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Solution - Absolute value equations

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

Other Ways to Solve

Absolute value equations

Step-by-step explanation

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

|x+2|+|x1|=0

Add |x1| to both sides of the equation:

|x+2|+|x1||x1|=|x1|

Simplify the arithmetic

|x+2|=|x1|

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
|x+2|=|x1|
without the absolute value bars:

|x|=|y||x+2|=|x1|
x=+y(x+2)=(x1)
x=y(x+2)=(x1)
+x=y(x+2)=(x1)
x=y(x+2)=(x1)

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||x+2|=|x1|
x=+y , +x=y(x+2)=(x1)
x=y , x=y(x+2)=(x1)

3. Solve the two equations for x

6 additional steps

(-x+2)=-(x-1)

Expand the parentheses:

(-x+2)=-x+1

Add to both sides:

(-x+2)+x=(-x+1)+x

Group like terms:

(-x+x)+2=(-x+1)+x

Simplify the arithmetic:

2=(-x+1)+x

Group like terms:

2=(-x+x)+1

Simplify the arithmetic:

2=1

The statement is false:

2=1

The equation is false so it has no solution.

12 additional steps

(-x+2)=-(-(x-1))

NT_MSLUS_MAINSTEP_RESOLVE_DOUBLE_MINUS:

(-x+2)=x-1

Subtract from both sides:

(-x+2)-x=(x-1)-x

Group like terms:

(-x-x)+2=(x-1)-x

Simplify the arithmetic:

-2x+2=(x-1)-x

Group like terms:

-2x+2=(x-x)-1

Simplify the arithmetic:

2x+2=1

Subtract from both sides:

(-2x+2)-2=-1-2

Simplify the arithmetic:

2x=12

Simplify the arithmetic:

2x=3

Divide both sides by :

(-2x)-2=-3-2

Cancel out the negatives:

2x2=-3-2

Simplify the fraction:

x=-3-2

Cancel out the negatives:

x=32

4. Graph

Each line represents the function of one side of the equation:
y=|x+2|
y=|x1|
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.