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# 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 $\frac{2x}{3-5x}$ and $\frac{9}{3-5x}$.

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

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

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{1-8x}{{x}^{2}-4}-\frac{7x+1}{{x}^{2}-4}=$$\frac{\left(1-8x\right)-\left(7x+1\right)}{{x}^{2}-4}=$$\frac{1-8x-7x-1}{{x}^{2}-4}=$$\frac{-15x}{{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}$$=\frac{12}{\left(x+3\right)\left(x-3\right)}+\frac{2}{x+3}$$=\frac{12+2·\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}$$=\frac{12+2x-6}{\left(x+3\right)\left(x-3\right)}$$=\frac{2x+6}{\left(x+3\right)\left(x-3\right)}$$=\frac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}$$=\frac{2}{\left(x-3\right)}$

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

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

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

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

$=\frac{5x·\left(x-1\right)+\left(x+1\right)-x·\left(x+3\right)}{\left(x+3\right)\left(x-1\right)}$

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

## 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 $\frac{{f}_{1}\left(x\right)}{{g}_{1}\left(x\right)}$ from $\frac{{f}_{2}\left(x\right)}{{g}_{2}\left(x\right)}$ means $\frac{{f}_{2}\left(x\right)}{{g}_{2}\left(x\right)}-\frac{{f}_{1}\left(x\right)}{{g}_{1}\left(x\right)}$.