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Solution - Absolute value equations

Exact form: a=34
a=\frac{3}{4}
Decimal form: a=0.75
a=0.75

Other Ways to Solve

Absolute value equations

Step-by-step explanation

1. Rewrite the equation with one absolute value terms on each side

|2a1||2a+2|=0

Add |2a+2| to both sides of the equation:

|2a1||2a+2|+|2a+2|=|2a+2|

Simplify the arithmetic

|2a1|=|2a+2|

2. 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
|2a1|=|2a+2|
without the absolute value bars:

|x|=|y||2a1|=|2a+2|
x=+y(2a1)=(2a+2)
x=y(2a1)=((2a+2))
+x=y(2a1)=(2a+2)
x=y(2a1)=(2a+2)

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||2a1|=|2a+2|
x=+y , +x=y(2a1)=(2a+2)
x=y , x=y(2a1)=((2a+2))

3. Solve the two equations for a

9 additional steps

(2a-1)=(-2a+2)

Add to both sides:

(2a-1)+2a=(-2a+2)+2a

Group like terms:

(2a+2a)-1=(-2a+2)+2a

Simplify the arithmetic:

4a-1=(-2a+2)+2a

Group like terms:

4a-1=(-2a+2a)+2

Simplify the arithmetic:

4a1=2

Add to both sides:

(4a-1)+1=2+1

Simplify the arithmetic:

4a=2+1

Simplify the arithmetic:

4a=3

Divide both sides by :

(4a)4=34

Simplify the fraction:

a=34

6 additional steps

(2a-1)=-(-2a+2)

Expand the parentheses:

(2a-1)=2a-2

Subtract from both sides:

(2a-1)-2a=(2a-2)-2a

Group like terms:

(2a-2a)-1=(2a-2)-2a

Simplify the arithmetic:

-1=(2a-2)-2a

Group like terms:

-1=(2a-2a)-2

Simplify the arithmetic:

1=2

The statement is false:

1=2

The equation is false so it has no solution.

4. List the solutions

a=34
(1 solution(s))

5. Graph

Each line represents the function of one side of the equation:
y=|2a1|
y=|2a+2|
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.