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Solution - Absolute value equations

Exact form: x=1,13
x=1 , \frac{1}{3}
Decimal form: x=1,0.333
x=1 , 0.333

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
|5x1|=|7x3|
without the absolute value bars:

|x|=|y||5x1|=|7x3|
x=+y(5x1)=(7x3)
x=y(5x1)=(7x3)
+x=y(5x1)=(7x3)
x=y(5x1)=(7x3)

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||5x1|=|7x3|
x=+y , +x=y(5x1)=(7x3)
x=y , x=y(5x1)=(7x3)

2. Solve the two equations for x

12 additional steps

(5x-1)=(7x-3)

Subtract from both sides:

(5x-1)-7x=(7x-3)-7x

Group like terms:

(5x-7x)-1=(7x-3)-7x

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

2x1=3

Add to both sides:

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

Simplify the arithmetic:

2x=3+1

Simplify the arithmetic:

2x=2

Divide both sides by :

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

Cancel out the negatives:

2x2=-2-2

Simplify the fraction:

x=-2-2

Cancel out the negatives:

x=22

Simplify the fraction:

x=1

12 additional steps

(5x-1)=-(7x-3)

Expand the parentheses:

(5x-1)=-7x+3

Add to both sides:

(5x-1)+7x=(-7x+3)+7x

Group like terms:

(5x+7x)-1=(-7x+3)+7x

Simplify the arithmetic:

12x-1=(-7x+3)+7x

Group like terms:

12x-1=(-7x+7x)+3

Simplify the arithmetic:

12x1=3

Add to both sides:

(12x-1)+1=3+1

Simplify the arithmetic:

12x=3+1

Simplify the arithmetic:

12x=4

Divide both sides by :

(12x)12=412

Simplify the fraction:

x=412

Find the greatest common factor of the numerator and denominator:

x=(1·4)(3·4)

Factor out and cancel the greatest common factor:

x=13

3. List the solutions

x=1,13
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5x1|
y=|7x3|
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.