1.25: Adding and Subtracting Rational Expressions

Think back to the chapter on rational expressions. Let's remember that a rational expression (also called an algebraic fraction) is one which can be written as a ratio of \(\dfrac{P}{Q}\) where \(\mathrm{P}\) and \(\mathrm{Q}\) are polynomials and \(\mathrm{Q} \neq 0 .\) We can add and subtract rational expressions in the same way we can add and subtract fractions.

Rational Expressions with “Like” Denominators

Adding and Subtracting Rational Expressions

Let \(P, Q\) and \(R\) be polynomials with \(R \neq 0\)

1. \(\dfrac{P}{R}+\dfrac{Q}{R}=\dfrac{P+Q}{R}\)
2. \(\dfrac{P}{R}-\dfrac{Q}{R}=\dfrac{P-Q}{R}\)

Example 23.1

Add and write the answer in simplest form:

\[\dfrac{3 y}{16}+\dfrac{5 y}{16}\nonumber\]

\[\dfrac{3 y}{16}+\dfrac{5 y}{16}\nonumber\]

\[\dfrac{3 y}{16}+\dfrac{5 y}{16}=\dfrac{3 y+5 y}{16}=\dfrac{8 y}{16}=\dfrac{y}{2}\nonumber\]

Example 23.2

Subtract and write the answer in simplest form:

\[\dfrac{5 x}{x-3}-\dfrac{15}{x-3}\nonumber\]

We begin by subtracting the numerators to get

\[\dfrac{5 x}{x-3}-\dfrac{15}{x-3}\nonumber\]

Next, we simplify the rational expression by using a method for we learned in the previous section. We factor the numerator and then cancel the common factor from the numerator and denominator

\[\dfrac{5 x-15}{x-3}=\dfrac{5\not{(x-3)}}{\not{(x-3)}}=5\nonumber\]

Rational Expressions with Unlike Denominators

Recall that in order to combine “unlike” fractions (those with different denominators) we first had to rewrite them with a common denominator. We chose to work with the least such common denominator because it streamlined the process. Combining unlike rational expressions requires us to do the same thing. We can find the LCD for rational expressions in exactly the same manner as we found the LCD for fractions, refer to Chapter 2 for the procedure. The only difference to keep in mind is that denominators of rational expressions may be polynomials. But this doesn’t hinder us. When we factor our denominators, we simply factor them as products of powers of prime numbers and polynomials. A prime (or irreducible) polynomial is a polynomial which cannot be factored any further.

Example 23.3

Combine \(\dfrac{5 x}{6}-\dfrac{2 x}{3}\).

Recall the steps to find the LCD (Chapter 2):

• Step 1 Factor (the denominators): 6 factors into \(2 \cdot 3\) and 3 factors into \(3 \cdot 1\)
• Step 2. List (the primes) 2,3
• Step 3. (Form the) LCD: \(2 \cdot 3=6\)

Now that we have found the LCD, we use it to rewrite each rational expression to transform our problem from

\[\dfrac{5 x}{6}-\dfrac{2 x}{3}\nonumber\]

to

\[\dfrac{5 x}{6}-\dfrac{2 \cdot 2 x}{2 \cdot 3} =\dfrac{5 x}{6}-\dfrac{4 x}{6}\nonumber\]

Now we can combine the rational expressions to get

\[\dfrac{5 x}{6}-\dfrac{4 x}{6}=\dfrac{x}{6}\nonumber\]

Example 23.4

Combine \(\dfrac{7}{x}+\dfrac{4}{3}\)

We encounter here our first "prime polynomial" which is \(x\). The LCD for this example is \(3 x\) and so we begin by rewriting the problem to read

\[\dfrac{7}{x}+\dfrac{4}{3}\nonumber\]

\[=\dfrac{3 \cdot 7}{3 \cdot x}+\dfrac{4 \cdot x}{3 \cdot x}\nonumber\]

\[=\dfrac{21}{3 x}+\dfrac{4 x}{3 x}\nonumber\]

Now, we combine the “like” rational expressions to yield the solution

\[\dfrac{21+4 x}{3 x}\nonumber\]

which cannot be simplified any further.

Example 23.5

Combine \(\dfrac{5}{4 a}-\dfrac{7 b}{6}\).

Let us begin by computing the LCD, keeping in mind that here \(a\) is a prime polynomial.

• Step 1. Factor: \(4 a\) factors into \(2^{2} \cdot a\) and 6 factors into \(2 \cdot 3\)
• Step 2. List: \(2,3, a\)
• Step 3. LCD: \(2^{2} \cdot 3 \cdot a=12 a\)

Next, we rewrite each fraction using this LCD:

\[\begin{align*} \dfrac{5}{4 a}-\dfrac{7 b}{6} &=\dfrac{3 \cdot 5}{3 \cdot 4 a}-\dfrac{7 b \cdot 2 a}{6 \cdot 2 a} \\[4pt] &=\dfrac{15}{12 a}-\dfrac{14 a b}{12 a}\end{align*}\]

Finally we can combine “like” rational expressions to get

\[\dfrac{15}{12 a}-\dfrac{14 a b}{12 a}=\dfrac{15-14 a b}{12 a}\nonumber\]

which cannot be reduced any further.

Exit Problem

Simplify: \(\dfrac{7}{12}-\dfrac{5}{8 b}\)