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Solution - Absolute value equations

Exact form: x=2
x=2

Other Ways to Solve

Absolute value equations

Step-by-step explanation

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

|x1|+|x3|=0

Add |x3| to both sides of the equation:

|x1|+|x3||x3|=|x3|

Simplify the arithmetic

|x1|=|x3|

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
|x1|=|x3|
without the absolute value bars:

|x|=|y||x1|=|x3|
x=+y(x1)=(x3)
x=y(x1)=(x3)
+x=y(x1)=(x3)
x=y(x1)=(x3)

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||x1|=|x3|
x=+y , +x=y(x1)=(x3)
x=y , x=y(x1)=(x3)

3. Solve the two equations for x

12 additional steps

(x-1)=-(x-3)

Expand the parentheses:

(x-1)=-x+3

Add to both sides:

(x-1)+x=(-x+3)+x

Group like terms:

(x+x)-1=(-x+3)+x

Simplify the arithmetic:

2x-1=(-x+3)+x

Group like terms:

2x-1=(-x+x)+3

Simplify the arithmetic:

2x1=3

Add to both sides:

(2x-1)+1=3+1

Simplify the arithmetic:

2x=3+1

Simplify the arithmetic:

2x=4

Divide both sides by :

(2x)2=42

Simplify the fraction:

x=42

Find the greatest common factor of the numerator and denominator:

x=(2·2)(1·2)

Factor out and cancel the greatest common factor:

x=2

6 additional steps

(x-1)=-(-(x-3))

NT_MSLUS_MAINSTEP_RESOLVE_DOUBLE_MINUS:

(x-1)=x-3

Subtract from both sides:

(x-1)-x=(x-3)-x

Group like terms:

(x-x)-1=(x-3)-x

Simplify the arithmetic:

-1=(x-3)-x

Group like terms:

-1=(x-x)-3

Simplify the arithmetic:

1=3

The statement is false:

1=3

The equation is false so it has no solution.

4. List the solutions

x=2
(1 solution(s))

5. Graph

Each line represents the function of one side of the equation:
y=|x1|
y=|x3|
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.