ORCCA Addition and Subtraction of Rational Expressions

Section 8.4 Addition and Subtraction of Rational Expressions

In the last section, we learned how to multiply and divide rational expressions. In this section, we will learn how to add and subtract rational expressions.

Subsection 8.4.1 Introduction

Example 8.4.1.

Julia is taking her family on a boat trip \(12\) miles down the river and back. The river flows at a speed of \(2\) miles per hour and she wants to drive the boat at a constant speed, \(b\) miles per hour downstream and back upstream. As explained in Figure 5.4.20 and Figure 5.4.21, due to the current of the river, the actual speed of travel is \(b+2\) miles per hour going downstream, and \(b-2\) miles per hour going upstream.

We need to review three forms of the formula for movement at a constant, or uniform, rate:

\begin{align*} d\amp=rt\amp r\amp=\frac{d}{t}\amp t\amp=\frac{d}{r} \end{align*}

where \(d\) stands for distance, \(r\) represents speed, and \(t\) stands for time. According to the third form, the time it takes the boat to travel downstream is \(\frac{12}{b+2}\text{,}\) and the time it takes to get back upstream is \(\frac{12}{b-2}\text{.}\)

The point of this section is to work with expressions like \(\frac{12}{b-2}+\frac{12}{b+2}\text{,}\) where two rational expressions are added (or subtracted). There are times when it is useful to combine them into a single fraction. We will learn that the expression \(\frac{12}{b-2}+\frac{12}{b+2}\) is equal to the expression \(\frac{24b}{b^2-4}\) and we will learn how to make that simplification.

Subsection 8.4.2 Addition and Subtraction of Rational Expressions with the Same Denominator

The process of adding and subtracting rational expressions will be very similar to the process of adding and subtracting purely numerical fractions.

If the two expressions have the same denominator, then we can rely on the property of adding and subtracting fractions and simplify that result.

Let's review how to add fractions with the same denominator:

\begin{align*} \frac{1}{10}+\frac{3}{10}\amp=\frac{1+3}{10}\\ \amp=\frac{4}{10}\\ \amp=\frac{2}{5} \end{align*}

We can add and subtract rational expressions in the same way:

\begin{align*} \frac{2}{3x}-\frac{5}{3x}\amp=\frac{2-5}{3x}\\ \amp=\frac{-3}{3x}\\ \amp=-\frac{1}{x} \end{align*}

Example 8.4.2.

Add the rational expressions: \(\dfrac{2x}{x+y}+\dfrac{2y}{x+y}\text{.}\)

Explanation

These expressions already have a common denominator:

\begin{align*} \frac{2x}{x+y}+\frac{2y}{x+y}\amp=\frac{2x+2y}{x+y}\\ \amp=\frac{2\cancelhighlight{(x+y)}}{\cancelhighlight{x+y}}\\ \amp=\frac{2}{1}\\ \amp=2 \end{align*}

Note that we didn't stop at \(\frac{2x+2y}{x+y}\text{.}\) If possible, we must simplify the numerator and denominator. Since this is a multivariable expression, this textbook ignores domain restrictions while canceling.

Subsection 8.4.3 Addition and Subtraction of Rational Expressions with Different Denominators

Below are the steps to follow when adding or subtracting rational expressions, in general.

List 8.4.3. Steps to Adding/Subtracting Rational Expressions

Identify the LCD: Determine the least common denominator of all of the denominators.
Build: If necessary, build each expression so that the denominators are the same.
Add/Subtract: Combine the numerators using the properties of adding and subtracting fractions.
Simplify: Simplify the resulting rational expression as much as possible. This may require factoring the numerator.

As you can see from the list above, to add rational expressions with different denominators, we'll need to build each fraction to the least common denominator. This is similar to the way we treat numerical fractions. Let's briefly review this process by adding \(\frac{3}{5}\) and \(\frac{1}{6}\text{:}\)

\begin{align*} \frac{3}{5}+\frac{1}{6}\amp=\frac{3}{5}\multiplyright{\frac{6}{6}}+\frac{1}{6}\multiplyright{\frac{5}{5}}\\ \amp=\frac{18}{30}+\frac{5}{30}\\ \amp=\frac{18+5}{30}\\ \amp=\frac{23}{30} \end{align*}

This exact method can be used when adding rational expressions containing variables. The key is that the expressions must have the same denominator before they can be added or subtracted. If they don't have this initially, then we'll identify the least common denominator and build each expression so that it has that denominator.

Finding the Least Common Denominator of Rational Expressions.

