Enter an equation or problem
Camera input is not recognized!

Solution - Absolute value equations

Exact form: x=12,12
x=\frac{1}{2} , \frac{1}{2}
Decimal form: x=0.5,0.5
x=0.5 , 0.5

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
|2x1|=|4x2|
without the absolute value bars:

|x|=|y||2x1|=|4x2|
x=+y(2x1)=(4x2)
x=y(2x1)=(4x2)
+x=y(2x1)=(4x2)
x=y(2x1)=(4x2)

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|=|4x2|
x=+y , +x=y(2x1)=(4x2)
x=y , x=y(2x1)=(4x2)

2. Solve the two equations for x

11 additional steps

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

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

12 additional steps

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

Expand the parentheses:

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

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

6x-1=(-4x+2)+4x

Group like terms:

6x-1=(-4x+4x)+2

Simplify the arithmetic:

6x1=2

Add to both sides:

(6x-1)+1=2+1

Simplify the arithmetic:

6x=2+1

Simplify the arithmetic:

6x=3

Divide both sides by :

(6x)6=36

Simplify the fraction:

x=36

Find the greatest common factor of the numerator and denominator:

x=(1·3)(2·3)

Factor out and cancel the greatest common factor:

x=12

3. List the solutions

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

4. Graph

Each line represents the function of one side of the equation:
y=|2x1|
y=|4x2|
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.