9.1: Rational Equations (2024)

  • Page ID
    45127
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)

    \( \newcommand{\vectorC}[1]{\textbf{#1}}\)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}}\)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}\)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)

    When solving rational equations, we can solve by using the same strategy we used to solve linear equations with fractions: clearing denominators. However, we first need to revisit excluded values.

    Excluded Values

    Note

    A rational expression is undefined where the denominator is zero. Recall, we cannot divide by zero, so it is critical we find these values and exclude them from the solution.

    Example 9.1.1

    Find the excluded value(s) of the expression: \(\dfrac{-3z}{z+5}\)

    Solution

    Step 1. Set the denominator of the rational expression equal to zero: \[z+5=0\nonumber\]

    Step 2. Solve the equation for \(z\): \[\begin{aligned}z+5&=0 \\ z&=-5\end{aligned}\]

    Step 3. The values found in the previous step are the values excluded from the expression. Hence, the excluded value is \(z = −5\).

    Example 9.1.2

    Find the excluded value(s) of the expression: \(\dfrac{x^2-1}{3x^2+5x}\)

    Solution

    Step 1. Set the denominator of the rational expression equal to zero: \[3x^2+5x=0\nonumber\]

    Step 2. Solve the equation for \(x\): \[\begin{aligned}3x^2+5x&=0 \\ x(3x+5)&=0 \\ x=0\quad &\text{or}\quad 3x+5=0 \\ x=0\quad &\text{or}\quad 3x=-5 \\ x=0\quad &\text{or}\quad x=-\dfrac{5}{3}\end{aligned}\]

    Step 3. The values found in the previous step are the values excluded from the expression. Hence, the excluded values are \(x = 0\) and \(x = −5\).

    Definition: Extraneous Solution

    Recall, the excluded values are values in which make the expression undefined. Hence, when solving a rational equation, the solution(s) is any value(s) except the excluded values. If we obtain a solution that is an excluded value, we call this an extraneous solution.

    Clearing Denominators Using the LCD

    Let’s recall an example from solving linear equations with fractions. Let’s be reminded of the process for clearing denominators when solving equations. In this section, we solve rational equations using the same process.

    Example 9.1.3

    Solve for \(x\): \(\dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{3}{4}\)

    Solution

    This is a similar problem from solving linear equations with fractions. We will clear denominators by multiplying each term by the LCD.

    \[\begin{array}{rl}\dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{3}{3}&\text{Multiply each term by LCD = }12 \\ \color{blue}{12}\color{black}{}\cdot\dfrac{2}{3}x-\color{blue}{12}\color{black}{}\cdot\dfrac{5}{6}=\color{blue}{12}\color{black}{}\cdot\dfrac{3}{4}&\text{Clear denominators} \\ 8x-10=9&\text{Isolate the variable term} \\ 8x=19&\text{Solve for }x \\ x=\dfrac{19}{8}&\text{Solution}\end{array}\nonumber\]

    Steps for solving rational equations

    Step 1. Determine the excluded values of the equation.

    Step 2. Clear denominators by multiplying each term by the lowest common denominator.

    Step 3. Solve the equation.

    Step 4. Verify that the solutions obtained are not an excluded value.

    Example 9.1.4

    Solve for \(x\): \(\dfrac{5x+5}{x+2}+3x=\dfrac{x^2}{x+2}\)

    Solution

    We can solve by following the above steps.

    Step 1. Determine the excluded values of the equation. \[\begin{aligned}x+2&=0 \\ x&=-2\end{aligned}\] The excluded value is \(x = −2\). This means we can obtain any solution except for \(x = −2\).

    Step 2. Clear denominators by multiplying each term by the lowest common denominator. \[\begin{array}{rl}\dfrac{5x+5}{x+2}+3x=\dfrac{x^2}{x+2}&\text{Multiply each term by LCD }=(x+2) \\ \color{blue}{(x+2)}\color{black}{}\cdot\dfrac{(5x+5)}{x+2}+\color{blue}{(x+2)}\color{black}{}\cdot 3x=\color{blue}{(x+2)}\color{black}{}\cdot\dfrac{x^2}{x+2}&\text{Clear denominators} \\ 5x+5+3x(x+2)=x^2\end{array}\nonumber\]

    Step 3. Solve the equation. \[\begin{array}{rl} 5x+5+3x(x+2)=x^2&\text{Distribute} \\ 5x+5+3x^2+6x=x^2&\text{Combine like terms} \\ 3x^2+11x+5=x^2&\text{Notice the term }x^2\text{; we solve by factoring} \\ 2x^2+11x+5=0&\text{Zero on one side and factor the other side} \\ (2x+1)(x+5)=0&\text{Apply the zero product rule} \\ 2x+1=0\text{ or }x+5=0&\text{Isolate variable terms} \\ 2x=-1\text{ or }x=-5&\text{Solve for }x \\ x=-\dfrac{1}{2}\text{ or }x=-5&\text{Solutions}\end{array}\nonumber\]

    Step 4. Verify that the solutions obtained are not an excluded value. Since the excluded value is \(x = −2\), and the solutions we obtained are \(x = −\dfrac{1}{2}\) and \(x = −5\), then we can conclude that \(x = −\dfrac{1}{2}\) and \(x = −5\) are, in fact, the solutions.

