# Rational Expressions

Rational expression can be defined as an algebraic expression in the form $\frac{P}{Q}$ where $Q\neq 0$ and both $P$ and $Q$ are polynomial expressions. A rational expression is nothing more than a fraction in which the numerator and the denominator are polynomials. When dealing with rational expressions we will always assume that whatever variable is it won’t give denominator zero. We rarely write these restrictions, but we will always need to keep them in mind.

Algebraic fractions are quotients of algebraic expression. If the expressions are polynomials, the fraction is called rational expression. It can be written in the form of $\frac{P}{Q}$, Q$\neq0$ and P, Q both are polynomials.

## Simplifying Rational Expressions

The rational expression becomes easier to handle if its written in the simplest form. Any rational expression can be simplified, by cancelling any common term that is present in both numerator and denominator. For that, its important to factor each expression in both numerator and denominator.

### Solved Example

Question: Simplify $\frac{125a^{2}b^{3}}{10a^{3}b}$
Solution:
Step 1:
First factorized, 125a2b3 = 5 * 5 * 5 * a * a *  b*  b * b,

and 10a3b = 2 * 5 * a * a  * a * b

Step 2:

$\frac{125a^{2}b^{3}}{10a^{3}b^{1}}$ = $\frac{5 * 5 * 5 * a * a * b * b * b} {2 * 5 * a * a * a * b}$

Step 3:
Cancelling the common terms

=> $\frac{125a^{2}b^{3}}{10a^{3}b^{1}}$=$\frac{25b^{2}}{2a^{2}}$

## Find Domain of Rational Expression

The domain of a rational expression is the set of all real numbers for which it is defined. All rational expressions will be defined for all real values except at the point where the denominator becomes zero. So, the domain of a rational expression will contain all points except at which denominator is zero.

### Solved Examples

Question 1: Find the domain of  $\frac{5}{y}$
Solution:
To define the domain, we need to check all real number for which the function exists. The domain will then be all other y-values. When is this denominator equal to zero

=> $y = 0$

Hence, domain of $\frac{5}{y}$ = $\left \{ y ; y\neq 0 \right \}$

Then the domain is "all y except zero".

Question 2: Find the domain of $\frac{3}{1 - x}$
Solution:
Here, domain of
$\frac{3}{1 - x}$ = $\left \{ x ; 1 - x\neq 0 \right \}$

Solve for x,

$1- x = 0$

=> $-x = -1$

So, $x = 1$
Hence domain of $\frac{3}{1 - x}$ = $\left \{ x ; x\neq 1 \right \}$

## Multiplying and Dividing Rational Expressions

Algebraic operations, as we have algebraic operation in real number, we use the same operations and rules for working out rational expression. Lets look at each operations and the rules needed to work them out. Multiplying and dividing of fractions is similar to the multiply and divide rational expressions.

The general form of performing the division operation,

$\frac{P}{Q}\div \frac{R}{S} = \frac{P}{Q}\ast \frac{S}{R}$

### Solved Examples

Question 1: Solve $\frac{14x^{2}}{3} * \frac{9x}{21x^{3}}$
Solution:
Given $\frac{14x^{2}}{3} * \frac{9x}{21x^{3}}$

Step 1:
Factor each term
14x2  = 2 * 7 * x * x
9x = 3 * 3 * x
21x3  = 3 * 7 *  x * x  * x

Step 2:
=> $\frac{14x^{2}}{3}*\frac{9x}{21x^{3}} = \frac{2 * 7 * x * x}{3}*\frac{3 * 3 * x}{3 * 7 * x * x * x}$

Step 3:
Cancel the common terms

$\frac{2*7*x*x}{3}*\frac{3*3*x}{3*7*x*x*x}$ = $\frac{2}{1}$

= $2$

Question 2: Solve $\frac{5x}{x-y}\div \frac{5y}{(2x-2y)}$
Solution:
Given $\frac{5x}{x-y}\div \frac{5y}{(2x-2y)}$

Step 1:
The division sign can be changed to multiplication by taking the reciprocal of the second term.

=> $\frac{5x}{x-y}\div \frac{5y}{(2x-2y)} = \frac{5x}{x-y}\times \frac{2x-2y}{5y}$

Step 2:
Now apply multiplication operations

=> $\frac{5x}{x-y}\times \frac{2x-2y}{5y}=\frac{5x}{x-y}\times\frac{2(x-y)}{5y}$

= $\frac{5x\times 2(x-y)}{(x-y)\times 5y}$

Step 3:
Cancelling the common term
=> $\frac{5x\times 2(x-y)}{(x-y)\times 5y}$

= $\frac{x\times 2}{y}$

= $\frac{2x}{y}$

=>  $\frac{5x}{x-y}\div \frac{5y}{(2x-2y)}$ = $\frac{2x}{y}$

## Adding and Subtracting Rational Expressions

Addition and subtracting with real number is simple but with rational expression, its not that easy. While adding and subtracting two algebraic expressions, its important that their denominators are the same.
Lets look at each operations and the rules needed to work them out:
If the denominators are same,

=> $\frac{P}{Q} \pm \frac{R}{Q} = \frac{P \pm R}{Q}$

If the denominators not be same,

=> $\frac{P}{Q} \pm \frac{R}{S} = \frac{PS \pm RQ}{QS}$

### Solved Examples

Question 1: Add $\frac{x-1}{x-y}+\frac{xy+2}{x-y}$
Solution:
$\frac{x-1}{x-y}+\frac{xy+2}{x-y} = \frac{(x-1)+(xy+2)}{x-y}$

= $\frac{x+xy+1}{x-y}$

Question 2: Solve $\frac{2x}{x - y} + \frac{y - 1}{2x - 1}$
Solution:
No common factor between $x - y$ and $2x - 1$

$\frac{2x}{x-y}+\frac{y-1}{2x-1}$=$\frac{(2x)(2x-1)+(x-y)(y-1)}{(x-y)(2x-1)}$

=$\frac{(2x^{2}-2x)+(xy-y^{2}-x+y)}{(x-y)(2x-1)}$

= $\frac{2x^{2}-3x+xy-y^{2}+y}{(x-y)(2x-1)}$

Question 3: Subtract  $\frac{y-1}{x}$  from  $\frac{2x}{y}$

Solution:
$\frac{2x}{y} - \frac{y-1}{x}$

Here, no common factor between $y$ and $x$

=> $\frac{2x}{y} - \frac{y-1}{x} = \frac{(2x)(x)-y(y-1)}{(x)(y)}$

= $\frac{2x^{2}-y^{2}+y}{xy}$