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

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

### Solved Example

**Question:**Simplify $\frac{125a^{2}b^{3}}{10a^{3}b}$

**Solution:**

**Step 1:**

First factorized, 125a

^{2}b

^{3 }= 5 * 5 * 5 * a * a * b* b * b,

and 10a

^{3}b = 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

**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 $\frac{5}{y}$ = $\left \{ y ; y\neq 0 \right \}$

=> $y = 0$

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 $\frac{3}{1 - x}$ = $\left \{ x ; x\neq 1 \right \}$

$\frac{3}{1 - x}$ = $\left \{ x ; 1 - x\neq 0 \right \}$

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

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

14x

9x = 3 * 3 * x

21x

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

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

= $2$

Step 1:

Factor each term

14x

^{2}= 2 * 7 * x * x9x = 3 * 3 * x

21x

^{3}= 3 * 7 * x * x * xStep 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} $

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

**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:**

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

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

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

**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)}$

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

=$ \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 $x$

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

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

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

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

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