    Example 9.1.5

    Solve for \(x\): \(\dfrac{x}{x+2}+\dfrac{1}{x+1}=\dfrac{5}{(x+1)(x+2)}\)

    Solution

    We can solve by following the above steps.

    Step 1. Determine the excluded values of the equation. \[\begin{array}{rl}x+2=0&x+1=0 \\ x=-2&x=-1\end{array}\nonumber\] The excluded values are \(x = −2\) and \(x = −1\). This means we can obtain any solution except for \(x = −2\) and \(x = −1\).

    Step 2. Clear denominators by multiplying each term by the lowest common denominator. \[\dfrac{x}{x+2}+\dfrac{1}{x+1}=\dfrac{5}{(x+1)(x+2)}\quad\text{Multiply each term by LCD }=(x+2)(x+1)\nonumber\] Clear denominators: \[\begin{aligned} \color{blue}{(x+2)(x+1)}\color{black}{}\cdot\dfrac{x}{x+2}+\color{blue}{(x+2)(x+1)}\color{black}{}\cdot\dfrac{1}{x+1}&=\color{blue}{(x+2)(x+1)}\color{black}{}\cdot\dfrac{5}{(x+1)(x+2)} \\ x(x+1)+1(x+2)&=5\end{aligned}\]

    Step 3. Solve the equation. \[\begin{array}{rl}x(x+1)+1(x+2)=5&\text{Distribute} \\ x^2+x+x+2=5&\text{Combine like terms} \\ x^2+2x+2=5&\text{Notice the term }x^2\text{; we solve by factoring} \\ x^2+2x-3=0&\text{Zero on one side and factor the other side} \\ (x+3)(x-1)=0&\text{Apply the zero product rule} \\ x+3=0\text{ or }x-1=0&\text{Isolate variable terms} \\ x=-3\text{ or }x=1&\text{Solutions}\end{array}\nonumber\]

    Step 4. Verify that the solutions obtained are not an excluded value. Since the excluded values are \(x = −2\) and \(x = −1\), and the solutions we obtained are \(x = −3\) and \(x = 1\), then we can conclude that \(x = −2\) and \(x = −1\) are, in fact, the solutions.

    Factoring Denominators

    In Example 9.1.5, the denominators are factored, but this is not always the case. Often we will need to factor denominators before finding the LCD.

    Example 9.1.6

    Solve for \(t\): \(\dfrac{t}{t-1}-\dfrac{1}{t-2}=\dfrac{11}{t^2-3t+2}\)

    Solution

    We can solve by following the above steps.

    Step 1. Determine the excluded values of the equation. Since we have three different denominators, we find excluded values for all different denominators. \[\begin{array}{rllr} t-1=0&t-2=0&\quad &t^2-3t+2=0 \\ t=1&t=2&\quad & (t-2)(t-1)=0 \\ &&& t-2=0\quad t-1=0 \\ &&&t=2\quad t=1\end{array}\nonumber\] The excluded values are \(t = 1\) and \(t = 2\). This means we can obtain any solution except for \(t = 1\) and \(t = 2\). Even though we obtained repeated values, we still must find the excluded values for each denominator to verify the solution(s) in the last step.

    Step 2. Clear denominators by multiplying each term by the lowest common denominator. \[\begin{array}{rl}\dfrac{t}{t-1}-\dfrac{1}{t-2}=\dfrac{11}{t^2-3t+2}&\text{Factor denominator} \\ \dfrac{t}{t-1}-\dfrac{1}{t-2}=\dfrac{11}{(t-2)(t-1)}&\text{Multiply each term by LCD }=(t-2)(t-1)\end{array}\nonumber\] Clear denominators: \[\begin{aligned}\color{blue}{(t-2)(t-1)}\color{black}{}\cdot\dfrac{t}{t-1}-\color{blue}{(t-2)(t-1)}\color{black}{}\cdot\dfrac{1}{t-2}&=\color{blue}{(t-2)(t-1)}\color{black}{}\cdot\dfrac{11}{(t-2)(t-1)} \\ t(t-2)-1(t-1)&=11\end{aligned}\]

