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Solution - Absolute value equations

Exact form: x=7,1
x=7 , -1

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
|6x+2|=|5x+9|
without the absolute value bars:

|x|=|y||6x+2|=|5x+9|
x=+y(6x+2)=(5x+9)
x=y(6x+2)=(5x+9)
+x=y(6x+2)=(5x+9)
x=y(6x+2)=(5x+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||6x+2|=|5x+9|
x=+y , +x=y(6x+2)=(5x+9)
x=y , x=y(6x+2)=(5x+9)

2. Solve the two equations for x

7 additional steps

(6x+2)=(5x+9)

Subtract from both sides:

(6x+2)-5x=(5x+9)-5x

Group like terms:

(6x-5x)+2=(5x+9)-5x

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

x+2=9

Subtract from both sides:

(x+2)-2=9-2

Simplify the arithmetic:

x=92

Simplify the arithmetic:

x=7

11 additional steps

(6x+2)=-(5x+9)

Expand the parentheses:

(6x+2)=-5x-9

Add to both sides:

(6x+2)+5x=(-5x-9)+5x

Group like terms:

(6x+5x)+2=(-5x-9)+5x

Simplify the arithmetic:

11x+2=(-5x-9)+5x

Group like terms:

11x+2=(-5x+5x)-9

Simplify the arithmetic:

11x+2=9

Subtract from both sides:

(11x+2)-2=-9-2

Simplify the arithmetic:

11x=92

Simplify the arithmetic:

11x=11

Divide both sides by :

(11x)11=-1111

Simplify the fraction:

x=-1111

Simplify the fraction:

x=1

3. List the solutions

x=7,1
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|6x+2|
y=|5x+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.