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Solution - Absolute value equations

Exact form: a=73
a=\frac{7}{3}
Mixed number form: a=213
a=2\frac{1}{3}
Decimal form: a=2.333
a=2.333

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
|3a+6|=|3a+8|
without the absolute value bars:

|x|=|y||3a+6|=|3a+8|
x=+y(3a+6)=(3a+8)
x=y(3a+6)=(3a+8)
+x=y(3a+6)=(3a+8)
x=y(3a+6)=(3a+8)

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||3a+6|=|3a+8|
x=+y , +x=y(3a+6)=(3a+8)
x=y , x=y(3a+6)=(3a+8)

2. Solve the two equations for a

5 additional steps

(-3a+6)=(-3a+8)

Add to both sides:

(-3a+6)+3a=(-3a+8)+3a

Group like terms:

(-3a+3a)+6=(-3a+8)+3a

Simplify the arithmetic:

6=(-3a+8)+3a

Group like terms:

6=(-3a+3a)+8

Simplify the arithmetic:

6=8

The statement is false:

6=8

The equation is false so it has no solution.

14 additional steps

(-3a+6)=-(-3a+8)

Expand the parentheses:

(-3a+6)=3a-8

Subtract from both sides:

(-3a+6)-3a=(3a-8)-3a

Group like terms:

(-3a-3a)+6=(3a-8)-3a

Simplify the arithmetic:

-6a+6=(3a-8)-3a

Group like terms:

-6a+6=(3a-3a)-8

Simplify the arithmetic:

6a+6=8

Subtract from both sides:

(-6a+6)-6=-8-6

Simplify the arithmetic:

6a=86

Simplify the arithmetic:

6a=14

Divide both sides by :

(-6a)-6=-14-6

Cancel out the negatives:

6a6=-14-6

Simplify the fraction:

a=-14-6

Cancel out the negatives:

a=146

Find the greatest common factor of the numerator and denominator:

a=(7·2)(3·2)

Factor out and cancel the greatest common factor:

a=73

3. Graph

Each line represents the function of one side of the equation:
y=|3a+6|
y=|3a+8|
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.