8.9: Complex Rational Expressions
- Page ID
- 60051
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Simple And Complex Fractions
Simple Fraction
In section 8.2 we saw that a simple fraction was a fraction of the form \(\dfrac{P}{Q}\), where \(P\) and \(Q\) are polynomials and \(Q \not = 0\).
Complex Fraction
A complex fraction is a fraction in which the numerator or denominator, or both, is a fraction. The fractions
\(\dfrac{\frac{8}{15}}{\frac{2}{3}}\) and \(\dfrac{1 - \frac{1}{x}}{1 - \frac{1}{x^2}}\)
are examples of complex fractions, or more generally, complex rational expressions.
There are two methods for simplifying complex rational expressions: the combine-divide method and the LCD-multiply-divide method.
The Combine-Divide Method
- If necessary, combine the terms of the numerator together.
- If necessary, combine the terms of the denominator together.
- Divide the numerator by the denominator.
Sample Set A
Simplify each complex rational expression.
\(\dfrac{\frac{x^3}{8}}{\frac{x^5}{12}}\)
Steps 1 and 2 are not necessary so we proceed with step 3:
\(\dfrac{\frac{x^3}{8}}{\frac{x^5}{12}} = \dfrac{x^3}{8} \cdot \dfrac{12}{x^5} = \dfrac{\cancel{x^3}}{^\cancel{8}_2} \cdot \dfrac{_\cancel{12}^3}{x^{\cancel{5}2}} = \dfrac{3}{2x^2}\)
\(\dfrac{1 - \frac{1}{x}}{1 - \frac{1}{x^2}}\)
Step 1: Combine the terms of the numerator: LCD = \(x\).
\(1 - \dfrac{1}{x} = \dfrac{x}{x} - \dfrac{1}{x} = \dfrac{x-1}{x}\)
Step 2: Combine the terms of the denominator: LCD = \(x^2\).
\(1 - \dfrac{1}{x^2} = \dfrac{x^2}{x^2} - \dfrac{1}{x^2} = \dfrac{x^2 - 1}{x^2}\)
Step 3: Divide the numerator by the denominator.
\(\begin{array}{flushleft}
\dfrac{\frac{x-1}{x}}{\frac{x^2-1}{x^2}} &= \dfrac{x-1}{x} \cdot \dfrac{x^2}{x^2-1}\\
&= \dfrac{\cancel{x-1}}{\cancel{x}} \dfrac{x^{\cancel{2}}}{(x+1)\cancel{(x+1)}}\\
&= \dfrac{x}{x+1}
\end{array}\)
Thus,
\(\dfrac{1 - \frac{1}{x}}{1 - \frac{1}{x^2}} = \dfrac{x}{x+1}\)
\(\dfrac{2 - \frac{13}{m} - \frac{7}{m^2}}{2 + \frac{3}{m} + \frac{1}{m^2}}\)
Step 1: Combine the terms of the numerator: LCD = \(m^2\).
\(2-\dfrac{13}{m}-\dfrac{7}{m^{2}}=\dfrac{2 m^{2}}{m^{2}}-\dfrac{13 m}{m^{2}}-\dfrac{7}{m^{2}}=\dfrac{2 m^{2}-13 m-7}{m^{2}}\)
Step 2: Combine the terms of the denominator: LCD = \(m^2\)
\(2+\dfrac{3}{m}+\dfrac{1}{m^{2}}=\dfrac{2 m^{2}}{m^{2}}+\dfrac{3 m}{m^{2}}+\dfrac{1}{m^{2}}=\dfrac{2 m^{2}+3 m+1}{m^{2}}\)
Step 3: Divide the numerator by the denominator:
\(\begin{array}{flushleft}
\dfrac{\frac{2 m^{2}-13 m-7}{m^{2}}}{\frac{2 m^{2}+3 m-1}{m^{2}}} &=\dfrac{2 m^{2}-13 m-7}{m^{2}} \cdot \frac{m^{2}}{2 m^{2}+3 m+1} \\
&=\dfrac{\cancel{(2 m+1)}(m-7)}{\cancel{m^2}} \cdot \dfrac{\cancel{m^2}}{\cancel{(2 m+1)}(m+1)} \\
&=\dfrac{m-7}{m+1}
\end{array}\)
Thus,
\(\dfrac{2 - \frac{13}{m} - \frac{7}{m^2}}{2 + \frac{3}{m} + \frac{1}{m^2}} = \dfrac{m - 7}{m + 1}\)
Practice Set A
Use the combine-divide method to simplify each expression.