    Step 3. Solve the equation. \[\begin{array}{rl}t(t-2)-1(t-1)=11&\text{Distribute} \\ t^2-2t-t+1=11&\text{Combine like terms} \\ t^2-3t+1=11&\text{Notice the term }t^2\text{; we solve by factoring} \\ t^2-3t-10=0&\text{Zero on one side and factor the other side}\end{array}\nonumber\]
    \[\begin{array}{rl}(t+2)(t-5)=0&\text{Apply the zero product rule} \\ t+2=0\text{ or }t-5=0&\text{Isolate variable terms} \\ t=-2\text{ or }t=5&\text{Solutions}\end{array}\nonumber\]

    Step 4. Verify that the solutions obtained are not an excluded value. Since the excluded values are \(t = 1\) and \(t = 2\), and the solutions we obtained are \(t = −2\) and \(t = 5\), then we can conclude that \(t = −2\) and \(t = 5\) are, in fact, the solutions.

    Note

    Rational functions are used to approximate or model more complex equations in science and engineering including physics, chemistry, biochemistry, optics and photography, and acoustics.

    Solving Rational Equations with Extraneous Solutions

    Example 9.1.7

    Solve for \(n\): \(\dfrac{n}{n+5}-\dfrac{2}{n-9}=\dfrac{-11n+15}{n^2-4n-45}\)

    Solution

    We can solve by following the above steps.

    Step 1. Determine the excluded values of the equation. Since \(n^2−4n−45\) factors into \((n+5)(n−9)\), which are the factors of the denominators on the left side, we take factors \((n+5)\) and \((n−9)\) and find the excluded values. \[\begin{array}{rl}n+5=0&n-9=0 \\ n=-5&n=9\end{array}\nonumber\] The excluded values are \(n = −5\) and \(n = 9\). This means we can obtain any solution except for \(n = −5\) and \(n = 9\).

    Step 2. Clear denominators by multiplying each term by the lowest common denominator. \[\begin{array}{rl}\dfrac{n}{n+5}-\dfrac{2}{n-9}=\dfrac{-11n+15}{n^2-4n-45}&\text{Factor denominator} \\ \dfrac{n}{n+5}-\dfrac{2}{n-9}=\dfrac{-11n+15}{(n+5)(n-9)}&\text{Multiply each term by LCD }=(n+5)(n-9)\end{array}\nonumber\] Clear denominators: \[\begin{aligned}\color{blue}{(n+5)(n-9)}\color{black}{}\cdot\dfrac{n}{n+5}-\color{blue}{(n+5)(n-9)}\color{black}{}\cdot\dfrac{2}{n-9}&=\color{blue}{(n+5)(n-9)}\color{black}{}\cdot\dfrac{-11n+15}{(n+5)(n-9)} \\ n(n-9)-2(n+5)&=-11n+15\end{aligned}\]

    Step 3. Solve the equation. \[\begin{array}{rl}n(n-9)-2(n+5)=-11n+15&\text{Distribute} \\ n^2-9n-2n-10=-11n+15&\text{Combine like terms} \\ n^2-11n-10=-11n+15&\text{Notice the term }n^2\text{; we solve by factoring} \\ n^2-25=0&\text{Zero on one side and factor the other side} \\ (n+5)(n-5)=0&\text{Apply the zero product rule} \\ n+5=0\text{ or }n-5=0&\text{Isolate variable terms} \\ n=-5\text{ or }n=5&\text{Solutions}\end{array}\nonumber\]

    Step 4. Verify that the solutions obtained are not an excluded value. Since the excluded values are \(n = −5\) and \(n = 9\), and the solutions we obtained are \(n = −5\) and \(n = 5\), then \(n = −5\) is an extraneous solution and we omit \(n = −5\). Hence, we can conclude the solution is \(n = 5\).

    Rational Equations Homework

    Solve. Be sure to verify all solutions.

    Exercise 9.1.1

    \(3x-\dfrac{1}{2}-\dfrac{1}{x}=0\)

    Exercise 9.1.2

    \(x+\dfrac{20}{x-4}=\dfrac{5x}{x-4}-2\)

    Exercise 9.1.3

    \(x+\dfrac{6}{x-3}=\dfrac{2x}{x-3}\)

    Exercise 9.1.4

    \(\dfrac{2x}{3x-4}=\dfrac{4x+5}{6x-1}-\dfrac{3}{3x-4}\)

    Exercise 9.1.5

    \(\dfrac{3m}{2m-5}-\dfrac{7}{3m+1}=\dfrac{3}{2}\)

    Exercise 9.1.6

    \(\dfrac{4-x}{1-x}=\dfrac{12}{3-x}\)

    Exercise 9.1.7

    \(\dfrac{7}{y-3}-\dfrac{1}{2}=\dfrac{y-2}{y-4}\)

    Exercise 9.1.8

    \(\dfrac{1}{x+2}-\dfrac{1}{2-x}=\dfrac{3x+8}{x^2-4}\)

