Add and Subtract Rational Expressions with Unlike Denominators

By the end of this section, you will be able to: find the least common denominator of rational expressions, find equivalent rational expressions, add rational expressions with different denominators, and subtract rational expressions with different denominators.

Find the least common denominator of rational expressions

When we add or subtract rational expressions with unlike denominators, we first need to get a common denominator. Reviewing how we did this with numerical fractions tells us exactly what to do with rational expressions.

Look at the numerical example $\tfrac{7}{12} + \tfrac{5}{18}$ . Because the denominators are not the same, the first step is to find the least common denominator (LCD) — the least common multiple of the denominators, the smallest number we can use as a common denominator.

To find the LCD of $12$ and $18$ , we factor each number into primes, lining up any common primes in columns. Then we “bring down” one prime from each column and multiply the factors:

\begin{array}{rcl} 12 &=& 2 \cdot 2 \cdot 3 \\ 18 &=& 2 \cdot 3 \cdot 3 \\[4pt] \text{LCD} &=& 2 \cdot 2 \cdot 3 \cdot 3 \\ \text{LCD} &=& 36 \end{array}

We do the same thing for rational expressions. However, we leave the LCD in factored form.

How to find the least common denominator of rational expressions.

Factor each expression completely.
List the factors of each expression. Match factors vertically when possible.
Bring down the columns.
Multiply the factors.

Remember, we always exclude values that would make a denominator zero.

Example. Find the LCD for $\tfrac{8}{x^2 - 2x - 3}$ and $\tfrac{3x}{x^2 + 4x + 3}$ .

Factor each denominator completely, lining up common factors. Then bring down the columns and multiply:

\begin{array}{rcl} x^2 - 2x - 3 &=& (x + 1)(x - 3) \\ x^2 + 4x + 3 &=& (x + 1)(x + 3) \\[4pt] \text{LCD} &=& (x + 1)(x - 3)(x + 3) \end{array}

The LCD is $(x + 1)(x - 3)(x + 3)$ .

Find the LCD for $\tfrac{2}{x^2 - x - 12}$ and $\tfrac{1}{x^2 - 16}$ . Leave your answer in factored form.

Find equivalent rational expressions

When we add numerical fractions, once we find the LCD we rewrite each fraction as an equivalent fraction with the LCD:

\frac{7}{12} + \frac{5}{18} = \frac{7 \cdot 3}{12 \cdot 3} + \frac{5 \cdot 2}{18 \cdot 2} = \frac{21}{36} + \frac{10}{36}

We do the same thing for rational expressions. To rewrite one with a new denominator, we multiply its numerator and denominator by whatever “missing” factors the LCD supplies.

Example. Rewrite $\tfrac{8}{x^2 - 2x - 3}$ and $\tfrac{3x}{x^2 + 4x + 3}$ as equivalent rational expressions with denominator $(x + 1)(x - 3)(x + 3)$ .

Factor each denominator, then multiply the numerator and denominator by the factor the LCD supplies that is missing from that denominator. Finally simplify each numerator:

\begin{array}{rcl} \tfrac{8}{x^2 - 2x - 3} &=& \tfrac{8}{(x + 1)(x - 3)} = \tfrac{8(x + 3)}{(x + 1)(x - 3)(x + 3)} = \tfrac{8x + 24}{(x + 1)(x - 3)(x + 3)} \\[8pt] \tfrac{3x}{x^2 + 4x + 3} &=& \tfrac{3x}{(x + 1)(x + 3)} = \tfrac{3x(x - 3)}{(x + 1)(x + 3)(x - 3)} = \tfrac{3x^2 - 9x}{(x + 1)(x - 3)(x + 3)} \end{array}

Rewrite $\tfrac{5}{x^2 - 2x - 8}$ with the LCD $(x - 4)(x + 2)(x + 3)$ . Enter just the new numerator.

Add rational expressions with different denominators

Now we have all the steps we need to add rational expressions with different denominators.

How to add rational expressions.

Determine if the expressions have a common denominator. If not, find the LCD and rewrite each rational expression as an equivalent rational expression with the LCD.
Add the rational expressions.
Simplify, if possible.

