ORCCA Addition and Subtraction of Rational Expressions

Section13.3Addition 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.

Figure13.3.1Alternative Video Lesson

Subsection13.3.1Introduction

Example13.3.2

Two towns are connected by a \(12\)-mile-long river, which flows from Town A to Town B at a speed of \(2\) miles per hour. A boat will travel at a constant speed, \(v\) miles per hour, from Town A to Town B, and then back to Town A. Due to the current of the river, the actual speed of travel is \(v+2\) miles per hour when going downstream from Town A to Town B, and \(v-2\) miles per hour going upstream back to Town A. If the boat driver plans to spend \(8\) hours for the whole trip, how fast should they drive the boat?

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

\begin{align*} d\amp=vt\amp v\amp=\frac{d}{t}\amp t\amp=\frac{d}{v} \end{align*}

where \(d\) stands for distance, \(v\) represents speed, and \(t\) stands for time. According to the third form, the time it takes the boat to travel from Town A to Town B is \(\frac{12}{v+2}\text{,}\) and the time it takes to get back is \(\frac{12}{v-2}\text{.}\)

The function to model the time of the whole trip is

\begin{equation*} t(v)=\frac{12}{v-2}+\frac{12}{v+2} \end{equation*}

where \(t\) stands for time in hours, and \(v\) is the boat's speed in miles per hour. Let's look at the graph of this function in Figure 13.3.3. Note that since the speed \(v\) and the time \(t(v)\) should be positive in context, it's only the first quadrant of Figure 13.3.3 that matters.

Figure13.3.3Graph of \(t(v)=\frac{12}{v-2}+\frac{12}{v+2}\) and \(t=8\)

To find the speed of the boat that would make the round trip last \(8\) hours we can use graphing technology to solve the equation

\begin{equation*} \frac{12}{v-2}+\frac{12}{v+2}=8 \end{equation*}

graphically and we see that \(v=4\text{.}\) This tells us that a speed of \(4\) miles per hour will give a total time of \(8\) hours to complete the trip. From Town A to Town B, it would take \(\frac{12}{v+2}=\frac{12}{4+2}=2\) hours; and from Town B to Town A, it would take \(\frac{12}{v-2}=\frac{12}{4-2}=6\) hours.

The point of this section is to work with expressions like \(\frac{12}{v-2}+\frac{12}{v+2}\text{,}\) 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 \(\frac{12}{v-2}+\frac{12}{v+2}\) is equal to the expression \(\frac{24v}{v^2-4}\text{,}\) and we will learn how to make that simplification.

Subsection13.3.2Addition 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*}

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.

List13.3.4Steps to Adding/Subtracting Rational Expressions

Example13.3.5

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

Solution

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, \text{ for } x\neq -y \end{align*}

Note that we didn't stop at \(\frac{2x+2y}{x+y}\text{.}\) If possible, we must simplify the numerator and denominator.

Subsection13.3.3Addition and Subtraction of Rational Expressions with Different Denominators

To add rational expressions with different denominators, we'll need to build each fraction to the least common denominator, in the same way we do with 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.

Let's apply this to adding the two expressions with denominators that are \(v-2\) and \(v+2\) from Example 13.3.2.

Example13.3.6

Add the rational expressions and fully simplify the function given by \(t(v)=\frac{12}{v-2}+\frac{12}{v+2}\text{.}\)

Solution

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

Example13.3.7

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

Solution

The least common denominator of \(5x^2y\) and \(20xy^2\) must include two \(x\)'s and two \(y\)'s, as well as \(20\text{.}\) Thus it is \(20x^2y^2\text{.}\) We will build both denominators to \(20x^2y^2\) before doing 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}{x^2y^2} \end{align*}

Let's look at a few more complicated examples.

Example13.3.8

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

Solution

To start, we'll make sure each denominator is factored. Then 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).

Warning13.3.9

In Example 13.3.8, 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.

Example13.3.10

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

Solution

\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.

Example13.3.11

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

Solution

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*}