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Solution - Absolute value equations

Exact form: x=143,-4
x=\frac{14}{3} , -4
Mixed number form: x=423,-4
x=4\frac{2}{3} , -4
Decimal form: x=4.667,4
x=4.667 , -4

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
|2x5|=|x+9|
without the absolute value bars:

|x|=|y||2x5|=|x+9|
x=+y(2x5)=(x+9)
x=y(2x5)=(x+9)
+x=y(2x5)=(x+9)
x=y(2x5)=(x+9)

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||2x5|=|x+9|
x=+y , +x=y(2x5)=(x+9)
x=y , x=y(2x5)=(x+9)

2. Solve the two equations for x

9 additional steps

(2x-5)=(-x+9)

Add to both sides:

(2x-5)+x=(-x+9)+x

Group like terms:

(2x+x)-5=(-x+9)+x

Simplify the arithmetic:

3x-5=(-x+9)+x

Group like terms:

3x-5=(-x+x)+9

Simplify the arithmetic:

3x5=9

Add to both sides:

(3x-5)+5=9+5

Simplify the arithmetic:

3x=9+5

Simplify the arithmetic:

3x=14

Divide both sides by :

(3x)3=143

Simplify the fraction:

x=143

8 additional steps

(2x-5)=-(-x+9)

Expand the parentheses:

(2x-5)=x-9

Subtract from both sides:

(2x-5)-x=(x-9)-x

Group like terms:

(2x-x)-5=(x-9)-x

Simplify the arithmetic:

x-5=(x-9)-x

Group like terms:

x-5=(x-x)-9

Simplify the arithmetic:

x5=9

Add to both sides:

(x-5)+5=-9+5

Simplify the arithmetic:

x=9+5

Simplify the arithmetic:

x=4

3. List the solutions

x=143,-4
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|2x5|
y=|x+9|
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.