We start by adding expressions whose denominators are monomials.

Example. Add $\tfrac{5}{12x^2 y} + \tfrac{4}{21xy^2}$ .

The denominators are not the same, so find the LCD. Factor each denominator, line up common factors, and multiply:

\begin{array}{rcl} 12x^2 y &=& 2 \cdot 2 \cdot 3 \cdot x \cdot x \cdot y \\ 21xy^2 &=& 3 \cdot 7 \cdot x \cdot y \cdot y \\[4pt] \text{LCD} &=& 2 \cdot 2 \cdot 3 \cdot 7 \cdot x \cdot x \cdot y \cdot y = 84x^2 y^2 \end{array}

Rewrite each rational expression with the LCD, then add the numerators over the common denominator:

\begin{array}{rcl} \tfrac{5}{12x^2 y} + \tfrac{4}{21xy^2} &=& \tfrac{5 \cdot 7y}{12x^2 y \cdot 7y} + \tfrac{4 \cdot 4x}{21xy^2 \cdot 4x} \\[6pt] &=& \tfrac{35y}{84x^2 y^2} + \tfrac{16x}{84x^2 y^2} \\[6pt] &=& \tfrac{16x + 35y}{84x^2 y^2} \end{array}

There are no factors common to the numerator and denominator, so the fraction cannot be simplified.

Now we are ready to tackle polynomial denominators.

Example. Add $\tfrac{3}{x - 3} + \tfrac{2}{x - 2}$ .

The expressions do not have a common denominator. The denominators $x - 3$ and $x - 2$ are already prime, so the LCD is $(x - 3)(x - 2)$ . Rewrite each expression with the LCD — keep the denominators factored — then add the numerators and simplify:

\begin{array}{rcl} \tfrac{3}{x - 3} + \tfrac{2}{x - 2} &=& \tfrac{3(x - 2)}{(x - 3)(x - 2)} + \tfrac{2(x - 3)}{(x - 2)(x - 3)} \\[6pt] &=& \tfrac{3x - 6}{(x - 3)(x - 2)} + \tfrac{2x - 6}{(x - 3)(x - 2)} \\[6pt] &=& \tfrac{3x - 6 + 2x - 6}{(x - 3)(x - 2)} \\[6pt] &=& \tfrac{5x - 12}{(x - 3)(x - 2)} \end{array}

Because $5x - 12$ cannot be factored, the answer is simplified.

Add: $\tfrac{2}{x - 2} + \tfrac{5}{x + 3}$ .

The next example has polynomial denominators that factor. Factor first, find the LCD, and be careful when a numerator does not simplify.

Example. Add $\tfrac{8}{x^2 - 2x - 3} + \tfrac{3x}{x^2 + 4x + 3}$ .

Factor each denominator: $x^2 - 2x - 3 = (x + 1)(x - 3)$ and $x^2 + 4x + 3 = (x + 1)(x + 3)$ , so the LCD is $(x + 1)(x - 3)(x + 3)$ . Rewrite each expression with the LCD, add the numerators, and simplify:

\begin{array}{rcl} \tfrac{8}{x^2 - 2x - 3} + \tfrac{3x}{x^2 + 4x + 3} &=& \tfrac{8(x + 3)}{(x + 1)(x - 3)(x + 3)} + \tfrac{3x(x - 3)}{(x + 1)(x + 3)(x - 3)} \\[6pt] &=& \tfrac{8x + 24}{(x + 1)(x - 3)(x + 3)} + \tfrac{3x^2 - 9x}{(x + 1)(x + 3)(x - 3)} \\[6pt] &=& \tfrac{3x^2 - x + 24}{(x + 1)(x - 3)(x + 3)} \end{array}

The numerator $3x^2 - x + 24$ is prime, so there are no common factors and the answer is simplified.

Subtract rational expressions with different denominators

The process we use to subtract rational expressions with different denominators is the same as for addition. We just have to be very careful of the signs when subtracting the numerators.

How to add or subtract rational expressions.

Determine if the expressions have a common denominator. If not, find the LCD and rewrite each rational expression as an equivalent rational expression with the LCD.
Add or subtract the rational expressions.
Simplify, if possible.

