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Solution - Absolute value equations

Exact form: a=72
a=\frac{7}{2}
Mixed number form: a=312
a=3\frac{1}{2}
Decimal form: a=3.5
a=3.5

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

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

2. Solve the two equations for a

5 additional steps

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

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

5=9

The statement is false:

5=9

The equation is false so it has no solution.

14 additional steps

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

Expand the parentheses:

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

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

-4a+5=(2a-9)-2a

Group like terms:

-4a+5=(2a-2a)-9

Simplify the arithmetic:

4a+5=9

Subtract from both sides:

(-4a+5)-5=-9-5

Simplify the arithmetic:

4a=95

Simplify the arithmetic:

4a=14

Divide both sides by :

(-4a)-4=-14-4

Cancel out the negatives:

4a4=-14-4

Simplify the fraction:

a=-14-4

Cancel out the negatives:

a=144

Find the greatest common factor of the numerator and denominator:

a=(7·2)(2·2)

Factor out and cancel the greatest common factor:

a=72

3. Graph

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