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Solution - Absolute value equations

Exact form: x=-12
x=-\frac{1}{2}
Decimal form: x=0.5
x=-0.5

Other Ways to Solve

Absolute value equations

Step-by-step explanation

1. Rewrite the equation with one absolute value terms on each side

|4x5|+|4x+9|=0

Add |4x+9| to both sides of the equation:

|4x5|+|4x+9||4x+9|=|4x+9|

Simplify the arithmetic

|4x5|=|4x+9|

2. 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
|4x5|=|4x+9|
without the absolute value bars:

|x|=|y||4x5|=|4x+9|
x=+y(4x5)=(4x+9)
x=y(4x5)=(4x+9)
+x=y(4x5)=(4x+9)
x=y(4x5)=(4x+9)

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||4x5|=|4x+9|
x=+y , +x=y(4x5)=(4x+9)
x=y , x=y(4x5)=(4x+9)

3. Solve the two equations for x

12 additional steps

(4x-5)=-(4x+9)

Expand the parentheses:

(4x-5)=-4x-9

Add to both sides:

(4x-5)+4x=(-4x-9)+4x

Group like terms:

(4x+4x)-5=(-4x-9)+4x

Simplify the arithmetic:

8x-5=(-4x-9)+4x

Group like terms:

8x-5=(-4x+4x)-9

Simplify the arithmetic:

8x5=9

Add to both sides:

(8x-5)+5=-9+5

Simplify the arithmetic:

8x=9+5

Simplify the arithmetic:

8x=4

Divide both sides by :

(8x)8=-48

Simplify the fraction:

x=-48

Find the greatest common factor of the numerator and denominator:

x=(-1·4)(2·4)

Factor out and cancel the greatest common factor:

x=-12

6 additional steps

(4x-5)=-(-(4x+9))

NT_MSLUS_MAINSTEP_RESOLVE_DOUBLE_MINUS:

(4x-5)=4x+9

Subtract from both sides:

(4x-5)-4x=(4x+9)-4x

Group like terms:

(4x-4x)-5=(4x+9)-4x

Simplify the arithmetic:

-5=(4x+9)-4x

Group like terms:

-5=(4x-4x)+9

Simplify the arithmetic:

5=9

The statement is false:

5=9

The equation is false so it has no solution.

4. List the solutions

x=-12
(1 solution(s))

5. Graph

Each line represents the function of one side of the equation:
y=|4x5|
y=|4x+9|
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.