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Solution - Absolute value equations

Exact form: x=2,0
x=-2 , 0

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
|x1|=|2x+1|
without the absolute value bars:

|x|=|y||x1|=|2x+1|
x=+y(x1)=(2x+1)
x=y(x1)=(2x+1)
+x=y(x1)=(2x+1)
x=y(x1)=(2x+1)

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|=|2x+1|
x=+y , +x=y(x1)=(2x+1)
x=y , x=y(x1)=(2x+1)

2. Solve the two equations for x

10 additional steps

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

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

x1=1

Add to both sides:

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

Simplify the arithmetic:

x=1+1

Simplify the arithmetic:

x=2

Multiply both sides by :

-x·-1=2·-1

Remove the one(s):

x=2·-1

Simplify the arithmetic:

x=2

9 additional steps

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

Expand the parentheses:

(x-1)=-2x-1

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

3x1=1

Add to both sides:

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

Simplify the arithmetic:

3x=1+1

Simplify the arithmetic:

3x=0

Divide both sides by the coefficient:

x=0

3. List the solutions

x=2,0
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|x1|
y=|2x+1|
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.