# What is the easiest way to simplify rational expressions?Currently, this is the area I am having the most trouble in; I want to be able to understand this better before my final next week. Any help...

What is the easiest way to simplify rational expressions?

Currently, this is the area I am having the most trouble in; I want to be able to understand this better before my final next week. Any help would be appreciated.

### 1 Answer | Add Yours

You need to focus on two chapters to understand how to simplify rational expressions.

The first chapter contains the addition/subtraction of rational expressions that do not have a common denominator.

Since you will understand better an worked example, you may consider the following example such that:

`1/(x-1) + (2x-5)/(x+1)`

Notice that the fractions do not have a common denominator, hence, you will need to identify the common denominator. In this case, since the denominators do not share a common factor, the common denominator will be the following product, such that:

`(x+1)(x-1) = x^2- 1`

You need to identify what is the missing factor to the first denominator and to multiply the fraction by the value identified, such that:

`1*(x+1)/((x-1)(x+1)) + ((2x-5)(x-1))/((x-1)(x+1)) `

Since the fractions share the same denominator, you may write the expression such taht:

`(x + 1 + (2x - 5)(x - 1))/(x^2 - 1)`

You need to open the brackets such that:

`(x + 1 + 2x^2 - 2x - 5x + 5)/(x^2 - 1) = (2x^2 - 6x + 6)/(x^2-1)`

Factoring out 2 to numerator yields:

`(2x^2 - 6x + 6)/(x^2-1) = 2(x^2 - 3x + 3)/(x^2-1)`

**Hence, simplifying the expression yields `1/(x-1) + (2x-5)/(x+1) = 2(x^2 - 3x + 3)/(x^2-1).` **

The first chapter contains the multiplication or division of rational expressions.

Selecting the following example yields:

`(5x-1)/(x+1)*(x-1)/(5x-1)`

You need to identify if there exists duplicate factors to numerator and denominators and cancel the factors, such that:

`1/(x+1)*(x-1)/1` ( notice that `5x-1` is the duplicate factor)

**Hence, simplifying the expression yields `(5x-1)/(x+1)*(x-1)/(5x-1) = (x-1)/(x+1).` **