The term “rational” in math is not used in the sense of “sane” or “sensible.” It is instead used to imply a ratio, or fraction.
Rational expression is a fraction whose numerator and denominator is a polynomial function in variables.
Example: $\frac{x + 2}{x^{2}-3}$ is a rational expression.
We know that division by zero is not allowed since it makes the value of rational expression undefined. So it is always assumed that the denominator of a rational expression is never zero. This condition puts restrictions to the possible values of the variable.
For example, in a rational expression $\frac{2x + 3 }{x - 4}$, $x$ can never take the value of 4 since $x = 4$ will make the denominator zero and the expression undefined.

## Solving Rational Expressions

We will see reducing rational expression to simplest form. Simplifying rational expression is quite similar to simplifying fractions.

First check for common factors (constants, variables or polynomials ) of numerator and denominator , cancel out the common terms. There are some common mistakes with theses problems.

In the expression $\frac{x}{(x - 4)}$ students may end up cancelling $x$ like $\frac{\not{x}}{\not{x} - 4}$ which is not valid since $x$ in the denominator is not times the whole number like $\frac{x}{4x}$.

In general, the rule is not to cancel something if there is + or on one side of the variable.
Example 1: $\frac{x^{2} + 2x}{6x + 3}$

First of all write the numerator and denominator as product of factors.

$\frac{x(x + 2)}{3(x + 2)}$

The polynomial $x + 2$ is common to both the numerator and denominator. We will cross them out.

$\frac{x (\not {x} + \not {2})}{3 (\not {x} + \not {2})}$ = $\frac{x}{3}$

We have to be careful while canceling out the common terms
Example 2: $\frac{x^{3} - x}{x^{2} - 5x + 4}$

Factorize the numerator and denominator.

$\frac{x^{3} - x}{x^{2} - 5x + 4}$ = $\frac{x(x^{2} - 1)}{(x - 4)(x - 1)}$

Let us factorize the numerator further.

$\frac{x^{3} - x}{x^{2} - 5x + 4}$ = $\frac{x(x - 1)(x + 1)}{(x - 4)(x - 1)}$

Here the binomial $x - 1$ is a common factor, cancelling it we get

$\frac{x^{3} - x}{x^{2} - 5x + 4}$ = $\frac{x(x + 1)}{(x - 4)}$
Example 3: Adding and subtracting rational expression.

Add $\frac{m - 3}{m^{2} + 5m + 6} + \frac{5}{m + 2}$

We treat the rational as fractions. While adding two rational expressions we need to have same denominators. Compute LCD (least common denominator) for the given expressions.

The denominators are $m^{2} + 5m + 6$ and $m + 2$.

To find common multiple let us first write factors for $m^{2} + 5m + 6$

$m^{2} + 5m + 6$ = $(m + 2)(m + 3)$

The common multiple would be $(m + 2)(m + 3)$.

We need to multiply the second expression by $\frac{(m+3)}{(m+3)}$.

So the rational expression would be

$\frac{m - 3}{m^{2} + 5m + 6}$ + $\frac{5}{m + 2}$ $\times$ $\frac{(m + 3)}{(m + 3)}$

= $\frac{m - 3}{(m + 2)(m + 3)} + \frac{5m + 15}{(m + 2)(m + 3)}$

= $\frac{m - 3 + 5m + 15}{(m + 2)(m + 3)}$

= $\frac{6m + 12}{(m + 2)(m + 3)}$

Let us check if we can simplify it further.

= $\frac{6(m + 12)}{(m + 2)(m + 3)}$

Cancelling $m + 2$ will give the final answer in reduced form.

= $\frac{6}{(m + 3)}$
Subtract:

$\frac{x}{x - 2}$ - $\frac{1}{x - 5}$

Get LCD,

LCD = $(x-2)(x-5)$

$\frac{x}{x - 2}$ $\times$ $\frac{x - 5}{x - 5}$ - $\frac{1}{x - 5}$ $\times$ $\frac{x - 2}{x - 2}$

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

= $\frac{x^{2} - 5x - x + 2}{(x - 2)(x - 5)}$

= $\frac{x^{2} - 6x + 2}{(x - 2)(x - 5)}$