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Solution - Absolute value equations

Exact form: b=-113,-13
b=-\frac{11}{3} , -13
Mixed number form: b=-323,-13
b=-3\frac{2}{3} , -13
Decimal form: b=3.667,13
b=-3.667 , -13

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+12|=|b1|
without the absolute value bars:

|x|=|y||2b+12|=|b1|
x=+y(2b+12)=(b1)
x=y(2b+12)=((b1))
+x=y(2b+12)=(b1)
x=y(2b+12)=(b1)

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+12|=|b1|
x=+y , +x=y(2b+12)=(b1)
x=y , x=y(2b+12)=((b1))

2. Solve the two equations for b

10 additional steps

(2b+12)=-(b-1)

Expand the parentheses:

(2b+12)=-b+1

Add to both sides:

(2b+12)+b=(-b+1)+b

Group like terms:

(2b+b)+12=(-b+1)+b

Simplify the arithmetic:

3b+12=(-b+1)+b

Group like terms:

3b+12=(-b+b)+1

Simplify the arithmetic:

3b+12=1

Subtract from both sides:

(3b+12)-12=1-12

Simplify the arithmetic:

3b=1-12

Simplify the arithmetic:

3b=-11

Divide both sides by :

(3b)3=-113

Simplify the fraction:

b=-113

8 additional steps

(2b+12)=-(-(b-1))

NT_MSLUS_MAINSTEP_RESOLVE_DOUBLE_MINUS:

(2b+12)=b-1

Subtract from both sides:

(2b+12)-b=(b-1)-b

Group like terms:

(2b-b)+12=(b-1)-b

Simplify the arithmetic:

b+12=(b-1)-b

Group like terms:

b+12=(b-b)-1

Simplify the arithmetic:

b+12=-1

Subtract from both sides:

(b+12)-12=-1-12

Simplify the arithmetic:

b=-1-12

Simplify the arithmetic:

b=-13

3. List the solutions

b=-113,-13
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|2b+12|
y=|b1|
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.