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Solution - Absolute value equations

Exact form: x=-4,-56
x=-4 , -\frac{5}{6}
Decimal form: x=4,0.833
x=-4 , -0.833

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
|4x3|=|8x+13|
without the absolute value bars:

|x|=|y||4x3|=|8x+13|
x=+y(4x3)=(8x+13)
x=y(4x3)=(8x+13)
+x=y(4x3)=(8x+13)
x=y(4x3)=(8x+13)

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||4x3|=|8x+13|
x=+y , +x=y(4x3)=(8x+13)
x=y , x=y(4x3)=(8x+13)

2. Solve the two equations for x

13 additional steps

(4x-3)=(8x+13)

Subtract from both sides:

(4x-3)-8x=(8x+13)-8x

Group like terms:

(4x-8x)-3=(8x+13)-8x

Simplify the arithmetic:

-4x-3=(8x+13)-8x

Group like terms:

-4x-3=(8x-8x)+13

Simplify the arithmetic:

4x3=13

Add to both sides:

(-4x-3)+3=13+3

Simplify the arithmetic:

4x=13+3

Simplify the arithmetic:

4x=16

Divide both sides by :

(-4x)-4=16-4

Cancel out the negatives:

4x4=16-4

Simplify the fraction:

x=16-4

Move the negative sign from the denominator to the numerator:

x=-164

Find the greatest common factor of the numerator and denominator:

x=(-4·4)(1·4)

Factor out and cancel the greatest common factor:

x=4

12 additional steps

(4x-3)=-(8x+13)

Expand the parentheses:

(4x-3)=-8x-13

Add to both sides:

(4x-3)+8x=(-8x-13)+8x

Group like terms:

(4x+8x)-3=(-8x-13)+8x

Simplify the arithmetic:

12x-3=(-8x-13)+8x

Group like terms:

12x-3=(-8x+8x)-13

Simplify the arithmetic:

12x3=13

Add to both sides:

(12x-3)+3=-13+3

Simplify the arithmetic:

12x=13+3

Simplify the arithmetic:

12x=10

Divide both sides by :

(12x)12=-1012

Simplify the fraction:

x=-1012

Find the greatest common factor of the numerator and denominator:

x=(-5·2)(6·2)

Factor out and cancel the greatest common factor:

x=-56

3. List the solutions

x=-4,-56
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|4x3|
y=|8x+13|
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.