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Solution - Absolute value equations

Exact form: x=12
x=\frac{1}{2}
Decimal form: x=0.5
x=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
|2x+1|=|2x3|
without the absolute value bars:

|x|=|y||2x+1|=|2x3|
x=+y(2x+1)=(2x3)
x=y(2x+1)=(2x3)
+x=y(2x+1)=(2x3)
x=y(2x+1)=(2x3)

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||2x+1|=|2x3|
x=+y , +x=y(2x+1)=(2x3)
x=y , x=y(2x+1)=(2x3)

2. Solve the two equations for x

5 additional steps

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

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

1=(2x-3)-2x

Group like terms:

1=(2x-2x)-3

Simplify the arithmetic:

1=3

The statement is false:

1=3

The equation is false so it has no solution.

12 additional steps

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

Expand the parentheses:

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

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

4x+1=3

Subtract from both sides:

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

Simplify the arithmetic:

4x=31

Simplify the arithmetic:

4x=2

Divide both sides by :

(4x)4=24

Simplify the fraction:

x=24

Find the greatest common factor of the numerator and denominator:

x=(1·2)(2·2)

Factor out and cancel the greatest common factor:

x=12

3. Graph

Each line represents the function of one side of the equation:
y=|2x+1|
y=|2x3|
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.