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Solution - Absolute value equations

Exact form: x=23,47
x=\frac{2}{3} , \frac{4}{7}
Decimal form: x=0.667,0.571
x=0.667 , 0.571

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
|5x3|=|2x1|
without the absolute value bars:

|x|=|y||5x3|=|2x1|
x=+y(5x3)=(2x1)
x=y(5x3)=(2x1)
+x=y(5x3)=(2x1)
x=y(5x3)=(2x1)

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||5x3|=|2x1|
x=+y , +x=y(5x3)=(2x1)
x=y , x=y(5x3)=(2x1)

2. Solve the two equations for x

9 additional steps

(5x-3)=(2x-1)

Subtract from both sides:

(5x-3)-2x=(2x-1)-2x

Group like terms:

(5x-2x)-3=(2x-1)-2x

Simplify the arithmetic:

3x-3=(2x-1)-2x

Group like terms:

3x-3=(2x-2x)-1

Simplify the arithmetic:

3x3=1

Add to both sides:

(3x-3)+3=-1+3

Simplify the arithmetic:

3x=1+3

Simplify the arithmetic:

3x=2

Divide both sides by :

(3x)3=23

Simplify the fraction:

x=23

10 additional steps

(5x-3)=-(2x-1)

Expand the parentheses:

(5x-3)=-2x+1

Add to both sides:

(5x-3)+2x=(-2x+1)+2x

Group like terms:

(5x+2x)-3=(-2x+1)+2x

Simplify the arithmetic:

7x-3=(-2x+1)+2x

Group like terms:

7x-3=(-2x+2x)+1

Simplify the arithmetic:

7x3=1

Add to both sides:

(7x-3)+3=1+3

Simplify the arithmetic:

7x=1+3

Simplify the arithmetic:

7x=4

Divide both sides by :

(7x)7=47

Simplify the fraction:

x=47

3. List the solutions

x=23,47
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5x3|
y=|2x1|
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.