**Adding and subtracting rational expressions** is very similar to adding and subtracting rational numbers, but one needs be a little bit more careful about the common denominator and what you multiply with what.

Example 1. Solve $\frac{x + 1}{x} + 1$.

The common denominator of these two numbers is obviously x.

$\frac{x + 1}{x} + 1 = \frac{x + 1 + x}{x} = \frac{2x + 1}{x}$

Example 2. Solve $\frac{1 + x}{1 – x} + \frac{1 – x}{1 + x}$.

Denominators of these expressions have no common factors, so their common denominator will be their product.

$\frac{1 + x}{1 – x} + \frac{1 – x}{1 + x} = \frac{(1 + x)(1 + x) + (1 – x)(1 – x)}{(1 – x)(1 + x)} = \frac{(1 + x)(1 + x) + (1 – x)(1 – x)}{(1 – x)(1 + x)}$

Example 3. Solve $\frac{1 + x^2}{1 – x^2} – \frac{1 – x}{1 + x}$.

First, we’ll factorize each of the denominators in order to conclude whether these two have common factors or not.

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

These two denominators have common factor 1 + x. this means that the common denominator will be 1 – x^2 because it contains both of the denominators.

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

Example 4. Solve $\frac{8 – b}{3} – \frac{2 + b}{3b} + \frac{1}{b}$.

The procedure of subtracting or adding three or more rational expressions is the same as that for two. If this will be too messy for you and you don’t want to mess with this, you can also add two of these and then add the third.

$\frac{8 – b}{3} – \frac{2 + b}{3b} + \frac{1}{b} = \frac{8b – b^2 – 2 – b + 3}{3b} = \frac{- b^2 + 7b – 1}{3b}$