In Section 1.2, a method for finding the least common of denominator of numerical fractions is suggested, which can also be applied to rational expressions.

To find the least common denominator (LCD) of two or more rational expressions with this method, we must first find the prime factors of each denominator. In other words, factor each denominator completely. These prime factors will be the building blocks we need to build the LCD of the rational expressions.

To build the LCD, we first collect all of the prime factors from the first denominator. Next, we collect only those prime factors from the second denominator that haven't already been collected from the first denominator.

If there is more than one rational expression, we continue collecting prime factors from each denominator that haven't already been collected. The last step then is to multiply together all of the prime factors that were collected in this process. This resulting product will be the LCD.

Let's see how this works with an example.

Example 8.4.4.

Find the least common denominator of the rational expressions \(\frac{1}{x^2-2x-15}\) and \(\frac{2x^2}{x^2-9}\text{.}\)

Explanation

The first step is to factor the denominators of each expression completely.

\begin{gather*} \frac{1}{x^2-2x-15}\\ =\frac{1}{(x+3)(x-5)} \end{gather*}

\begin{gather*} \frac{2x^2}{x^2-9}\\ =\frac{2x^2}{(x+3)(x-3)} \end{gather*}

Now, we collect all of the factors from the first denominator, which are \((x+3)\) and \((x-5)\text{.}\) Next, we collect only the factors of the second denominator that we haven't already taken from the first denominator. Notice that we have already taken a factor of \((x+3)\text{.}\) Therefore, we need only collect the factor of \((x-3)\) from the second denominator. This gives us three factors: \((x+3)\text{,}\) \((x-5)\text{,}\) and \((x-3)\text{.}\) Multiplying these together gives us our LCD:

LCD \(=(x+3)(x-5)(x-3)\)

Example 8.4.5.

Find the least common denominator of the rational expressions \(\frac{5y^2}{6xy+30y^2}\) and \(\frac{y}{4x^2+12xy-40y^2}\text{.}\)

Explanation

The first step is to factor the denominators of each expression completely.

\begin{gather*} \frac{5y^2}{6xy+30y^2}\\ =\frac{5y^2}{2\cdot 3\cdot\ y(x+5y)} \end{gather*}

\begin{gather*} \frac{y}{4x^2+12xy-40y^2}\\ =\frac{y}{2\cdot 2(x+5y)(x-2y)} \end{gather*}

Now, we collect all of the factors from the first denominator, which are \(2\text{,}\) \(3\text{,}\) \(y\text{,}\) and \((x+5y)\text{.}\) Next, we collect only the factors of the second denominator that we haven't already taken from the first denominator. Notice that we have already taken factors of \(2\) and \((x+5y)\text{.}\) Therefore, we need only collect the second factor of \(2\text{,}\) as well as the factor of \((x-2y)\) from the second denominator. This gives us the following factors to build the LCD: \(2\text{,}\) \(2\text{,}\) \(3\text{,}\) \(y\text{,}\) \((x+5y)\) and \((x-2y)\text{.}\) Multiplying these together gives us:

LCD \(=2\cdot 2\cdot 3\cdot y(x+5y)(x-2y)=12y(x+5y)(x-2y)\)

Before we apply this new skill to adding or subtracting rational expressions, let's go over the next step of building each expression to an equivalent rational expression with the LCD, by looking again at the expressions from Example 8.4.4.

Example 8.4.6.

Now that we found the LCD of rational expressions \(\frac{1}{x^2-2x-15}\) and \(\frac{2x^2}{x^2-9}\) to be \((x+3)(x-5)(x-3)\text{.}\) Let's build each of these rational expressions up to an equivalent rational expression that has \((x+3)(x-5)(x-3)\) as its denominator.

To do that, we will need to multiply each expression by a fraction equivalent to one. That fraction will need to include, in both the numerator and denominator, all factors of the LCD that are missing from the denominator of the original expression. To see which factors are missing, we need to look at each denominator in factored form.

\begin{align*} \frac{1}{x^2-2x-15}\amp=\frac{1}{(x+3)(x-5)}\multiplyright{\frac{x-3}{x-3}}\\ \amp=\frac{1(x-3)}{(x+3)(x-5)(x-3)}\\ \amp=\frac{x-3}{(x+3)(x-5)(x-3)} \end{align*}

\begin{align*} \frac{2x^2}{x^2-9}\amp=\frac{2x^2}{(x+3)(x-3)}\multiplyright{\frac{x-5}{x-5}}\\ \amp=\frac{2x^2\left(x-5\right)}{(x+3)(x-5)(x-3)}\\ \amp=\frac{2x^3-10x^2}{(x+3)(x-5)(x-3)} \end{align*}

Notice that the denominator of the first rational expression was missing a factor of \(x-3\text{,}\) so we multiplied by a fraction whose numerator and denominator are both \(x-3\text{.}\) Similarly, the second rational expression was missing a factor of \(x-5\text{,}\) so we used that factor to make a fraction equivalent to one to multiply the second rational expression by. Now, we have two rational expressions with the same denominator, namely the LCD.