\(\dfrac{\frac{27x^2}{6}}{\frac{15x^3}{8}}\)
- Answer
-
\(\dfrac{12}{5x}\)
\(\dfrac{3 - \frac{1}{x}}{3 + \frac{1}{x}}\)
- Answer
-
\(\dfrac{3x - 1}{3x + 1}\)
\(\dfrac{1 + \frac{x}{y}}{x - \frac{y^2}{x}}\)
- Answer
-
\(\dfrac{x}{y(x-y)}\)
\(\dfrac{m - 3 + \frac{2}{m}}{m - 4 + \frac{3}{m}}\)
- Answer
-
\(\dfrac{m-2}{m-3}\)
\(\dfrac{1 + \frac{1}{x-1}}{1 - \frac{1}{x-1}}\)
- Answer
-
\(\dfrac{x}{x-2}\)
The LCD-Multiply-Divide Method
- Find the LCD of all the terms.
- Multiply the numerator and denominator by the LCD.
- Reduce if necessary.
Sample Set B
Simplify each complex fraction.
\(\dfrac{1 - \frac{4}{a^2}}{1 + \frac{2}{a}}\)
Step 1: The LCD \(=a^2\).
Step 2: Multiply both the numerator and denominator by \(a^2\).
\(\begin{array}{flushleft}
\dfrac{a^2(1 - \frac{4}{a^2})}{a^2(1 + \frac{2}{a})} &= \dfrac{a^2 \cdot 1-a^2 \cdot \frac{4}{a^2}}{a^2 \cdot 1+a^2\cdot\frac{2}{a}}\\
&= \dfrac{a^2-4}{a^2 + 2a}
\end{array}\).
Step 3: Reduce:
\(\begin{array}{flushleft}
\frac{a^{2}-4}{a^{2}+2 a} &=\frac{\cancel{(a+2)}(a-2)}{a\cancel{(a+2)}} \\
&=\frac{a-2}{a}
\end{array}\)
Thus,
\(\dfrac{1-\frac{4}{a^2}}{1 + \frac{2}{a}} = \dfrac{a-2}{a}\)
\(\dfrac{1 - \frac{5}{x} - \frac{6}{x^2}}{1 + \frac{6}{x} + \frac{5}{x^2}}\)
Step 1: The LCD is \(x^2\).
Step 2: Multiply the numerator and denominator by \(x^2\).
\(\begin{array}{flushleft}
\dfrac{x^{2}(1-\frac{5}{x}-\frac{6}{x^{2}})}{x^{2}(1+\frac{6}{x}+\frac{5}{x^{2}})} &= \dfrac{x^{2} \cdot 1-x^{\cancel{2}} \cdot \frac{5}{\cancel{x}}-\cancel{x^{2}} \cdot \frac{6}{\cancel{x^{2}}}}{x^{2} \cdot 1+x^{\cancel{2}} \cdot \frac{6}{\cancel{x}}+\cancel{x^2} \cdot \frac{5}{\cancel{x^2}}} \\
&=\dfrac{x^{2}-5 x-6}{x^{2}+6 x+5}
\end{array}\)
Step 3: Reduce:
\(\begin{array}{flushleft}
\dfrac{x^{2}-5 x-6}{x^{2}+6 x+5} &=\dfrac{(x-6)(x+1)}{(x+5)(x+1)} \\
&=\dfrac{x-6}{x+5}
\end{array}\)
Thus,
\(\dfrac{1 - \frac{5}{x} - \frac{6}{x^2}}{1 + \frac{6}{x} + \frac{5}{x^2}} = \dfrac{x-6}{x+5}\)
Practice Set B
The following problems are the same problems as the problems in Practice Set A. Simplify these expressions using the LCD-multiply-divide method. Compare the answers to the answers produced in Practice Set A.
\(\dfrac{\frac{27x^2}{6}}{\frac{15x^3}{8}}\)
- Answer
-
\(\dfrac{12}{5x}\)
\(\dfrac{3 - \frac{1}{x}}{3 + \frac{1}{x}}\)
- Answer
-
\(\dfrac{3x - 1}{3x + 1}\)
\(\dfrac{1 + \frac{x}{y}}{x - \frac{y^2}{x}}\)
- Answer
-
\(\dfrac{x}{y(x-y)}\)
\(\dfrac{m - 3 + \frac{2}{m}}{m - 4 + \frac{3}{m}}\)
- Answer
-
\(\dfrac{m-2}{m-3}\)
\(\dfrac{1 + \frac{1}{x-1}}{1 - \frac{1}{x-1}}\)
- Answer
-
\(\dfrac{x}{x-2}\)
Exercises
For the following problems, simplify each complex rational expression.
