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Solution - Absolute value equations

Exact form: x=4
x=4

Other Ways to Solve

Absolute value equations

Step-by-step explanation

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

|x3|+|x5|=0

Add |x5| to both sides of the equation:

|x3|+|x5||x5|=|x5|

Simplify the arithmetic

|x3|=|x5|

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
|x3|=|x5|
without the absolute value bars:

|x|=|y||x3|=|x5|
x=+y(x3)=(x5)
x=y(x3)=(x5)
+x=y(x3)=(x5)
x=y(x3)=(x5)

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||x3|=|x5|
x=+y , +x=y(x3)=(x5)
x=y , x=y(x3)=(x5)

3. Solve the two equations for x

12 additional steps

(x-3)=-(x-5)

Expand the parentheses:

(x-3)=-x+5

Add to both sides:

(x-3)+x=(-x+5)+x

Group like terms:

(x+x)-3=(-x+5)+x

Simplify the arithmetic:

2x-3=(-x+5)+x

Group like terms:

2x-3=(-x+x)+5

Simplify the arithmetic:

2x3=5

Add to both sides:

(2x-3)+3=5+3

Simplify the arithmetic:

2x=5+3

Simplify the arithmetic:

2x=8

Divide both sides by :

(2x)2=82

Simplify the fraction:

x=82

Find the greatest common factor of the numerator and denominator:

x=(4·2)(1·2)

Factor out and cancel the greatest common factor:

x=4

6 additional steps

(x-3)=-(-(x-5))

NT_MSLUS_MAINSTEP_RESOLVE_DOUBLE_MINUS:

(x-3)=x-5

Subtract from both sides:

(x-3)-x=(x-5)-x

Group like terms:

(x-x)-3=(x-5)-x

Simplify the arithmetic:

-3=(x-5)-x

Group like terms:

-3=(x-x)-5

Simplify the arithmetic:

3=5

The statement is false:

3=5

The equation is false so it has no solution.

4. List the solutions

x=4
(1 solution(s))

5. Graph

Each line represents the function of one side of the equation:
y=|x3|
y=|x5|
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.