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Solution - Absolute value equations

Exact form: a=75
a=\frac{7}{5}
Mixed number form: a=125
a=1\frac{2}{5}
Decimal form: a=1.4
a=1.4

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
|5a+5|=|5a+9|
without the absolute value bars:

|x|=|y||5a+5|=|5a+9|
x=+y(5a+5)=(5a+9)
x=y(5a+5)=(5a+9)
+x=y(5a+5)=(5a+9)
x=y(5a+5)=(5a+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||5a+5|=|5a+9|
x=+y , +x=y(5a+5)=(5a+9)
x=y , x=y(5a+5)=(5a+9)

2. Solve the two equations for a

5 additional steps

(-5a+5)=(-5a+9)

Add to both sides:

(-5a+5)+5a=(-5a+9)+5a

Group like terms:

(-5a+5a)+5=(-5a+9)+5a

Simplify the arithmetic:

5=(-5a+9)+5a

Group like terms:

5=(-5a+5a)+9

Simplify the arithmetic:

5=9

The statement is false:

5=9

The equation is false so it has no solution.

14 additional steps

(-5a+5)=-(-5a+9)

Expand the parentheses:

(-5a+5)=5a-9

Subtract from both sides:

(-5a+5)-5a=(5a-9)-5a

Group like terms:

(-5a-5a)+5=(5a-9)-5a

Simplify the arithmetic:

-10a+5=(5a-9)-5a

Group like terms:

-10a+5=(5a-5a)-9

Simplify the arithmetic:

10a+5=9

Subtract from both sides:

(-10a+5)-5=-9-5

Simplify the arithmetic:

10a=95

Simplify the arithmetic:

10a=14

Divide both sides by :

(-10a)-10=-14-10

Cancel out the negatives:

10a10=-14-10

Simplify the fraction:

a=-14-10

Cancel out the negatives:

a=1410

Find the greatest common factor of the numerator and denominator:

a=(7·2)(5·2)

Factor out and cancel the greatest common factor:

a=75

3. Graph

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