Enter an equation or problem
Camera input is not recognized!

Solution - Absolute value equations

Exact form: a=-52,-116
a=-\frac{5}{2} , -\frac{11}{6}
Mixed number form: a=-212,-156
a=-2\frac{1}{2} , -1\frac{5}{6}
Decimal form: a=2.5,1.833
a=-2.5 , -1.833

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+3|=|4a+8|
without the absolute value bars:

|x|=|y||2a+3|=|4a+8|
x=+y(2a+3)=(4a+8)
x=y(2a+3)=(4a+8)
+x=y(2a+3)=(4a+8)
x=y(2a+3)=(4a+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||2a+3|=|4a+8|
x=+y , +x=y(2a+3)=(4a+8)
x=y , x=y(2a+3)=(4a+8)

2. Solve the two equations for a

11 additional steps

(2a+3)=(4a+8)

Subtract from both sides:

(2a+3)-4a=(4a+8)-4a

Group like terms:

(2a-4a)+3=(4a+8)-4a

Simplify the arithmetic:

-2a+3=(4a+8)-4a

Group like terms:

-2a+3=(4a-4a)+8

Simplify the arithmetic:

2a+3=8

Subtract from both sides:

(-2a+3)-3=8-3

Simplify the arithmetic:

2a=83

Simplify the arithmetic:

2a=5

Divide both sides by :

(-2a)-2=5-2

Cancel out the negatives:

2a2=5-2

Simplify the fraction:

a=5-2

Move the negative sign from the denominator to the numerator:

a=-52

10 additional steps

(2a+3)=-(4a+8)

Expand the parentheses:

(2a+3)=-4a-8

Add to both sides:

(2a+3)+4a=(-4a-8)+4a

Group like terms:

(2a+4a)+3=(-4a-8)+4a

Simplify the arithmetic:

6a+3=(-4a-8)+4a

Group like terms:

6a+3=(-4a+4a)-8

Simplify the arithmetic:

6a+3=8

Subtract from both sides:

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

Simplify the arithmetic:

6a=83

Simplify the arithmetic:

6a=11

Divide both sides by :

(6a)6=-116

Simplify the fraction:

a=-116

3. List the solutions

a=-52,-116
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|2a+3|
y=|4a+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.