(3x-1)-3x

Simplify the arithmetic:

-2x-5=(3x-1)-3x

Group like terms:

-2x-5=(3x-3x)-1

Simplify the arithmetic:

2x5=1

Add to both sides:

(-2x-5)+5=-1+5

Simplify the arithmetic:

2x=1+5

Simplify the arithmetic:

2x=4

Divide both sides by :

(-2x)-2=4-2

Cancel out the negatives:

2x2=4-2

Simplify the fraction:

x=4-2

Move the negative sign from the denominator to the numerator:

x=-42

Find the greatest common factor of the numerator and denominator:

x=(-2·2)(1·2)

Factor out and cancel the greatest common factor:

x=2

12 additional steps

(x-5)=-(3x-1)

Expand the parentheses:

(x-5)=-3x+1

Add to both sides:

(x-5)+3x=(-3x+1)+3x

Group like terms:

(x+3x)-5=(-3x+1)+3x

Simplify the arithmetic:

4x-5=(-3x+1)+3x

Group like terms:

4x-5=(-3x+3x)+1

Simplify the arithmetic:

4x5=1

Add to both sides:

(4x-5)+5=1+5

Simplify the arithmetic:

4x=1+5

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

3. List the solutions

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

4. Graph

Each line represents the function of one side of the equation:
y=|x5|
y=|3x1|
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.