Students learn how to add and subtract whole numbers before they are taught multiplication and division. However, with fractions and rational expressions, multiplication and division are sometimes taught first because these operations are easier to perform than addition and subtraction. Addition and subtraction of rational expressions are not as easy to perform as multiplication because, as with fractions, the process involves finding common denominators.

Adding and Subtracting Rational Expressions with Like Denominators

Recall that to add or subtract fractions with like denominators, we keep the common denominator and perform the operations on the numerators. Then we simplify the final answer if possible. For example,

[latex]\require{color}\displaystyle\frac{2}{9}+\frac{5}{9}-\frac{4}{9} = \frac{2+5-4}{9} = \frac{3}{9} = \frac{{\color{red}\cancel{\color{black}{3}}}\cdot 1}{{\color{red}\cancel{\color{black}{3}}}\cdot 3} = \frac{1}{3}[/latex]

We proceed similarly with rational expressions with like denominators, remembering to identify the restricted values of the variable in the expressions.

The Least Common Multiple of Two or More Polynomials

When we add or subtract fractions with unlike denominators, the first step is to identify a common denominator. The most convenient one is the least common denominator which is the least common multiple (LCM) of all denominators. Recall that to find the LCM of a collection of whole numbers, we first find the prime factorization of each number (in concise form using exponents), and then note all distinct prime factors of all the numbers in the collection. The LCM is a product of all prime factors with each factor raised to the highest power that appears in all factorizations of the numbers in the collection. For example, for the [latex] LCM (6,15,20) [/latex], we first present [latex] 6 [/latex] as [latex] 2\cdot 3 = 2^1\cdot 3^1 [/latex], [latex] 15 [/latex] as [latex] 3\cdot 5 = 3^1\cdot 5^1 [/latex], and [latex] 20 [/latex] as [latex] 2^2\cdot 5 = 2^2\cdot 5^1 [/latex]. We note that [latex] 2 [/latex], [latex] 3 [/latex], and [latex] 5 [/latex] are all distinct prime factors of all the numbers in the collection with the highest power of [latex] 2 [/latex] in the factorizations being [latex] 2 [/latex], and the highest powers of [latex] 3 [/latex] and [latex] 5 [/latex] in the factorizations being [latex] 1 [/latex]. Hence, [latex] LCM (6,15,20) = 2^2\cdot 3^1\cdot 5^1 = 60 [/latex].

We will also start with finding the least common denominator when adding or subtracting rational expressions. The procedure of finding the LCM of a collection of polynomials is analogous to the one used for a collection of whole numbers. We first find the prime factorization of each polynomial (in concise form using exponents), and then note all distinct prime factors of all the polynomials in the collection. The LCM is a product of all prime factors with each factor raised to the highest power that appears in all factorizations of the polynomials in the collection. Consider the following examples.

You might have noticed that we didn’t multiply out and simplify the answers in the examples above. The factored form will be best for possible simplifications and unless specified otherwise, we will leave the numerator and denominator of the final answer for adding and subtracting rational expressions in factored form.

Adding and Subtracting Rational Expressions with Unlike Denominators

We’re ready now to look into the general case of adding and subtracting rational expressions. Let’s start with another example of operations on fractions to review what we do once we find the least common denominator. The idea is to rewrite each fraction in equivalent form with that least common denominator. To do that, we multiply each fraction that does not have the common denominator by a specific “form of 1.” This form of 1 is a fraction having identical numerator and denominator consisting of the factors missing when compared to the least common denominator. This form of 1 is typically different for each fraction involved. For example, to compute

[latex] \displaystyle\frac{1}{15}-\frac{1}{6}+\frac{1}{20} [/latex]

we use the [latex] LCM(6,15,20) [/latex] of [latex] 60=2\cdot 2\cdot 3\cdot 5 [/latex] that we computed earlier in this section to rewrite each fraction in equivalent form with [latex] 60 [/latex] as the denominator. The factored form of the least common denominator (LCD) together with factored forms of each denominator in our fractions will help identify the missing factors. Since [latex] 15 = 3\cdot 5 [/latex], the denominator of [latex] \dfrac{1}{15} [/latex] is missing the factors [latex] 2\cdot 2 [/latex] of the LCD. Hence, we multiply [latex] \dfrac{1}{15} [/latex] by [latex] \dfrac{2\cdot 2}{2\cdot 2} [/latex] to obtain equivalent form [latex] \dfrac{4}{60} [/latex]. In similar way, since [latex] 6 = 2\cdot 3 [/latex], we multiply [latex] \dfrac{1}{6} [/latex] by [latex] \dfrac{2\cdot 5}{2\cdot 5} [/latex] to obtain [latex] \dfrac{10}{60} [/latex], and since [latex] 20 = 2\cdot 2\cdot 5 [/latex], we multiply [latex] \dfrac{1}{20} [/latex] by [latex] \dfrac{3}{3} [/latex] to obtain [latex] \dfrac{3}{60} [/latex]. Once all fractions have a like denominator, we perform the operations on the numerators and simplify the final answer if possible.

[latex] \begin{align} \frac{1}{15}-\frac{1}{6}+\frac{1}{20} &= \frac{1{\color{red}{} \cdot 2\cdot 2}}{15{\color{red}{} \cdot 2\cdot 2}}-\frac{1{\color{red}{} \cdot 2\cdot 5}}{6{\color{red}{} \cdot 2\cdot 5}}+\frac{1{\color{red}{} \cdot 3}}{20{\color{red}{} \cdot 3}}\\[5pt] &= \frac{4}{60}-\frac{10}{60}+\frac{3}{60}\\[5pt] &= \frac{4-10+3}{60}\\[5pt] &= \frac{-3}{60}\\[5pt] &= \frac{{\color{red}\cancel{\color{black}{3}}}\cdot(-1)}{{\color{red}\cancel{\color{black}{3}}}\cdot 20}\\[5pt] &= -\frac{1}{20} \end{align} [/latex]

Let’s try it on rational expressions, remembering that we keep the LCD in factored form.

Before moving on to the next example, recall that every integer [latex] a [/latex] can be presented as [latex] \dfrac{a}{1} [/latex].

CAUTION!

When subtracting a fraction while operating on fractions with like denominators, make sure to properly distribute the negative in the numerator of the fraction being  subtracted. Note that in the example above, once we simplified the numerators, since the fraction [latex] \dfrac{x^2-4}{(x+2)(x-2)} [/latex] was being subtracted, when we moved to operate on the numerators, we put the numerator of that fraction being subtracted, [latex] x^2-4 [/latex], in parentheses to properly distribute the negative in the next step. This is commonly missed, so please pay attention.

In the video that follows, we present an example of adding two rational expression whose denominators are binomials with no common factors.

This next video contains an example of subtracting rational expressions.

We will conclude with one more example to practice finding the LCD, subtracting, and simplifying rational expressions.