\(\dfrac{1+\frac{1}{4}}{1-\frac{1}{4}}\)
- Answer
-
\(\dfrac{5}{3}\)
\(\dfrac{1-\frac{1}{3}}{1+\frac{1}{3}}\)
\(\dfrac{1-\frac{1}{y}}{1+\frac{1}{y}}\)
- Answer
-
\(\dfrac{y-1}{y+1}\)
\(\dfrac{a+\frac{1}{x}}{a-\frac{1}{x}}\)
\(\dfrac{\frac{a}{b}+\frac{c}{b}}{\frac{a}{b}-\frac{c}{b}}\)
- Answer
-
\(\dfrac{a+c}{a-c}\)
\(\dfrac{\frac{5}{m}+\frac{4}{m}}{\frac{5}{m}-\frac{4}{m}}\)
\(\dfrac{3+\frac{1}{x}}{\frac{3 x+1}{x^{2}}}\)
- Answer
-
\(x\)
\(\dfrac{1+\frac{x}{x+y}}{1-\frac{x}{x+y}}\)
\(\dfrac{2+\frac{5}{a+1}}{2-\frac{5}{a+1}}\)
- Answer
-
\(\dfrac{2a + 7}{2a - 3}\)
\(\dfrac{1-\frac{1}{a-1}}{1+\frac{1}{a-1}}\)
\(\dfrac{4-\frac{1}{m^{2}}}{2+\frac{1}{m}}\)
- Answer
-
\(\dfrac{2m - 1}{m}\)
\(\dfrac{9-\frac{1}{x^{2}}}{3-\frac{1}{x}}\)
\(\dfrac{k-\frac{1}{k}}{\frac{k+1}{k}}\)
- Answer
-
\(k-1\)
\(\dfrac{\frac{m}{m+1}-1}{\frac{m+1}{2}}\)
\(\dfrac{\frac{2 x y}{2 x-y}-y}{\frac{2 x-y}{3}}\)
- Answer
-
\(\dfrac{3y^2}{(2x - y)^2}\)
\(\dfrac{\frac{1}{a+b}-\frac{1}{a-b}}{\frac{1}{a+b}+\frac{1}{a-b}}\)
\(\dfrac{\frac{5}{x+3}-\frac{5}{x-3}}{\frac{5}{x+3}+\frac{5}{x-3}}\)
- Answer
-
\(\dfrac{-3}{x}\)
\(\dfrac{2+\frac{1}{y+1}}{\frac{1}{y}+\frac{2}{3}}\)
\(\dfrac{\frac{1}{x^{2}}-\frac{1}{y^{2}}}{\frac{1}{x}+\frac{1}{y}}\)
- Answer
-
\(\dfrac{y-x}{xy}\)
\(\dfrac{1+\frac{5}{x}+\frac{6}{x^{2}}}{1-\frac{1}{x}-\frac{12}{x^{2}}}\)
\(\dfrac{1+\frac{1}{y}-\frac{2}{y^{2}}}{1+\frac{7}{y}+\frac{10}{y^{2}}}\)
- Answer
-
\(\dfrac{y-1}{y+5}\)
\(\dfrac{\frac{3 n}{m}-2-\frac{m}{n}}{\frac{3 n}{m}+4+\frac{m}{n}}\)
- Answer
-
\(3x−4\)
\(\dfrac{\frac{y}{x+y}-\frac{x}{x-y}}{\frac{x}{x+y}+\frac{y}{x-y}}\)
\(\dfrac{\frac{a}{a-2}-\frac{a}{a+2}}{\frac{2 a}{a-2}+\frac{a^{2}}{a+2}}\)
- Answer
-
\(\dfrac{4}{a^2 + 4}\)
\(3 - \dfrac{2}{1 - \frac{1}{m+1}}\)
\(\dfrac{x-\frac{1}{1-\frac{1}{x}}}{x+\frac{1}{1+\frac{1}{x}}}\)
- Answer
-
\(\dfrac{(x-2)(x+1)}{(x-1)(x+2)}\)
In electricity theory, when two resistors of resistance \(R_1\) and \(R_2\) ohms are connected in parallel, the total resistance \(R\) is:
\(R = \dfrac{1}{\frac{1}{R_1} + \frac{1}{R_2}}\)
Write this complex fraction as a simple fraction.
According to Einstein's theory of relativity, two velocities \(v_1\) and \(v_2\) are not added according to \(v = v_1 + v_2\), but rather by
\(v = \dfrac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}}\)
Write this complex fraction as a simple fraction.
Einstein's formula is really only applicable for velocities near the speed of light (\(c=186,000\) miles per second). At very much lower velocities, such as 500 miles per hour, the formula \(v=v_1+v_2\) provides an extremely good approximation.
- Answer
-
\(\dfrac{c^2(V_1 + V_2)}{c^2 + V_1V_2}\)
Exercises For Review
Supply the missing word. Absolute value speaks to the question of how ____ and not “which way.”
Find the product. \((3x + 4)^2\)
- Answer
-
\(9x^2 + 24x + 16\)
Factor \(x^4 - y^4\)
Solve the equation \(\dfrac{3}{x-1} - \dfrac{5}{x+3} = 0\).
- Answer
-
\(x=7\)
One inlet pipe can fill a tank in 10 minutes. Another inlet pipe can fill the same tank in 4 minutes. How long does it take both pipes working together to fill the tank?