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Section 7.4 Rational Functions

Rational functions are algebraic fractions, so in this lesson we review the basic operations with algebraic fractions.

Subsection 1. Multiply and divide fractions

Subsubsection Examples

Example 7.39.

Multiply \(~\dfrac{4y^2-1}{4-y^2} \cdot \dfrac{y^2-2y}{4y+2}\)

Solution

We factor each numerator and denominator, and look for common factors.

\begin{align*} \amp \dfrac{4y^2-1}{4-y^2} \cdot \dfrac{y^2-2y}{4y+2} \amp \\ \amp\qquad=\dfrac{(2y-1)\cancel{(2y+1)}}{\cancel{(2-y)}(2+y)} \cdot \dfrac{y(-1)\cancel{(y-2)}}{2\cancel{(2y+1)}} \amp \amp \blert{\text{Divide out common factors.}}\\ \amp\qquad = \dfrac{-y(2y-1)}{2(y+2)} \amp \amp \blert{\text{Note:}~~y-2=-(2-y)} \end{align*}
Example 7.40.

Divide \(~\dfrac{6ab}{2a+b} \div (4a^2b)\)

Solution

We multiply the first fraction by the reciprocal of the second fraction.

\begin{align*} \dfrac{6ab}{2a+b} \div (4a^2b) \amp = \dfrac{\blert{\cancel{2} \cdot 3}\cancel{ab}}{2a+b} \cdot \dfrac{1}{\blert{\cancel{2} \cdot 2}a \cdot \cancel{ab}} \amp \amp \blert{\text{Divide out common factors.}}\\ \amp = \dfrac{3}{2a(2a+b)} \end{align*}

Subsubsection Exercises

Multiply \(~\dfrac{3xy}{4xy-6y^2} \cdot \dfrac{2x-3y}{12x}\)

Answer
\(\dfrac{1}{8}\)

Multiply \(~\dfrac{9x^2-25}{2x-2} \cdot \dfrac{x^2-1}{6x-10}\)

Answer
\(\dfrac{(3x+5)(x+1)}{4}\)

Divide \(~(x^2-9) \div \dfrac{x^2-6x+9}{3x}\)

Answer
\(\dfrac{3x(x+3)}{x-3}\)

Divide \(~\dfrac{x^2-1}{x+3} \div \dfrac{x^2-x-2}{x^2+5x+6}\)

Answer
\(\dfrac{(x-1)(x+2)}{x-2}\)

Subsection 2. Add and subtract fractions

Subsubsection Examples

Example 7.45.

Subtract \(~\dfrac{3x}{x+2} - \dfrac{2x}{x-3}\)

Solution

Step 1: The LCD for the fractions is \((x+2)(x-3)\text{.}\)

Step 2: We build each fraction to an equivalent one with the LCD.

\begin{equation*} \dfrac{3x\alert{(x-3)}}{(x+2)\alert{(x-3)}} = \dfrac{3x^2-9x}{x^2-x-6}~~~~\text{and}~~~~\dfrac{2x\alert{(x+2)}}{(x-3)\alert{(x+2)}} = \dfrac{2x^2+4x}{x^2-x-6} \end{equation*}

Step 3: Combine the numerators over the same denominator.

\begin{align*} \dfrac{3x}{x+2} - \dfrac{2x}{x-3} \amp = \dfrac{3x^2-9x}{x^2-x-6}-\dfrac{2x^2+4x}{x^2-x-6} \\ \amp = \dfrac{(3x^2-9x)-(2x^2+4x)}{x^2-x-6} \amp \amp \blert{\text{Subtract the numerators.}}\\ \amp = \dfrac{x^2-13x}{x^2-x-6} \end{align*}

Step 4: Reduce the result, if possible. We factor numerator and denominator to find

\begin{equation*} \dfrac{x(x-13)}{(x-3)(x+2)} \end{equation*}

The fraction cannot be reduced.

Example 7.46.

Write as a single fraction \(~1+\dfrac{2}{a} - \dfrac{a^2+2}{a^2+a}\)

Solution

Step 1: By factoring each denominator, we find that the LCD for the fractions is \(a(a+1)\text{.}\)

Step 2: We build each fraction to an equivalent one with the LCD.

\begin{align*} 1=\dfrac{1\cdot\alert{a(a+1)}}{1\cdot\alert{a(a+1)}} \amp = \dfrac{a^2+a}{a(a+1)}\\ \dfrac{2}{a}=\dfrac{2\cdot\alert{(a+1)}}{a\cdot\alert{(a+1)}} \amp = \dfrac{2a+2}{a(a+1)}\\ \dfrac{a^2+2}{a^2+a} \amp = \dfrac{a^2+2}{a(a+1)} \end{align*}

Step 3: Combine the numerators over the same denominator.

\begin{align*} 1+\dfrac{2}{a}-\dfrac{a^2+2}{a^2+a} \amp =\dfrac{a^2+a}{a(a+1)}+\dfrac{2a+2}{a(a+1)}-\dfrac{a^2+2}{a(a+1)}\\ \amp = \dfrac{(a^2+a)+(2a+2)-(a^2+2)}{a(a+1)} =\dfrac{3a}{a(a+1)} \end{align*}

Step 4: Reduce the fraction to find

\begin{equation*} \dfrac{3\cancel{a}}{\cancel{a}(a+1)} = \dfrac{3}{a+1} \end{equation*}

Subsubsection Exercises

Subtract \(~\dfrac{x+1}{x^2+2x} - \dfrac{x-1}{x^2-3x}\)

Answer
\(\dfrac{-3x-1}{x(x+2)(x-3)}\)

Write as a single fraction \(~y-\dfrac{2}{y^2-1} + \dfrac{3}{y+1}\)

Answer
\(\dfrac{y^3+2y-5}{y^2-1}\)

Subsection 3. Simplify complex fractions

Subsubsection Example

Example 7.49.

Simplify \(~\dfrac{x+\dfrac{3}{4}}{x-\dfrac{1}{2}}\)

Solution

Consider all the simple fractions that appear in the complex fraction; in this case their LCD is 4. We multiply each term of the numerator and each term of the denominator by 4.

\begin{equation*} \dfrac{\alert{4}\left(x+\dfrac{3}{4}\right)} {\alert{4}\left(x-\dfrac{1}{2}\right)} =\dfrac{\alert{4}(x)+\alert{4}\left(\dfrac{3}{4}\right)}{\alert{4}(x)-\alert{4}\left(\dfrac{1}{2}\right)} = \dfrac{4x+3}{4x-2} \end{equation*}

The original complex fraction is equivalent tp the simple fraction \(\dfrac{4x+3}{4x-2}\text{.}\)

Subsubsection Exercises

Write the complex fraction as a simple fraction in lowest terms:

\begin{equation*} \dfrac{\dfrac{2}{y}+\dfrac{1}{2y}}{y+\dfrac{y}{2}} \end{equation*}
Answer
\(\dfrac{5}{3y^2}\)

Write the complex fraction as a simple fraction in lowest terms:

\begin{equation*} \dfrac{H-T}{\dfrac{H}{T}-\dfrac{T}{H}} \end{equation*}
Answer
\(\dfrac{TH}{H+T}\)