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Solution - Absolute value equations

Exact form: x=-72,-14
x=-\frac{7}{2} , -\frac{1}{4}
Mixed number form: x=-312,-14
x=-3\frac{1}{2} , -\frac{1}{4}
Decimal form: x=3.5,0.25
x=-3.5 , -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
|5x2|=|7x+5|
without the absolute value bars:

|x|=|y||5x2|=|7x+5|
x=+y(5x2)=(7x+5)
x=y(5x2)=(7x+5)
+x=y(5x2)=(7x+5)
x=y(5x2)=(7x+5)

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||5x2|=|7x+5|
x=+y , +x=y(5x2)=(7x+5)
x=y , x=y(5x2)=(7x+5)

2. Solve the two equations for x

11 additional steps

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

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

2x2=5

Add to both sides:

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

Simplify the arithmetic:

2x=5+2

Simplify the arithmetic:

2x=7

Divide both sides by :

(-2x)-2=7-2

Cancel out the negatives:

2x2=7-2

Simplify the fraction:

x=7-2

Move the negative sign from the denominator to the numerator:

x=-72

12 additional steps

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

Expand the parentheses:

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

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

12x-2=(-7x-5)+7x

Group like terms:

12x-2=(-7x+7x)-5

Simplify the arithmetic:

12x2=5

Add to both sides:

(12x-2)+2=-5+2

Simplify the arithmetic:

12x=5+2

Simplify the arithmetic:

12x=3

Divide both sides by :

(12x)12=-312

Simplify the fraction:

x=-312

Find the greatest common factor of the numerator and denominator:

x=(-1·3)(4·3)

Factor out and cancel the greatest common factor:

x=-14

3. List the solutions

x=-72,-14
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5x2|
y=|7x+5|
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.