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Solution - Absolute value equations

Exact form: x=103,2
x=\frac{10}{3} , 2
Mixed number form: x=313,2
x=3\frac{1}{3} , 2
Decimal form: x=3.333,2
x=3.333 , 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
|2x6|=|x+4|
without the absolute value bars:

|x|=|y||2x6|=|x+4|
x=+y(2x6)=(x+4)
x=y(2x6)=(x+4)
+x=y(2x6)=(x+4)
x=y(2x6)=(x+4)

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||2x6|=|x+4|
x=+y , +x=y(2x6)=(x+4)
x=y , x=y(2x6)=(x+4)

2. Solve the two equations for x

9 additional steps

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

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

3x-6=(-x+4)+x

Group like terms:

3x-6=(-x+x)+4

Simplify the arithmetic:

3x6=4

Add to both sides:

(3x-6)+6=4+6

Simplify the arithmetic:

3x=4+6

Simplify the arithmetic:

3x=10

Divide both sides by :

(3x)3=103

Simplify the fraction:

x=103

8 additional steps

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

Expand the parentheses:

(2x-6)=x-4

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

x-6=(x-4)-x

Group like terms:

x-6=(x-x)-4

Simplify the arithmetic:

x6=4

Add to both sides:

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

Simplify the arithmetic:

x=4+6

Simplify the arithmetic:

x=2

3. List the solutions

x=103,2
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|2x6|
y=|x+4|
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.