Enter an equation or problem
Camera input is not recognized!

Solution - Absolute value equations

Exact form: x=3,13
x=3 , \frac{1}{3}
Decimal form: x=3,0.333
x=3 , 0.333

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

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

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

2. Solve the two equations for x

12 additional steps

(x+1)=2·(x-1)

Expand the parentheses:

(x+1)=2x+2·-1

Simplify the arithmetic:

(x+1)=2x-2

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

x+1=2

Subtract from both sides:

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

Simplify the arithmetic:

x=21

Simplify the arithmetic:

x=3

Multiply both sides by :

-x·-1=-3·-1

Remove the one(s):

x=-3·-1

Simplify the arithmetic:

x=3

14 additional steps

(x+1)=2·(-(x-1))

Expand the parentheses:

(x+1)=2·(-x+1)

(x+1)=2·-x+2·1

Group like terms:

(x+1)=(2·-1)x+2·1

Multiply the coefficients:

(x+1)=-2x+2·1

Simplify the arithmetic:

(x+1)=-2x+2

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

3x+1=2

Subtract from both sides:

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

Simplify the arithmetic:

3x=21

Simplify the arithmetic:

3x=1

Divide both sides by :

(3x)3=13

Simplify the fraction:

x=13

3. List the solutions

x=3,13
(2 solution(s))

4. Graph

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