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Precalculus for Team-Based Inquiry Learning 2025 Edition — Instructor Version

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Section 4.6 Rational Equations (PR6)

Subsection 4.6.1 Activities

Definition 4.6.1.

An algebraic expression is called a rational expression if it can be written as the ratio of two polynomials, \(p\) and \(q\text{.}\)
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An equation is called a rational equation if it consists of only rational expressions and constants.
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Observation 4.6.2.

Technically, linear and quadratic equations are also rational equations. They are a special case where the denominator of the rational expressions is \(1\text{.}\) We will focus in this section on cases where the denominator is not a constant; that is, rational equations where there are variables in the denominator.
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With variables in the denominator, there will often be values that cause the denominator to be zero. This is a problem because division by zero is undefined. Thus, we need to be sure to exclude any values that would make those denominators equal to zero.
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Remark 4.6.3.

For this section, it might be helpful to refer back to Activity 1.1.3, where you solved a linear equation with fractions. When revisiting that activity, think about how you had to carefully choose a number to get the denominators to cancel out. Keep this in mind as you work through the next activity.
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Activity 4.6.4.

Which value(s) should be excluded as possible solutions to the following rational equations? Select all that apply.
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(a)
\begin{equation*} \dfrac{2}{x+5}=\dfrac{x-3}{x-8}-7 \end{equation*}
  1. \(\displaystyle -7\)
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  2. \(\displaystyle -5\)
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  3. \(\displaystyle 2\)
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  4. \(\displaystyle 3\)
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  5. \(\displaystyle 8\)
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Answer.
B and E
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(b)
\begin{equation*} \dfrac{x^2-6x+8}{x^2-4x+3}=0 \end{equation*}
  1. \(\displaystyle 0\)
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  2. \(\displaystyle 1\)
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  3. \(\displaystyle 2\)
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  4. \(\displaystyle 3\)
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  5. \(\displaystyle 4\)
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Answer.
B and D
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Activity 4.6.5.

Consider the rational equation
\begin{equation*} 5 = -\dfrac{6}{x-2} \end{equation*}
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(a)
What value should be excluded as a possible solution?
  1. \(\displaystyle 5\)
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  2. \(\displaystyle 6\)
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  3. \(\displaystyle -6\)
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  4. \(\displaystyle 2\)
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  5. \(\displaystyle -2\)
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Answer.
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(b)
To solve, we begin by clearing out the fraction involved. What can we multiply each term by that will clear the fraction?
  1. \(\displaystyle x-5\)
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  2. \(\displaystyle x-6\)
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  3. \(\displaystyle x+6\)
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  4. \(\displaystyle x-2\)
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  5. \(\displaystyle x+2\)
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Answer.
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(c)
Multiply each term by the expression you chose and simplify. Which of the following linear equations does the rational equation simplify to?
  1. \(\displaystyle 5(x-5) = -6\)
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  2. \(\displaystyle 5(x-6) = -6\)
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  3. \(\displaystyle 5(x+6) = -6\)
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  4. \(\displaystyle 5(x-2) = -6\)
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  5. \(\displaystyle 5(x+2) = -6\)
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Answer.
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(d)
Solve the linear equation. Check your answer using the original rational equation.
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Answer.
\(x=\dfrac{4}{5}\)
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Activity 4.6.6.

Consider the rational equation
\begin{equation*} \dfrac{4}{x+1} = -\dfrac{2}{x+6} \end{equation*}
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(a)
What values should be excluded as possible solutions?
  1. \(2\) and \(4\)
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  2. \(1\) and \(6\)
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  3. \(-1\) and \(-6\)
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  4. \(1\) and \(4\)
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  5. \(2\) and \(6\)
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Answer.
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(b)
To solve, we’ll once again begin by clearing out the fraction involved. Which of the following should we multiply each term by to clear out all of the fractions?
  1. \(x-2\) and \(x-4\)
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  2. \(x-1\) and \(x-6\)
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  3. \(x+1\) and \(x+6\)
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  4. \(x-1\) and \(x-4\)
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  5. \(x-2\) and \(x-6\)
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Answer.
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(c)
Multiply each term by the expressions you chose and simplify. Which of the following linear equations does the rational equation simplify to?
  1. \(\displaystyle 4(x+1) = -2(x+6)\)
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  2. \(\displaystyle 4(x+6) = -2(x+1)\)
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  3. \(\displaystyle 4(x+1)(x+6) = -2(x+1)(x+6)\)
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  4. \(\displaystyle 4(x+1) = -2(-x-6)\)
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  5. \(\displaystyle 4(x+6) = -2(-x-1)\)
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Answer.
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(d)
Solve the linear equation. Check your answer using the original rational equation.
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Answer.
\(x=-\dfrac{13}{3}\)
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Observation 4.6.7.