Let's apply what we've learned so far to adding the two expressions with denominators that are \(b-2\) and \(b+2\) from Example 8.4.1.

Example 8.4.7.

Add the rational expressions and fully simplify the function given by \(t(b)=\frac{12}{b-2}+\frac{12}{b+2}\text{,}\) where the function name, \(t\text{,}\) stands for time.

Explanation

Note that both denominators are factored completely and, when we follow the process of collecting factors to build the LCD, we obtain an LCD of \((b-2)(b+2)\text{.}\) Next, we build each expression up to an equivalent rational expression that has the LCD as its denominator by multiplying both the numerator and denominator by the missing factor needed. Once we obtain common denominators, we are ready to add, as you can see below.

\begin{align*} t(b)\amp=\frac{12}{b-2}+\frac{12}{b+2}\\ t(b)\amp=\frac{12}{b-2}\multiplyright{\frac{b+2}{b+2}}+\frac{12}{b+2}\multiplyright{\frac{b-2}{b-2}}\\ t(b)\amp=\frac{12b+24}{(b-2)(b+2)}+\frac{12b-24}{(b+2)(b-2)}\\ t(b)\amp=\frac{(12b+24)+(12b-24)}{(b+2)(b-2)}\\ t(b)\amp=\frac{24b}{(b+2)(b-2)} \end{align*}

Example 8.4.8.

Add the rational expressions: \(\dfrac{2}{5x^2y}+\dfrac{3}{20xy^2}\)

Explanation

By collecting prime factors of each denominator, using the method we've discussed previously, we see that the least common denominator of \(5x^2y\) and \(20xy^2\) must include two \(x\)'s and two \(y\)'s, as well as \(5\cdot 2\cdot 2=20\text{.}\) Thus, the LCD is \(20x^2y^2\text{.}\) We will build both denominators to \(20x^2y^2\) before doing the addition.

\begin{align*} \frac{2}{5x^2y}+\frac{3}{20xy^2}\amp=\frac{2}{5x^2y}\multiplyright{\frac{4y}{4y}}+\frac{3}{20xy^2}\multiplyright{\frac{x}{x}}\\ \amp=\frac{8y}{20x^2y^2}+\frac{3x}{20x^2y^2}\\ \amp=\frac{8y+3x}{20x^2y^2} \end{align*}

Let's look at a few more complicated examples.

Example 8.4.9.

Subtract the rational expressions: \(\dfrac{y}{y-2}-\dfrac{8y-8}{y^2-4}\)

Explanation

To start, we'll make sure each denominator is factored. Next, we'll find the least common denominator and build each expression to that denominator. Then, we will be able to combine the numerators and simplify the expression.

\begin{align*} \frac{y}{y-2}-\frac{8y-8}{y^2-4}\amp=\frac{y}{y-2}-\frac{8y-8}{(y+2)(y-2)}\\ \amp=\frac{y}{y-2}\multiplyright{\frac{y+2}{y+2}}-\frac{8y-8}{(y+2)(y-2)}\\ \amp=\frac{y^2+2y}{(y+2)(y-2)}-\frac{8y-8}{(y+2)(y-2)}\\ \amp=\frac{y^2+2y-\highlight{\attention{(}}8y-8\highlight{\attention{)}}}{(y+2)(y-2)}\\ \amp=\frac{y^2+2y-8y+8}{(y+2)(y-2)}\\ \amp=\frac{y^2-6y+8}{(y+2)(y-2)}\\ \amp=\frac{\cancelhighlight{(y-2)}(y-4)}{(y+2)\cancelhighlight{(y-2)}}\\ \amp=\frac{y-4}{y+2}, \text{ for } y\neq 2 \end{align*}

Note that we must factor the numerator in \(\frac{y^2-6y+8}{(y+2)(y-2)}\) and try to reduce the fraction (which we did).

Warning 8.4.10.

In Example 8.4.9, be careful to subtract the entire numerator of \(8y-8\text{.}\) When this expression is in the numerator of \(\frac{8y-8}{(y+2)(y-2)}\text{,}\) it's implicitly grouped and doesn't need parentheses. But once \(8y-8\) is subtracted from \(y^2+2y\text{,}\) we need to add parentheses so the entire expression is subtracted.

