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Solution - Absolute value equations

Exact form: b=6,2
b=6 , 2

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
4|b3|=|2b|
without the absolute value bars:

|x|=|y|4|b3|=|2b|
x=+y4(b3)=(2b)
x=y4(b3)=(2b)
+x=y4(b3)=(2b)
x=y4((b3))=(2b)

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|4|b3|=|2b|
x=+y , +x=y4(b3)=(2b)
x=y , x=y4(b3)=(2b)

2. Solve the two equations for b

12 additional steps

4·(b-3)=2b

Expand the parentheses:

4b+4·-3=2b

Simplify the arithmetic:

4b-12=2b

Subtract from both sides:

(4b-12)-2b=(2b)-2b

Group like terms:

(4b-2b)-12=(2b)-2b

Simplify the arithmetic:

2b-12=(2b)-2b

Simplify the arithmetic:

2b-12=0

Add to both sides:

(2b-12)+12=0+12

Simplify the arithmetic:

2b=0+12

Simplify the arithmetic:

2b=12

Divide both sides by :

(2b)2=122

Simplify the fraction:

b=122

Find the greatest common factor of the numerator and denominator:

b=(6·2)(1·2)

Factor out and cancel the greatest common factor:

b=6

12 additional steps

4·(b-3)=-(2b)

Expand the parentheses:

4b+4·-3=-(2b)

Simplify the arithmetic:

4b-12=-(2b)

Add to both sides:

(4b-12)+2b=(-2b)+2b

Group like terms:

(4b+2b)-12=(-2b)+2b

Simplify the arithmetic:

6b-12=(-2b)+2b

Simplify the arithmetic:

6b-12=0

Add to both sides:

(6b-12)+12=0+12

Simplify the arithmetic:

6b=0+12

Simplify the arithmetic:

6b=12

Divide both sides by :

(6b)6=126

Simplify the fraction:

b=126

Find the greatest common factor of the numerator and denominator:

b=(2·6)(1·6)

Factor out and cancel the greatest common factor:

b=2

3. List the solutions

b=6,2
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=4|b3|
y=|2b|
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.