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Solution - Absolute value equations

Exact form: b=32,3
b=\frac{3}{2} , 3
Mixed number form: b=112,3
b=1\frac{1}{2} , 3
Decimal form: b=1.5,3
b=1.5 , 3

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
|2b+3|=|2b3|
without the absolute value bars:

|x|=|y||2b+3|=|2b3|
x=+y(2b+3)=(2b3)
x=y(2b+3)=(2b3)
+x=y(2b+3)=(2b3)
x=y(2b+3)=(2b3)

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||2b+3|=|2b3|
x=+y , +x=y(2b+3)=(2b3)
x=y , x=y(2b+3)=(2b3)

2. Solve the two equations for b

13 additional steps

(-2b+3)=(2b-3)

Subtract from both sides:

(-2b+3)-2b=(2b-3)-2b

Group like terms:

(-2b-2b)+3=(2b-3)-2b

Simplify the arithmetic:

-4b+3=(2b-3)-2b

Group like terms:

-4b+3=(2b-2b)-3

Simplify the arithmetic:

-4b+3=-3

Subtract from both sides:

(-4b+3)-3=-3-3

Simplify the arithmetic:

-4b=-3-3

Simplify the arithmetic:

-4b=-6

Divide both sides by :

(-4b)-4=-6-4

Cancel out the negatives:

4b4=-6-4

Simplify the fraction:

b=-6-4

Cancel out the negatives:

b=64

Find the greatest common factor of the numerator and denominator:

b=(3·2)(2·2)

Factor out and cancel the greatest common factor:

b=32

5 additional steps

(-2b+3)=-(2b-3)

Expand the parentheses:

(-2b+3)=-2b+3

Add to both sides:

(-2b+3)+2b=(-2b+3)+2b

Group like terms:

(-2b+2b)+3=(-2b+3)+2b

Simplify the arithmetic:

3=(-2b+3)+2b

Group like terms:

3=(-2b+2b)+3

Simplify the arithmetic:

3=3

3. List the solutions

b=32,3
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|2b+3|
y=|2b3|
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.