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Solution - Absolute value equations

Exact form: x=6,-285
x=6 , -\frac{28}{5}
Mixed number form: x=6,-535
x=6 , -5\frac{3}{5}
Decimal form: x=6,5.6
x=6 , -5.6

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
|4x+5|=|x+23|
without the absolute value bars:

|x|=|y||4x+5|=|x+23|
x=+y(4x+5)=(x+23)
x=y(4x+5)=(x+23)
+x=y(4x+5)=(x+23)
x=y(4x+5)=(x+23)

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||4x+5|=|x+23|
x=+y , +x=y(4x+5)=(x+23)
x=y , x=y(4x+5)=(x+23)

2. Solve the two equations for x

11 additional steps

(4x+5)=(x+23)

Subtract from both sides:

(4x+5)-x=(x+23)-x

Group like terms:

(4x-x)+5=(x+23)-x

Simplify the arithmetic:

3x+5=(x+23)-x

Group like terms:

3x+5=(x-x)+23

Simplify the arithmetic:

3x+5=23

Subtract from both sides:

(3x+5)-5=23-5

Simplify the arithmetic:

3x=235

Simplify the arithmetic:

3x=18

Divide both sides by :

(3x)3=183

Simplify the fraction:

x=183

Find the greatest common factor of the numerator and denominator:

x=(6·3)(1·3)

Factor out and cancel the greatest common factor:

x=6

10 additional steps

(4x+5)=-(x+23)

Expand the parentheses:

(4x+5)=-x-23

Add to both sides:

(4x+5)+x=(-x-23)+x

Group like terms:

(4x+x)+5=(-x-23)+x

Simplify the arithmetic:

5x+5=(-x-23)+x

Group like terms:

5x+5=(-x+x)-23

Simplify the arithmetic:

5x+5=23

Subtract from both sides:

(5x+5)-5=-23-5

Simplify the arithmetic:

5x=235

Simplify the arithmetic:

5x=28

Divide both sides by :

(5x)5=-285

Simplify the fraction:

x=-285

3. List the solutions

x=6,-285
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|4x+5|
y=|x+23|
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.