Enter an equation or problem
Camera input is not recognized!

Solution - Absolute value equations

Exact form: a=4,-83
a=4 , -\frac{8}{3}
Mixed number form: a=4,-223
a=4 , -2\frac{2}{3}
Decimal form: a=4,2.667
a=4 , -2.667

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
|2a+2|=|a+6|
without the absolute value bars:

|x|=|y||2a+2|=|a+6|
x=+y(2a+2)=(a+6)
x=y(2a+2)=(a+6)
+x=y(2a+2)=(a+6)
x=y(2a+2)=(a+6)

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||2a+2|=|a+6|
x=+y , +x=y(2a+2)=(a+6)
x=y , x=y(2a+2)=(a+6)

2. Solve the two equations for a

7 additional steps

(2a+2)=(a+6)

Subtract from both sides:

(2a+2)-a=(a+6)-a

Group like terms:

(2a-a)+2=(a+6)-a

Simplify the arithmetic:

a+2=(a+6)-a

Group like terms:

a+2=(a-a)+6

Simplify the arithmetic:

a+2=6

Subtract from both sides:

(a+2)-2=6-2

Simplify the arithmetic:

a=62

Simplify the arithmetic:

a=4

10 additional steps

(2a+2)=-(a+6)

Expand the parentheses:

(2a+2)=-a-6

Add to both sides:

(2a+2)+a=(-a-6)+a

Group like terms:

(2a+a)+2=(-a-6)+a

Simplify the arithmetic:

3a+2=(-a-6)+a

Group like terms:

3a+2=(-a+a)-6

Simplify the arithmetic:

3a+2=6

Subtract from both sides:

(3a+2)-2=-6-2

Simplify the arithmetic:

3a=62

Simplify the arithmetic:

3a=8

Divide both sides by :

(3a)3=-83

Simplify the fraction:

a=-83

3. List the solutions

a=4,-83
(2 solution(s))

4. Graph

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