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Solution - Absolute value equations

Exact form: x=6,14
x=6 , \frac{1}{4}
Decimal form: x=6,0.25
x=6 , 0.25

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
|3x+5|=|5x7|
without the absolute value bars:

|x|=|y||3x+5|=|5x7|
x=+y(3x+5)=(5x7)
x=y(3x+5)=(5x7)
+x=y(3x+5)=(5x7)
x=y(3x+5)=(5x7)

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||3x+5|=|5x7|
x=+y , +x=y(3x+5)=(5x7)
x=y , x=y(3x+5)=(5x7)

2. Solve the two equations for x

13 additional steps

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

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

-2x+5=(5x-7)-5x

Group like terms:

-2x+5=(5x-5x)-7

Simplify the arithmetic:

2x+5=7

Subtract from both sides:

(-2x+5)-5=-7-5

Simplify the arithmetic:

2x=75

Simplify the arithmetic:

2x=12

Divide both sides by :

(-2x)-2=-12-2

Cancel out the negatives:

2x2=-12-2

Simplify the fraction:

x=-12-2

Cancel out the negatives:

x=122

Find the greatest common factor of the numerator and denominator:

x=(6·2)(1·2)

Factor out and cancel the greatest common factor:

x=6

12 additional steps

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

Expand the parentheses:

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

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

8x+5=(-5x+7)+5x

Group like terms:

8x+5=(-5x+5x)+7

Simplify the arithmetic:

8x+5=7

Subtract from both sides:

(8x+5)-5=7-5

Simplify the arithmetic:

8x=75

Simplify the arithmetic:

8x=2

Divide both sides by :

(8x)8=28

Simplify the fraction:

x=28

Find the greatest common factor of the numerator and denominator:

x=(1·2)(4·2)

Factor out and cancel the greatest common factor:

x=14

3. List the solutions

x=6,14
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|3x+5|
y=|5x7|
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.