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Solution - Absolute value equations

Exact form: x=1,53
x=1 , \frac{5}{3}
Mixed number form: x=1,123
x=1 , 1\frac{2}{3}
Decimal form: x=1,1.667
x=1 , 1.667

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
|5x7|=|x3|
without the absolute value bars:

|x|=|y||5x7|=|x3|
x=+y(5x7)=(x3)
x=y(5x7)=(x3)
+x=y(5x7)=(x3)
x=y(5x7)=(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||5x7|=|x3|
x=+y , +x=y(5x7)=(x3)
x=y , x=y(5x7)=(x3)

2. Solve the two equations for x

10 additional steps

(5x-7)=(x-3)

Subtract from both sides:

(5x-7)-x=(x-3)-x

Group like terms:

(5x-x)-7=(x-3)-x

Simplify the arithmetic:

4x-7=(x-3)-x

Group like terms:

4x-7=(x-x)-3

Simplify the arithmetic:

4x7=3

Add to both sides:

(4x-7)+7=-3+7

Simplify the arithmetic:

4x=3+7

Simplify the arithmetic:

4x=4

Divide both sides by :

(4x)4=44

Simplify the fraction:

x=44

Simplify the fraction:

x=1

12 additional steps

(5x-7)=-(x-3)

Expand the parentheses:

(5x-7)=-x+3

Add to both sides:

(5x-7)+x=(-x+3)+x

Group like terms:

(5x+x)-7=(-x+3)+x

Simplify the arithmetic:

6x-7=(-x+3)+x

Group like terms:

6x-7=(-x+x)+3

Simplify the arithmetic:

6x7=3

Add to both sides:

(6x-7)+7=3+7

Simplify the arithmetic:

6x=3+7

Simplify the arithmetic:

6x=10

Divide both sides by :

(6x)6=106

Simplify the fraction:

x=106

Find the greatest common factor of the numerator and denominator:

x=(5·2)(3·2)

Factor out and cancel the greatest common factor:

x=53

3. List the solutions

x=1,53
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5x7|
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.