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Solution - Absolute value equations

Exact form: x=8,2
x=8 , 2

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
|1.6x5|=|x0.2|
without the absolute value bars:

|x|=|y||1.6x5|=|x0.2|
x=+y(1.6x5)=(x0.2)
x=y(1.6x5)=(x0.2)
+x=y(1.6x5)=(x0.2)
x=y(1.6x5)=(x0.2)

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||1.6x5|=|x0.2|
x=+y , +x=y(1.6x5)=(x0.2)
x=y , x=y(1.6x5)=(x0.2)

2. Solve the two equations for x

10 additional steps

(1.6x-5)=(x-0.2)

Subtract from both sides:

(1.6x-5)-x=(x-0.2)-x

Group like terms:

(1.6x-x)-5=(x-0.2)-x

Simplify the arithmetic:

0.6x-5=(x-0.2)-x

Group like terms:

0.6x-5=(x-x)-0.2

Simplify the arithmetic:

0.6x5=0.2

Add to both sides:

(0.6x-5)+5=-0.2+5

Simplify the arithmetic:

0.6x=0.2+5

Simplify the arithmetic:

0.6x=4.8

Divide both sides by :

(0.6x)0.6=4.80.6

Simplify the arithmetic:

x=4.80.6

Simplify the arithmetic:

x=8

11 additional steps

(1.6x-5)=-(x-0.2)

Expand the parentheses:

(1.6x-5)=-x+0.2

Add to both sides:

(1.6x-5)+x=(-x+0.2)+x

Group like terms:

(1.6x+x)-5=(-x+0.2)+x

Simplify the arithmetic:

2.6x-5=(-x+0.2)+x

Group like terms:

2.6x-5=(-x+x)+0.2

Simplify the arithmetic:

2.6x5=0.2

Add to both sides:

(2.6x-5)+5=0.2+5

Simplify the arithmetic:

2.6x=0.2+5

Simplify the arithmetic:

2.6x=5.2

Divide both sides by :

(2.6x)2.6=5.22.6

Simplify the arithmetic:

x=5.22.6

Simplify the arithmetic:

x=2

3. List the solutions

x=8,2
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|1.6x5|
y=|x0.2|
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.