In the next example, we'll look at adding a rational expression to a polynomial. Much like adding a fraction and an integer, we'll rely on writing that expression as itself over one in order to build its denominator.

Example 8.4.11.

Add the expressions: \(-\dfrac{2}{r-1}+r\)

Explanation

\begin{align*} -\frac{2}{r-1}+r\amp=-\frac{2}{r-1}+\frac{r}{1}\\ \amp=-\frac{2}{r-1}+\frac{r}{1}\multiplyright{\frac{r-1}{r-1}}\\ \amp=\frac{-2}{r-1}+\frac{r^2-r}{r-1}\\ \amp=\frac{-2+r^2-r}{r-1}\\ \amp=\frac{r^2-r-2}{r-1}\\ \amp=\frac{(r-2)(r+1)}{r-1} \end{align*}

Note that we factored the numerator to reduce the fraction if possible. Even though it was not possible in this case, leaving it in factored form makes it easier to see that it is reduced.

Example 8.4.12.

Subtract the expressions: \(\dfrac{6}{x^2-2x-8}-\dfrac{1}{x^2+3x+2}\)

Explanation

To start, we'll need to factor each of the denominators. After that, we'll identify the LCD and build each denominator accordingly. Then we can combine the numerators and simplify the resulting expression.

\begin{align*} \frac{6}{x^2-2x-8}-\frac{1}{x^2+3x+2}\amp=\frac{6}{(x-4)(x+2)}-\frac{1}{(x+2)(x+1)}\\ \amp=\frac{6}{(x-4)(x+2)}\multiplyright{\frac{x+1}{x+1}}-\frac{1}{(x+2)(x+1)}\multiplyright{\frac{x-4}{x-4}}\\ \amp=\frac{6x+6}{(x-4)(x+2)(x+1)}-\frac{x-4}{(x+2)(x+1)(x-4)}\\ \amp=\frac{6x+6-(x-4)}{(x-4)(x+2)(x+1)}\\ \amp=\frac{6x+6-x+4}{(x-4)(x+2)(x+1)}\\ \amp=\frac{5x+10}{(x-4)(x+2)(x+1)}\\ \amp=\frac{5\cancelhighlight{(x+2)}}{(x-4)\cancelhighlight{(x+2)}(x+1)}\\ \amp=\frac{5}{(x-4)(x+1)}, \text{ for }x\neq -2 \end{align*}

Exercises 8.4.4 Exercises

Review and Warmup

1.

Add: \(\displaystyle{\frac{7}{12} + \frac{7}{12}}\)

2.

Add: \(\displaystyle{\frac{11}{9} + \frac{1}{9}}\)

3.

Add: \(\displaystyle{\frac{5}{6} + \frac{7}{10}}\)

4.

Add: \(\displaystyle{\frac{7}{10} + \frac{4}{5}}\)

5.

Subtract: \(\displaystyle{\frac{13}{28} - \frac{9}{28}}\)

6.

Subtract: \(\displaystyle{\frac{13}{32} - \frac{9}{32}}\)

7.

Subtract: \(\displaystyle{\frac{4}{7} - \frac{15}{28}}\)

8.

Subtract: \(\displaystyle{\frac{4}{7} - \frac{13}{28}}\)

9.

Factor the given polynomial.

\({t^{2}-64}=\)

10.

Factor the given polynomial.

\({t^{2}-16}=\)

11.

Factor the given polynomial.

\({x^{2}+13x+30}=\)

12.

Factor the given polynomial.

\({x^{2}+13x+42}=\)

13.

Factor the given polynomial.

\({y^{2}-4y+3}=\)

14.

Factor the given polynomial.

\({y^{2}-14y+45}=\)

15.

Factor the given polynomial.

\({4r^{2}-24r+20}=\)

16.

Factor the given polynomial.

\({7r^{2}-28r+21}=\)

17.

Add the polynomials.

\(\displaystyle{\left({x-3}\right)+\left({7x+6}\right)}\)

18.

Add the polynomials.

\(\displaystyle{\left({-6x^{2}+9x+7}\right)+\left({5x^{2}-6x-1}\right)}\)

19.

Subtract the polynomials.

\(\displaystyle{\left({x-3}\right)-\left({7x+6}\right)}\)

20.

Subtract the polynomials.

\(\displaystyle{\left({x^{2}-3x}\right)-\left({7x^{2}+6x}\right)}\)