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Solution - Absolute value equations

Exact form: b=2,25
b=2 , \frac{2}{5}
Decimal form: b=2,0.4
b=2 , 0.4

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
|4b4|=|b+2|
without the absolute value bars:

|x|=|y||4b4|=|b+2|
x=+y(4b4)=(b+2)
x=y(4b4)=(b+2)
+x=y(4b4)=(b+2)
x=y(4b4)=(b+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||4b4|=|b+2|
x=+y , +x=y(4b4)=(b+2)
x=y , x=y(4b4)=(b+2)

2. Solve the two equations for b

11 additional steps

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

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

3b-4=2

Add to both sides:

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

Simplify the arithmetic:

3b=2+4

Simplify the arithmetic:

3b=6

Divide both sides by :

(3b)3=63

Simplify the fraction:

b=63

Find the greatest common factor of the numerator and denominator:

b=(2·3)(1·3)

Factor out and cancel the greatest common factor:

b=2

10 additional steps

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

Expand the parentheses:

(4b-4)=-b-2

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

5b-4=(-b-2)+b

Group like terms:

5b-4=(-b+b)-2

Simplify the arithmetic:

5b-4=-2

Add to both sides:

(5b-4)+4=-2+4

Simplify the arithmetic:

5b=-2+4

Simplify the arithmetic:

5b=2

Divide both sides by :

(5b)5=25

Simplify the fraction:

b=25

3. List the solutions

b=2,25
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|4b4|
y=|b+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.