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Solution - Absolute value equations

Exact form: x=6,13
x=6 , \frac{1}{3}
Decimal form: x=6,0.333
x=6 , 0.333

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
|2x+5|=|4x7|
without the absolute value bars:

|x|=|y||2x+5|=|4x7|
x=+y(2x+5)=(4x7)
x=y(2x+5)=(4x7)
+x=y(2x+5)=(4x7)
x=y(2x+5)=(4x7)

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||2x+5|=|4x7|
x=+y , +x=y(2x+5)=(4x7)
x=y , x=y(2x+5)=(4x7)

2. Solve the two equations for x

13 additional steps

(2x+5)=(4x-7)

Subtract from both sides:

(2x+5)-4x=(4x-7)-4x

Group like terms:

(2x-4x)+5=(4x-7)-4x

Simplify the arithmetic:

-2x+5=(4x-7)-4x

Group like terms:

-2x+5=(4x-4x)-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

(2x+5)=-(4x-7)

Expand the parentheses:

(2x+5)=-4x+7

Add to both sides:

(2x+5)+4x=(-4x+7)+4x

Group like terms:

(2x+4x)+5=(-4x+7)+4x

Simplify the arithmetic:

6x+5=(-4x+7)+4x

Group like terms:

6x+5=(-4x+4x)+7

Simplify the arithmetic:

6x+5=7

Subtract from both sides:

(6x+5)-5=7-5

Simplify the arithmetic:

6x=75

Simplify the arithmetic:

6x=2

Divide both sides by :

(6x)6=26

Simplify the fraction:

x=26

Find the greatest common factor of the numerator and denominator:

x=(1·2)(3·2)

Factor out and cancel the greatest common factor:

x=13

3. List the solutions

x=6,13
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|2x+5|
y=|4x7|
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.