Algebra 2 - Adding and Subtracting Rational Expressions

Introduction

Addition and subtraction of rational expressions can only be performed if they have common denominators.
If the two rational expressions to be added don't have a common denominator, we need to modify the rational expressions so that equivalent rational expressions have the same denominators.

Adding & Subtracting Rational Expressions

For simplicity's sake, we classify rational expressions' addition and subtraction into two cases.

1. When rational expressions have a common (same) denominator:

Keep the denominator as it is, and add or subtract the numerators.
Simplify (cancel out common factors) the remaining expression if possible.

2. When rational expressions have different denominators:

Find the LCM (least common multiple) of all the denominators.
Modify the rational expression into an equivalent rational expression with the same denominator (as we do while adding or subtracting fractions with unlike denominators).
Simplify the resulting rational expression if possible.

Solved Examples

Example 1: Add $\frac{2 x}{3 - 5 x}$ and $\frac{9}{3 - 5 x}$ .

Solution: Here, we see that the denominators of the rational expressions to be added are the same. Hence, we can directly add the numerator terms.

Thus, $\frac{2 x}{3 - 5 x} +$ $\frac{9}{3 - 5 x}$ $= \frac{2 x + 9}{3 - 5 x}$

Example 2: Subtract $\frac{7 x + 1}{x^{2} - 4}$ from $\frac{1 - 8 x}{x^{2} - 4}$ .

Solution: In this problem, the denominators of the rational expressions to be subtracted are the same. Hence, we can directly perform subtraction operations.

Thus, $\frac{1 - 8 x}{x^{2} - 4} - \frac{7 x + 1}{x^{2} - 4} =$ $\frac{(1 - 8 x) - (7 x + 1)}{x^{2} - 4} =$ $\frac{1 - 8 x - 7 x - 1}{x^{2} - 4} =$ $\frac{- 15 x}{x^{2} - 4}$

Example 3: Simplify $\frac{12}{x^{2} - 9} + \frac{2}{x + 3}$ .

Solution: $\frac{12}{x^{2} - 9} + \frac{2}{x + 3}$ $= \frac{12}{(x + 3) (x - 3)} + \frac{2}{x + 3}$ $= \frac{12 + 2 \cdot (x - 3)}{(x + 3) (x - 3)}$ $= \frac{12 + 2 x - 6}{(x + 3) (x - 3)}$ $= \frac{2 x + 6}{(x + 3) (x - 3)}$ $= \frac{2 (x + 3)}{(x + 3) (x - 3)}$ $= \frac{2}{(x - 3)}$

Example 4: Simplify $\frac{x - 1}{x} - \frac{y}{y - 1}$ .

Solution: $\frac{x - 1}{x} - \frac{y}{y - 1}$ $= \frac{(x - 1) \cdot (y - 1) - x y}{x \cdot (y - 1)}$ $= \frac{x y - x - y + 1 - x y}{x \cdot (y - 1)}$ $= \frac{- x - y + 1}{x \cdot (y - 1)}$ $= \frac{1 - (x + y)}{x \cdot (y - 1)}$

Example 5: Simplify $\frac{5 x}{(x + 3)} + \frac{x + 1}{x^{2} + 2 x - 3} - \frac{x}{(x - 1)}$

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

$= \frac{5 x \cdot (x - 1) + (x + 1) - x \cdot (x + 3)}{(x + 3) (x - 1)}$

$= \frac{5 x^{2} - 5 x + x + 1 - x^{2} - 3 x}{(x + 3) (x - 1)}$ $= \frac{4 x^{2} - 7 x + 1}{(x + 3) (x - 1)}$

Cheat Sheet

To add or subtract rational expressions, they must have common denominators.
If the denominators of rational expressions to be added are different, then we must first express them into an equivalent rational expression with common denominators.

Blunder Areas

Subtracting $\frac{f_{1} (x)}{g_{1} (x)}$ from $\frac{f_{2} (x)}{g_{2} (x)}$ means $\frac{f_{2} (x)}{g_{2} (x)} - \frac{f_{1} (x)}{g_{1} (x)}$ .

Abhishek Tiwari
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