1. **Stating the problem:** We want to learn how to add and subtract algebraic fractions, which are fractions that contain variables in the numerator, denominator, or both. 2. **Formula and rules:** To add or subtract algebraic fractions, the key is to have a common denominator. The general formula is: $$\frac{A}{B} \pm \frac{C}{D} = \frac{AD \pm BC}{BD}$$ where $A$, $B$, $C$, and $D$ are algebraic expressions and $B \neq 0$, $D \neq 0$. 3. **Important rules:** - Find the least common denominator (LCD) of the fractions. - Rewrite each fraction with the LCD as the denominator. - Add or subtract the numerators. - Simplify the resulting fraction if possible. 4. **Example:** Add $\frac{2}{x}$ and $\frac{3}{x+1}$. - Step 1: Find the LCD, which is $x(x+1)$. - Step 2: Rewrite each fraction: $$\frac{2}{x} = \frac{2(x+1)}{x(x+1)}$$ $$\frac{3}{x+1} = \frac{3x}{x(x+1)}$$ - Step 3: Add the numerators: $$\frac{2(x+1)}{x(x+1)} + \frac{3x}{x(x+1)} = \frac{2(x+1) + 3x}{x(x+1)}$$ - Step 4: Simplify the numerator: $$2(x+1) + 3x = 2x + 2 + 3x = 5x + 2$$ - Final answer: $$\frac{5x + 2}{x(x+1)}$$ 5. **Example:** Subtract $\frac{5}{x-2}$ from $\frac{3}{x+3}$. - Step 1: Find the LCD, which is $(x-2)(x+3)$. - Step 2: Rewrite each fraction: $$\frac{3}{x+3} = \frac{3(x-2)}{(x-2)(x+3)}$$ $$\frac{5}{x-2} = \frac{5(x+3)}{(x-2)(x+3)}$$ - Step 3: Subtract the numerators: $$\frac{3(x-2)}{(x-2)(x+3)} - \frac{5(x+3)}{(x-2)(x+3)} = \frac{3(x-2) - 5(x+3)}{(x-2)(x+3)}$$ - Step 4: Simplify the numerator: $$3(x-2) - 5(x+3) = 3x - 6 - 5x - 15 = -2x - 21$$ - Final answer: $$\frac{-2x - 21}{(x-2)(x+3)}$$