Simplification of Rational Expressions

Expressions that do not contain surds and can be expressed in the form of p/q where q is not equal to 0 are called rational expressions. Simplification of Rational Expressions is only possible if the denominators of the given expressions are identical or same or equal.

Rational Expressions:

An expression in the form of p/q where q is not equal to 0 is called a rational expression. But, the surds can not be a rational expression.

Some examples of rational expressions are:

• $\dfrac{a}{b}$
• $\dfrac{a²}{a-2}$
• $\dfrac{a+3}{a-b}$

And, expressions consisting surds can not be rational expressions such as:

• $\sqrt{\dfrac{a}{b}}$
• $\sqrt{\dfrac{a-b}{a+b}}$

Simplification of Rational Expressions:

We know, rational expressions always have a denominator that is not equal to zero. When we want to add or subtract two different rational expressions then, we need to remember the following points:

• The denominator of both the expressions should be same.
• If the denominators of both the expressions are not same, we need to take the LCM to make them same. 
• After this, you can add or subtract the terms in the numerator. 
• You will combine the expressions and write only single denominator. Meaning, out of the two same denominators of the two expressions, you will only write one.
• Then, write the expressions in the numerator as well as in the denominator in factor form. Now, divide the equal factors, if any.
• Whatever remains is your answer. 

SOLVED EXAMPLES

EXAMPLE 1 | Simplify: $\dfrac{x \ – \ 2}{x^2 \ – \ 1} \ – \ \dfrac{x \ + \ 1}{x^2 \ – \ 2x \ + \ 1}$

Solution:

In this question, we have two terms in the expression having different denominators. Our aim is to subtract the second term from the first term. Our first approach has to be to make the denominator same. For this, we will first try to simplify the term in the denominator of first term and second term then take LCM for further operation.

Denominator of first term: $(x^2 \ – \ 1)$. This can be factorized as $(x \ + \ 1)(x \ – \ 1)$. So, we have simplified the first term as: $$\dfrac{x \ – \ 2}{(x \ + \ 1)(x \ – \ 1)}$$

Denominator of second term: $(x^2 \ – \ 2x \ + 1)$. This is the expression of $(a \ – \ b)^2$ so, it can be written as $(x \ – \ 1)^2$. So, we have simplified the second term as: $$\dfrac{x \ + \ 1}{(x \ – \ 1)^2}$$

Now, we have simplified the overall expression as$$\dfrac{x \ – \ 2}{(x \ + \ 1)(x \ – \ 1)} \ – \ \dfrac{x \ + \ 1}{(x \ – \ 1)^2}$$

$$= \ \dfrac{x \ – \ 2}{(x \ + \ 1)(x \ – \ 1)} \ – \ \dfrac{x \ + \ 1}{(x \ – \ 1)(x \ – \ 1)}$$

Now, we will take the LCM and simplify further

$$= \ \dfrac{(x \ – \ 2)(x \ – \ 1) \ – \ (x \ + \ 1)(x \ + \ 1)}{(x \ – \ 1)(x \ – \ 1)(x \ + \ 1)}$$

$$= \ \dfrac{(x^2 \ – \ 3x \ + \ 2) \ – \ (x^2 \ + \ 2x \ + \ 1)}{(x – 1)^2(x+1)}$$

$$= \ \dfrac{x^2 \ – \ 3x \ + \ 2 \ – \ x^2 \ – \ 2x \ – \ 1 }{(x-1)^2(x+1)}$$

$$= \ \dfrac{- 5x \ + \ 1}{(x + 1)(x^2 – 1)}$$

$$= \ \dfrac{1 \ – \ 5x}{(x+1)(x^2 – 1)}$$

Hence, $\dfrac{x \ – \ 2}{x^2 \ – \ 1} \ – \ \dfrac{x \ + \ 1}{x^2 \ – \ 2x \ + \ 1} \ = \ \dfrac{1 \ – \ 5x}{(x+1)(x^2 – 1)}$