Adding and Subtracting Rational Expressions

To add or subtract fractions, you need a common denominator, and rational expressions are no different. This is usually the hardest of the rational expression skills, because it combines factoring, expanding, and careful sign work, so take it one step at a time.

Key ideas

Start with numbers

$$\frac{1}{6} + \frac{3}{4} = \frac{2}{12} + \frac{9}{12} = \frac{11}{12}$$

The lowest common denominator (LCD) is $12$, the smallest number both $6$ and $4$ divide into. Each fraction is rewritten with that denominator, then the numerators are added.

Same denominator

Add or subtract the numerators and keep the denominator:

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

Put the second numerator in brackets when subtracting, so the minus sign reaches every term.

Different denominators

1. Factor each denominator, and state the restrictions.
2. Find the LCD: use every factor that appears, each as many times as it appears in any one denominator.
3. Rewrite each fraction over the LCD by multiplying its numerator and denominator by the missing factors.
4. Combine the numerators (brackets for subtraction), then expand and simplify the numerator.
5. Factor the numerator if possible and cancel any common factor.

For example, for denominators $x(x + 1)$ and $(x + 1)^2$, the LCD is $x(x + 1)^2$.

Worked examples

Example 1: Monomial denominators

Simplify $\dfrac{2}{3x} + \dfrac{5}{6x^2}$.

Solution. The LCD is $6x^2$. The first fraction needs to be multiplied by $\dfrac{2x}{2x}$:

$$\frac{2}{3x} + \frac{5}{6x^2} = \frac{4x}{6x^2} + \frac{5}{6x^2} = \frac{4x + 5}{6x^2}, \qquad x \ne 0$$

Example 2: Binomial denominators

Simplify $\dfrac{3}{x + 2} - \dfrac{1}{x - 1}$.

Solution. The LCD is $(x + 2)(x - 1)$.

$$\begin{aligned} \frac{3}{x + 2} - \frac{1}{x - 1} &= \frac{3(x - 1)}{(x + 2)(x - 1)} - \frac{1(x + 2)}{(x + 2)(x - 1)} \\ &= \frac{3x - 3 - x - 2}{(x + 2)(x - 1)} \\ &= \frac{2x - 5}{(x + 2)(x - 1)}, \qquad x \ne -2, 1 \end{aligned}$$

Example 3: Factor the denominators first

Simplify $\dfrac{x}{x^2 - 9} + \dfrac{2}{x + 3}$.

Solution. $x^2 - 9 = (x - 3)(x + 3)$, so the LCD is $(x - 3)(x + 3)$. Only the second fraction needs changing:

$$\begin{aligned} \frac{x}{(x - 3)(x + 3)} + \frac{2(x - 3)}{(x - 3)(x + 3)} &= \frac{x + 2x - 6}{(x - 3)(x + 3)} \\ &= \frac{3x - 6}{(x - 3)(x + 3)} \\ &= \frac{3(x - 2)}{(x - 3)(x + 3)}, \qquad x \ne 3, -3 \end{aligned}$$

Example 4: Simplify at the end

Simplify $\dfrac{x + 4}{x^2 - x - 2} - \dfrac{2}{x - 2}$.

Solution. $x^2 - x - 2 = (x - 2)(x + 1)$. The LCD is $(x - 2)(x + 1)$, and the restrictions are $x \ne 2, -1$.

$$\begin{aligned} \frac{x + 4}{(x - 2)(x + 1)} - \frac{2(x + 1)}{(x - 2)(x + 1)} &= \frac{(x + 4) - (2x + 2)}{(x - 2)(x + 1)} \\ &= \frac{2 - x}{(x - 2)(x + 1)} \end{aligned}$$

The numerator $2 - x$ is the opposite of $x - 2$, so it cancels to $-1$:

$$\frac{-(x - 2)}{(x - 2)(x + 1)} = \frac{-1}{x + 1}, \qquad x \ne 2, -1$$

Common mistakes

Adding the denominators. $\dfrac{1}{x} + \dfrac{1}{y}$ is not $\dfrac{2}{x + y}$. You need a common denominator: $\dfrac{y + x}{xy}$.

Not distributing the subtraction. In Example 4, $(x + 4) - (2x + 2) = -x + 2$. Writing $x + 4 - 2x + 2$ gives the wrong answer.