In Activity 4.6.6, you may have noticed that the resulting linear equation looked like the result of cross-multiplying. This is no coincidence! Cross-multiplying is a method of clearing out fractions that works specifically when the equation is in proportional form: \(\dfrac{a}{b}=\dfrac{c}{d}\text{.}\)
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Activity 4.6.8.

Consider the rational equation
\begin{equation*} \dfrac{x}{x+2} = -\dfrac{2}{x+2}-\dfrac{2}{5} \end{equation*}
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(a)
What value(s) should be excluded as possible solutions?
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Answer.
\(x=-2\)
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(b)
To solve, we’ll once again begin by clearing out the fraction involved. Which of the following should we multiply each term by to clear out all of the fractions?
  1. \(x+2\text{,}\) \(x+2\text{,}\) and \(5\)
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  2. \(x+2\) and \(5\)
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  3. \(\displaystyle x+2\)
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  4. \(\displaystyle 5\)
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Answer.
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(c)
Multiply each term by the expressions you chose and simplify. You should end up with a linear equation.
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Answer.
\(5x=-10-2(x+2)\)
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(d)
Solve the linear equation. Check your answer using the original rational equation.
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Answer.
No solution
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Observation 4.6.9.

Activity 4.6.8 demonstrates why it is so important to determine excluded values and check our answers when solving rational equations. Just because a number is a solution to the linear equation we found, it doesn’t mean it is automatically a solution to the rational equation we started with.
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Activity 4.6.10.

Consider the rational equation
\begin{equation*} \dfrac{2x}{x-1}- \dfrac{3}{x-3} = \dfrac{x^2-11x+18}{x^2-4x+3} \end{equation*}
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(a)
What values should be excluded as possible solutions? Select all that apply.
  1. \(\displaystyle 0\)
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  2. \(\displaystyle 1\)
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  3. \(\displaystyle 2\)
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  4. \(\displaystyle 3\)
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  5. \(\displaystyle 9\)
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Answer.
B and D
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(b)
To solve, we’ll begin by clearing out any fractions involved. Which of the following should we multiply each term by to clear out all of the fractions?
  1. \(\displaystyle x-1\)
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  2. \(x-1\) and \(x-3\)
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  3. \(x-1\text{,}\) \(x-3\text{,}\) and \(x^2-4x+3\)
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  4. \(x-1\) and \(x^2-4x+3\)
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  5. \(x-3\) and \(x^2-4x+3\)
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Answer.
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(c)
Multiply each term by the expressions you chose and simplify. Notice that the result is a quadratic equation. Which of the following quadratic equations does the rational equation simplify to?
  1. \(\displaystyle x^2+2x-15=0\)
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  2. \(\displaystyle x^2-11x+18=0\)
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  3. \(\displaystyle x^2-9x-9=0\)
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  4. \(\displaystyle x^2-13x+21=0\)
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Answer.
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(d)
Solve the quadratic equation. Check your answer using the original rational equation. What are the solutions to the rational equation?
  1. \(x=3\) and \(x=-5\)
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  2. \(x=-3\) and \(x=5\)
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  3. \(\displaystyle x=3\)
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  4. \(\displaystyle x=-5\)
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  5. \(\displaystyle x=-3\)
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  6. \(\displaystyle x=5\)
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Answer.
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Activity 4.6.11.

Consider the rational equation
\begin{equation*} \dfrac{2x}{x-2}- \dfrac{x^2+21x-15}{x^2+3x-10} = \dfrac{-6}{x+5} \end{equation*}
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(a)
What values should be excluded as possible solutions?
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Answer.
\(x=2\) and \(x=-5\)
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(b)
What expression(s) should we multiply by to clear out all of the fractions?
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Answer.
\(x-2\) and \(x+5\)
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(c)
Multiply each term by the expressions you chose and simplify. Your result should be a quadratic equation.
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Answer.
\(x^{2}-5x+3=0\)
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(d)
Solve the quadratic equation. Check your answer using the original rational equation. What are the solutions to the rational equation?
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Answer.
\(x=\dfrac{5\pm \sqrt{13}}{2}\)
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Activity 4.6.12.

Solve the following rational equations.
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(a)
\(\displaystyle \dfrac{4}{x}+9=16\)
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Answer.
\(x=\dfrac{4}{7}\)
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(b)
\(\displaystyle -5=\dfrac{2}{x-4}\)
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Answer.
\(x=\dfrac{18}{5}\)
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(c)
\(\displaystyle \dfrac{-3}{x-10}=\dfrac{x}{x-6}\)
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Answer.
\(x=9\) and \(x=-2\)
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(d)
\(\displaystyle \dfrac{x+2}{x-3}+\dfrac{x}{2x-1}=6\)
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Answer.
\(x=\dfrac{7\pm\sqrt{29}}{3}\)
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Subsection 4.6.2 Exercises

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