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Solution - Absolute value equations

Exact form: x=-1,15
x=-1 , \frac{1}{5}
Decimal form: x=1,0.2
x=-1 , 0.2

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
|2x1|=3|x|
without the absolute value bars:

|x|=|y||2x1|=3|x|
x=+y(2x1)=3(x)
x=y(2x1)=3((x))
+x=y(2x1)=3(x)
x=y(2x1)=3(x)

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||2x1|=3|x|
x=+y , +x=y(2x1)=3(x)
x=y , x=y(2x1)=3((x))

2. Solve the two equations for x

9 additional steps

(2x-1)=3x

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Simplify the arithmetic:

x1=0

Add to both sides:

(-x-1)+1=0+1

Simplify the arithmetic:

x=0+1

Simplify the arithmetic:

x=1

Multiply both sides by :

-x·-1=1·-1

Remove the one(s):

x=1·-1

Remove the one(s):

x=1

10 additional steps

(2x-1)=3·-x

Group like terms:

(2x-1)=(3·-1)x

Multiply the coefficients:

(2x-1)=-3x

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Simplify the arithmetic:

5x1=0

Add to both sides:

(5x-1)+1=0+1

Simplify the arithmetic:

5x=0+1

Simplify the arithmetic:

5x=1

Divide both sides by :

(5x)5=15

Simplify the fraction:

x=15

3. List the solutions

x=-1,15
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|2x1|
y=3|x|
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.