Example. Subtract $\tfrac{x}{x - 3} - \tfrac{x - 2}{x + 3}$ .

The denominators $x - 3$ and $x + 3$ are prime, so the LCD is $(x - 3)(x + 3)$ . Rewrite each expression with the LCD, then subtract the numerators — distributing the sign carefully — and simplify:

\begin{array}{rcl} \tfrac{x}{x - 3} - \tfrac{x - 2}{x + 3} &=& \tfrac{x(x + 3)}{(x - 3)(x + 3)} - \tfrac{(x - 2)(x - 3)}{(x + 3)(x - 3)} \\[6pt] &=& \tfrac{x^2 + 3x}{(x - 3)(x + 3)} - \tfrac{x^2 - 5x + 6}{(x - 3)(x + 3)} \\[6pt] &=& \tfrac{x^2 + 3x - (x^2 - 5x + 6)}{(x - 3)(x + 3)} \\[6pt] &=& \tfrac{x^2 + 3x - x^2 + 5x - 6}{(x - 3)(x + 3)} \\[6pt] &=& \tfrac{8x - 6}{(x - 3)(x + 3)} = \tfrac{2(4x - 3)}{(x - 3)(x + 3)} \end{array}

The numerator and denominator have no factors in common, so the answer is simplified.

The next example has a denominator that factors, letting the final fraction simplify.

Example. Subtract $\tfrac{8y}{y^2 - 16} - \tfrac{4}{y - 4}$ .

Factor $y^2 - 16 = (y - 4)(y + 4)$ , so the LCD is $(y - 4)(y + 4)$ . Rewrite the second fraction with the LCD, subtract the numerators, then factor and remove the common factor:

\begin{array}{rcl} \tfrac{8y}{y^2 - 16} - \tfrac{4}{y - 4} &=& \tfrac{8y}{(y - 4)(y + 4)} - \tfrac{4(y + 4)}{(y - 4)(y + 4)} \\[6pt] &=& \tfrac{8y - (4y + 16)}{(y - 4)(y + 4)} \\[6pt] &=& \tfrac{4y - 16}{(y - 4)(y + 4)} \\[6pt] &=& \tfrac{4(y - 4)}{(y - 4)(y + 4)} \\[6pt] &=& \tfrac{4}{y + 4} \end{array}

When one expression is not in fraction form, we can write it as a fraction with denominator $1$ .

Example. Subtract $\tfrac{5c + 4}{c - 2} - 3$ .

Write $3$ as $\tfrac{3}{1}$ so there are two rational expressions. The LCD of $c - 2$ and $1$ is $c - 2$ ; rewrite $\tfrac{3}{1}$ with that denominator, subtract, and check for common factors:

\begin{array}{rcl} \tfrac{5c + 4}{c - 2} - 3 &=& \tfrac{5c + 4}{c - 2} - \tfrac{3(c - 2)}{c - 2} \\[6pt] &=& \tfrac{5c + 4 - (3c - 6)}{c - 2} \\[6pt] &=& \tfrac{2c + 10}{c - 2} = \tfrac{2(c + 5)}{c - 2} \end{array}

There are no common factors, so the rational expression is simplified.

Subtract: $\tfrac{2x}{x^2 - 4} - \tfrac{1}{x + 2}$ .

Key terms

least common denominator (LCD) — the least common multiple of the denominators of two or more rational expressions; found by factoring each denominator completely and multiplying together each factor the greatest number of times it appears in any one denominator. equivalent rational expression — a rational expression rewritten with a new denominator by multiplying its numerator and denominator by the same nonzero factor, so its value is unchanged.

This section is adapted from Elementary Algebra 2e, Section 8.4: Add and Subtract Rational Expressions with Unlike Denominators by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recast the worked examples as prose with display equality chains, condensed the numerical warm-ups and the “How To” procedures, and merged the separate add/subtract procedure boxes into one; omitted the Be Prepared quiz, Self Check checklist, media links, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.

Add and Subtract Rational Expressions with a Common Denominator Simplify Complex Rational Expressions