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Solution - Absolute value equations

Exact form: x=0,27
x=0 , \frac{2}{7}
Decimal form: x=0,0.286
x=0 , 0.286

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
2|3x1|=|8x2|
without the absolute value bars:

|x|=|y|2|3x1|=|8x2|
x=+y2(3x1)=(8x2)
x=y2(3x1)=(8x2)
+x=y2(3x1)=(8x2)
x=y2((3x1))=(8x2)

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|2|3x1|=|8x2|
x=+y , +x=y2(3x1)=(8x2)
x=y , x=y2(3x1)=(8x2)

2. Solve the two equations for x

11 additional steps

2·(3x-1)=(8x-2)

Expand the parentheses:

2·3x+2·-1=(8x-2)

Multiply the coefficients:

6x+2·-1=(8x-2)

Simplify the arithmetic:

6x-2=(8x-2)

Subtract from both sides:

(6x-2)-8x=(8x-2)-8x

Group like terms:

(6x-8x)-2=(8x-2)-8x

Simplify the arithmetic:

-2x-2=(8x-2)-8x

Group like terms:

-2x-2=(8x-8x)-2

Simplify the arithmetic:

2x2=2

Add to both sides:

(-2x-2)+2=-2+2

Simplify the arithmetic:

2x=2+2

Simplify the arithmetic:

2x=0

Divide both sides by the coefficient:

x=0

15 additional steps

2·(3x-1)=-(8x-2)

Expand the parentheses:

2·3x+2·-1=-(8x-2)

Multiply the coefficients:

6x+2·-1=-(8x-2)

Simplify the arithmetic:

6x-2=-(8x-2)

Expand the parentheses:

6x2=8x+2

Add to both sides:

(6x-2)+8x=(-8x+2)+8x

Group like terms:

(6x+8x)-2=(-8x+2)+8x

Simplify the arithmetic:

14x-2=(-8x+2)+8x

Group like terms:

14x-2=(-8x+8x)+2

Simplify the arithmetic:

14x2=2

Add to both sides:

(14x-2)+2=2+2

Simplify the arithmetic:

14x=2+2

Simplify the arithmetic:

14x=4

Divide both sides by :

(14x)14=414

Simplify the fraction:

x=414

Find the greatest common factor of the numerator and denominator:

x=(2·2)(7·2)

Factor out and cancel the greatest common factor:

x=27

3. List the solutions

x=0,27
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=2|3x1|
y=|8x2|
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.