    Exercise 9.1.9

    \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{5}{6}\)

    Exercise 9.1.10

    \(\dfrac{3}{2x+1}+\dfrac{2x+1}{1-2x}=1-\dfrac{8x^2}{4x^2-1}\)

    Exercise 9.1.11

    \(\dfrac{x-2}{x+3}-\dfrac{1}{x-2}=\dfrac{1}{x^2+x-6}\)

    Exercise 9.1.12

    \(\dfrac{3}{x+2}+\dfrac{x-1}{x+5}=\dfrac{5x+20}{6x+24}\)

    Exercise 9.1.13

    \(\dfrac{x}{x-1}-\dfrac{2}{x+1}=\dfrac{4x^2}{x^2-1}\)

    Exercise 9.1.14

    \(\dfrac{2x}{x+1}-\dfrac{3}{x+5}=\dfrac{-8x^2}{x^2+6x+5}\)

    Exercise 9.1.15

    \(\dfrac{x-5}{x-9}+\dfrac{x+3}{x-3}=\dfrac{-4x^2}{x^2-12x+27}\)

    Exercise 9.1.16

    \(\dfrac{x-3}{x-6}+\dfrac{x+5}{x+3}=\dfrac{-2x^2}{x^2-3x-18}\)

    Exercise 9.1.17

    \(\dfrac{4x+1}{x+3}+\dfrac{5x-3}{x-1}=\dfrac{8x^2}{x^2+2x-3}\)

    Exercise 9.1.18

    \(\dfrac{6x+5}{2x^2-2x}-\dfrac{2}{1-x^2}=\dfrac{3x}{x^2-1}\)

    Exercise 9.1.19

    \(x+1=\dfrac{4}{x+1}\)

    Exercise 9.1.20

    \(\dfrac{x^2+6}{x-1}+\dfrac{x-2}{x-1}=2x\)

    Exercise 9.1.21

    \(\dfrac{x-4}{x-1}=\dfrac{12}{3-x}+1\)

    Exercise 9.1.22

    \(\dfrac{4x}{2x-6}-\dfrac{4}{5x-15}=\dfrac{1}{2}\)

    Exercise 9.1.23

    \(\dfrac{7}{3-x}+\dfrac{1}{2}=\dfrac{3}{4-x}\)

    Exercise 9.1.24

    \(\dfrac{2}{3-x}-\dfrac{6}{8-x}=1\)

    Exercise 9.1.25

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

    Exercise 9.1.26

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

    Exercise 9.1.27

    \(\dfrac{3x-5}{5x-5}+\dfrac{5x-1}{7x-7}-\dfrac{x-4}{1-x}=2\)

    Exercise 9.1.28

    \(\dfrac{x-1}{x-2}+\dfrac{x+4}{2x+1}=\dfrac{1}{2x^2-3x-2}\)

    Exercise 9.1.29

    \(\dfrac{x}{x+3}-\dfrac{4}{x-2}=\dfrac{-5x^2}{x^2+x-6}\)

    Exercise 9.1.30

    \(\dfrac{2x}{x+2}+\dfrac{2}{x-4}=\dfrac{3x}{x^2-2x-8}\)

    Exercise 9.1.31

    \(\dfrac{x}{x+1}-\dfrac{3}{x+3}=\dfrac{-2x^2}{x^2+4x+3}\)

    Exercise 9.1.32

    \(\dfrac{x-3}{x+6}+\dfrac{x-2}{x-3}=\dfrac{x^2}{x^2+3x-18}\)

    Exercise 9.1.33

    \(\dfrac{x+3}{x-2}+\dfrac{x-2}{x+1}=\dfrac{9x^2}{x^2-x-2}\)

    Exercise 9.1.34

    \(\dfrac{3x-1}{x+6}-\dfrac{2x-3}{x-3}=\dfrac{-3x^2}{x^2+3x-18}\)

    9.1: Rational Equations (2024)

    References

    Top Articles
    Latest Posts
    Article information

    Author: Fredrick Kertzmann

    Last Updated:

    Views: 6721

    Rating: 4.6 / 5 (66 voted)

    Reviews: 89% of readers found this page helpful

    Author information

    Name: Fredrick Kertzmann

    Birthday: 2000-04-29

    Address: Apt. 203 613 Huels Gateway, Ralphtown, LA 40204

    Phone: +2135150832870

    Job: Regional Design Producer

    Hobby: Nordic skating, Lacemaking, Mountain biking, Rowing, Gardening, Water sports, role-playing games

    Introduction: My name is Fredrick Kertzmann, I am a gleaming, encouraging, inexpensive, thankful, tender, quaint, precious person who loves writing and wants to share my knowledge and understanding with you.