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Solution - Absolute value equations

Exact form: x=9.85
x=9.85

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
|x3.5|=|x+16.2|
without the absolute value bars:

|x|=|y||x3.5|=|x+16.2|
x=+y(x3.5)=(x+16.2)
x=y(x3.5)=(x+16.2)
+x=y(x3.5)=(x+16.2)
x=y(x3.5)=(x+16.2)

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||x3.5|=|x+16.2|
x=+y , +x=y(x3.5)=(x+16.2)
x=y , x=y(x3.5)=(x+16.2)

2. Solve the two equations for x

10 additional steps

(x-3.5)=(-x+16.2)

Add to both sides:

(x-3.5)+x=(-x+16.2)+x

Group like terms:

(x+x)-3.5=(-x+16.2)+x

Simplify the arithmetic:

2x-3.5=(-x+16.2)+x

Group like terms:

2x-3.5=(-x+x)+16.2

Simplify the arithmetic:

2x3.5=16.2

Add to both sides:

(2x-3.5)+3.5=16.2+3.5

Simplify the arithmetic:

2x=16.2+3.5

Simplify the arithmetic:

2x=19.7

Divide both sides by :

(2x)2=19.72

Simplify the fraction:

x=19.72

Simplify the arithmetic:

x=9.85

6 additional steps

(x-3.5)=-(-x+16.2)

Expand the parentheses:

(x-3.5)=x-16.2

Subtract from both sides:

(x-3.5)-x=(x-16.2)-x

Group like terms:

(x-x)-3.5=(x-16.2)-x

Simplify the arithmetic:

-3.5=(x-16.2)-x

Group like terms:

-3.5=(x-x)-16.2

Simplify the arithmetic:

3.5=16.2

The statement is false:

3.5=16.2

The equation is false so it has no solution.

3. List the solutions

x=9.85
(1 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|x3.5|
y=|x+16.2|
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.