Math Core

To add $\dfrac{1}{4} + \dfrac{1}{6}$ you rewrite both fractions over a common denominator, $12$ . Adding rational expressions is the same process with polynomial denominators, and it is the key skill for solving rational equations in the next lesson.

Like denominators

When the denominators already match, add or subtract the numerators and keep the denominator:

\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}, \qquad \frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}.

Then simplify if you can.

Worked example: Same denominator

Simplify $\dfrac{3x}{x^2 - 4} - \dfrac{x + 4}{x^2 - 4}$ .

Subtract the entire second numerator. Parentheses keep the signs straight:

\frac{3x - (x + 4)}{x^2 - 4} = \frac{2x - 4}{x^2 - 4}.

Now factor and cancel:

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

Common mistake

When subtracting, the minus sign applies to every term of the second numerator. Writing $3x - x + 4$ instead of $3x - x - 4$ is the single most common error in this topic. Always put the second numerator in parentheses before you combine.

Finding the least common denominator

With numbers, the least common denominator (LCD) of $\dfrac{1}{12}$ and $\dfrac{1}{18}$ comes from prime factorizations: $12 = 2^2 \cdot 3$ and $18 = 2 \cdot 3^2$ , so the LCD is $2^2 \cdot 3^2 = 36$ . Polynomials work the same way, with factors playing the role of primes.

Finding the LCD

Factor each denominator completely.
List every different factor that appears.
Raise each factor to the highest power it has in any one denominator.
The LCD is the product of those factors.

For example, the denominators $x^2 - 4 = (x - 2)(x + 2)$ and $x^2 + 4x + 4 = (x + 2)^2$ have LCD $(x - 2)(x + 2)^2$ . The factor $(x + 2)$ appears squared in the second denominator, so it is squared in the LCD.

Unlike denominators

To add fractions with different denominators, rewrite each one over the LCD by multiplying its numerator and denominator by whatever factors it is missing. Then combine numerators.

Worked example: Two linear denominators

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

The LCD is $(x + 1)(x - 2)$ . The first fraction is missing $(x - 2)$ ; the second is missing $(x + 1)$ :

\begin{aligned} \frac{2}{x + 1} + \frac{3}{x - 2} &= \frac{2(x - 2)}{(x + 1)(x - 2)} + \frac{3(x + 1)}{(x + 1)(x - 2)} \\ &= \frac{2x - 4 + 3x + 3}{(x + 1)(x - 2)} \\ &= \frac{5x - 1}{(x + 1)(x - 2)}. \end{aligned}

The numerator $5x - 1$ doesn't share a factor with the denominator, so this is fully simplified. It's usual to leave the denominator factored.

Worked example: Factor first, then subtract

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

Factor the first denominator: $x^2 - x - 2 = (x - 2)(x + 1)$ . That already contains $(x - 2)$ , so the LCD is $(x - 2)(x + 1)$ . Only the second fraction needs rewriting:

\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{-x + 2}{(x - 2)(x + 1)}. \end{aligned}

The numerator $-x + 2 = -(x - 2)$ cancels with the denominator:

= \frac{-1}{x + 1}, \qquad x \ne 2, -1.

Always check the final numerator for a factor that cancels. It happens more often than you'd expect.

Complex fractions

A complex fraction has fractions inside its numerator or denominator. One clean method: multiply the top and bottom by the LCD of all the small fractions. That clears every inner denominator at once.

Worked example: Simplifying a complex fraction

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

The inner denominators are $x$ and $x^2$ , so multiply top and bottom by $x^2$ :

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

Tip

Check a sum by plugging in an allowed number. For $\dfrac{2}{x + 1} + \dfrac{3}{x - 2}$ at $x = 3$ : the original gives $\dfrac{2}{4} + \dfrac{3}{1} = 3.5$ , and $\dfrac{5(3) - 1}{(4)(1)} = \dfrac{14}{4} = 3.5$ . Match.

Practice

Practice 1

Simplify $\dfrac{3}{x} + \dfrac{5}{x}$ .

Practice 2

What is the least common denominator of $\dfrac{1}{x^2 - 9}$ and $\dfrac{1}{x^2 + 6x + 9}$ ?

Practice 3

Simplify $\dfrac{4}{x + 3} + \dfrac{1}{x - 1}$ .

Practice 4

The difference $\dfrac{x}{x - 3} - \dfrac{3}{x - 3}$ simplifies to a constant for every allowed $x$ . What is the constant?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Simplify $\dfrac{x + 12}{x^2 - x - 6} - \dfrac{3}{x - 3}$ .

Practice 6

Simplify the complex fraction $\dfrac{\dfrac{1}{x} - \dfrac{1}{4}}{x - 4}$ .

Practice 7

One hose fills a tank in $3$ hours, so it fills $\dfrac{1}{3}$ of the tank per hour. A second hose fills it in $6$ hours. Working together, what fraction of the tank do they fill per hour?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.