Forgetting to multiply the numerator. When you multiply a denominator by a missing factor, multiply its numerator by the same factor.

Stopping too early. Always check whether the final numerator factors and cancels, as in Example 4.

Cancelling terms across the plus sign. In $\dfrac{x + 2}{x + 5}$, nothing cancels.

Losing restrictions. In Example 4, $x \ne 2$ is still a restriction even though $(x - 2)$ cancelled.

Practice

1. (Warm-up) Simplify $\dfrac{3}{x} + \dfrac{7}{x}$.

Solution

$$\frac{10}{x}, \qquad x \ne 0$$

2. (Warm-up) Simplify $\dfrac{2x + 1}{x - 4} - \dfrac{x - 3}{x - 4}$.

Solution

$$\frac{(2x + 1) - (x - 3)}{x - 4} = \frac{2x + 1 - x + 3}{x - 4} = \frac{x + 4}{x - 4}, \qquad x \ne 4$$

3. (Warm-up) Simplify $\dfrac{1}{2a} + \dfrac{3}{5a}$.

Solution

The LCD is $10a$:

$$\frac{5}{10a} + \frac{6}{10a} = \frac{11}{10a}, \qquad a \ne 0$$

4. (Core) Simplify $\dfrac{4}{x + 1} + \dfrac{3}{x - 2}$.

Solution

$$\frac{4(x - 2) + 3(x + 1)}{(x + 1)(x - 2)} = \frac{4x - 8 + 3x + 3}{(x + 1)(x - 2)} = \frac{7x - 5}{(x + 1)(x - 2)}, \qquad x \ne -1, 2$$

5. (Core) Simplify $\dfrac{5}{x - 3} - \dfrac{2}{x + 4}$.

Solution

$$\frac{5(x + 4) - 2(x - 3)}{(x - 3)(x + 4)} = \frac{5x + 20 - 2x + 6}{(x - 3)(x + 4)} = \frac{3x + 26}{(x - 3)(x + 4)}, \qquad x \ne 3, -4$$

6. (Core) Simplify $\dfrac{2}{x^2 - 1} + \dfrac{1}{x + 1}$.

Solution

The LCD is $(x - 1)(x + 1)$:

$$\frac{2 + (x - 1)}{(x - 1)(x + 1)} = \frac{x + 1}{(x - 1)(x + 1)} = \frac{1}{x - 1}, \qquad x \ne 1, -1$$

7. (Core) Simplify $\dfrac{x}{x^2 + 5x + 6} - \dfrac{2}{x^2 + 4x + 3}$.

Solution

Factor: $x^2 + 5x + 6 = (x + 2)(x + 3)$ and $x^2 + 4x + 3 = (x + 1)(x + 3)$. The LCD is $(x + 1)(x + 2)(x + 3)$.

$$\begin{aligned} \frac{x(x + 1) - 2(x + 2)}{(x + 1)(x + 2)(x + 3)} &= \frac{x^2 + x - 2x - 4}{(x + 1)(x + 2)(x + 3)} \\ &= \frac{x^2 - x - 4}{(x + 1)(x + 2)(x + 3)}, \qquad x \ne -1, -2, -3 \end{aligned}$$

The numerator doesn’t factor (no two integers multiply to $-4$ and add to $-1$), so this is fully simplified.

8. (Challenge) Simplify $\dfrac{1}{x} + \dfrac{1}{x + 1} - \dfrac{2}{x(x + 1)}$.

Solution

The LCD is $x(x + 1)$:

$$\frac{(x + 1) + x - 2}{x(x + 1)} = \frac{2x - 1}{x(x + 1)}, \qquad x \ne 0, -1$$

9. (Challenge) A cyclist rides $30$ km at a speed of $v$ km/h, then $20$ km at $(v + 5)$ km/h.

• (a) Write a single simplified expression for the total time.
• (b) Find the total time when $v = 20$.

Solution

(a) Time = distance ÷ speed:

$$\frac{30}{v} + \frac{20}{v + 5} = \frac{30(v + 5) + 20v}{v(v + 5)} = \frac{50v + 150}{v(v + 5)} = \frac{50(v + 3)}{v(v + 5)}$$

(b) $\dfrac{50(23)}{20(25)} = \dfrac{1150}{500} = 2.3$ hours. Check: $\dfrac{30}{20} + \dfrac{20}{25} = 1.5 + 0.8 = 2.3$. ✓