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Section 18.5 Going Deeper: Adding and Subtracting Rational Expressions

In some ways, the exercises in this section give a false impression of working with rational expressions. It is not often the case that we're adding and subtracting the types of monomial expressions that were presented. The reason the exercises look that way is because the emphasis is on learning to recognize when the denominators have common factors and when they don't, which leads to knowing what to multiply by to create the common denominators. By using different variables, it's much easier for students to train themselves to recognize the common factors.

It's likely that you will only ever encounter single variable rational expressions. But the skill of identifying different terms will still come into play because you will need to recognize when polynomial factors are the same or different. For example, instead of working with fractions of the form

\begin{equation*} \frac{1}{a} + \frac{1}{b} \end{equation*}

you will be working with fractions of the form

\begin{equation*} \frac{1}{x + 2} + \frac{1}{x - 3}. \end{equation*}

In the previous "Going Deeper" section, we emphasized the importance of multiplicative factors when reducing rational expressions. The same idea holds for adding and subtracting rational expressions. We need to think in terms of multiplicative factors. The way we accomplish this is to think of the whole binomial as a single object. Basically, we want to envision a set of parentheses around each binomial.

\begin{equation*} \frac{1}{x + 2} + \frac{1}{x - 3} = \frac{1}{(x + 2)} + \frac{1}{(x - 3)} \end{equation*}

And once we have done this, all of the experience that we were developing in this section can come into play. All of the previous logic applies to this calculation.

\begin{equation*} \begin{aligned} \frac{1}{x + 2} + \frac{1}{x - 3} \amp = \frac{1 \cdot (x - 3)}{(x + 2) \cdot (x - 3)} + \frac{1 \cdot (x + 2)}{(x - 3) \cdot (x + 2)} \amp \eqnspacer \amp \text{Common denominator} \\ \amp = \frac{x - 3}{(x + 2)(x - 3)} + \frac{x + 2}{(x + 2)(x - 3)} \\ \amp = \frac{(x - 3) + (x + 2)}{(x + 2)(x - 3)} \\ \amp = \frac{2x - 1}{(x + 2)(x - 3)} \end{aligned} \end{equation*}

When subtracting rational expressions, it's extremely important to keep the parentheses around terms when subtracting. This did not come up in the section because we were only working with monomial terms. But when working with binomials, those parentheses are critical to avoid errors.

\begin{equation*} \begin{aligned} \frac{x}{x - 1} - \frac{3}{x + 3} \amp = \frac{x \cdot (x + 3)}{(x - 1) \cdot (x + 3)} - \frac{3 \cdot (x - 1)}{(x + 3) \cdot (x - 1)} \amp \eqnspacer \amp \text{Common denominator} \\ \amp = \frac{x^2 + 3x}{(x - 1)(x + 3)} - \frac{3x - 3}{(x - 1)(x + 3)} \\ \amp = \frac{(x^2 + 3x) - (3x - 3)}{(x - 1)(x + 3)} \\ \amp = \frac{x^2 + 3x - 3x + 3}{(x - 1)(x + 3)} \\ \amp = \frac{x^2 + 3}{(x - 1)(x + 3)} \end{aligned} \end{equation*}

In some expressions, you may find that after adding or subtracting, it may be possible to factor the numerator. When you can see a way to factor, it's best to do it. The reason is that sometimes these factorization steps will reveal that the fraction can be reduced or simplified, which can be very helpful.

\begin{equation*} \begin{aligned} \frac{1}{x - 2} + \frac{x - 5}{(x - 2)(x + 1)} \amp = \frac{1 \cdot (x + 1)}{(x - 2) \cdot (x + 1)} + \frac{x - 5}{(x - 2)(x + 1)} \amp \eqnspacer \amp \text{Common denominator} \\ \amp = \frac{x + 1}{(x - 2)(x + 1)} + \frac{x - 5}{(x - 2)(x + 1)} \\ \amp = \frac{(x + 1) + (x - 5)}{(x - 2)(x + 1)} \\ \amp = \frac{2x - 4}{(x - 2)(x + 1)} \\ \amp = \frac{2(x - 2)}{(x - 2)(x + 1)} \\ \amp = \frac{2 \cdot \cancel{(x - 2)}}{(x + 1) \cdot \cancel{(x - 2)}} \\ \amp = \frac{2}{x + 1} \end{aligned} \end{equation*}

It is hard to know in advance which calculations will lead to fractions that reduce and which ones don't, so you will just want to get into the habit of factoring the numerator when you can.