Rational Equations
Learning Objectives
- Methods for solving rational equations
- Solve rational equations by clearing denominators
- Identify extraneous solutions in a rational equation
[latex]\begin{eqnarray*} \frac{2}{3}\,x - \frac{5}{6}\,= \frac{3}{4}\,& & \text{ Multiply}\,\text{ each} \text{ term}\,\text{ by}\,\text{ LCD}, 12\\ & & \\ \frac{2 (12)}{3}\,x - \frac{5 (12)}{6}\,= \frac{3 (12)}{4}\,& & \text{ Reduce}\,\text{ fractions}\\ & & \\ 2 (4) x - 5 (2) = 3 (3) & & \text{ Multiply}\\ 8 x - 10 = 9 & & \text{ Solve}\\ \underline{+ 10 + 10}\,& & \text{ Add}\,10 \text{ to}\,\text{ both} \text{ sides}\\ 8 x = 19 & & \text{ Divide}\,\text{ both}\,\text{ sides}\,\text{ by}\,8\\ \overline{8}\, \overline{8}\,& & \\ x = \frac{19}{8}\,& & \text{ Our}\,\text{ Solution} \end{eqnarray*}[/latex]
We could have found a common denominator and worked with fractions, but that often leads to more mistakes. We can apply the same idea to solving rational equations. The difference between a linear equation and a rational equation is that rational equations can have polynomials in the numerator and denominator of the fractions. This means that clearing the denominator may sometimes mean multiplying the whole rational equation by a polynomial. In the next example, we will clear the denominators of a rational equation with a terms that has a polynomial in the numerator.Example
Solve the equation: [latex]\displaystyle 3 x - \frac{1}{2}\,= \frac{1}{x}[/latex]Answer: [latex]\begin{eqnarray*} 3 x - \frac{1}{2}\,= \frac{1}{x}\,& & \text{ Multiply}\,\text{ each} \text{ term}\,\text{ by}\,\text{ LCD}, (2 x)\\ 3 x (2 x) - \frac{(2 x)}{2}\,= \frac{(2 x)}{x}\,& & \text{ Reduce} \text{ fractions}\\ 6 x^2 - x = 2 & & \text{ Distribute}\\ \underline{- 2 - 2}\,& & \text{ Subtract}\,2 \text{ from}\,\text{ both} \text{ sides}\\ 6 x^2 - x - 2 = 0 & & \text{ Factor}\\ (3 x - 2) (2 x + 1) = 0 & & \text{ Set}\,\text{ each}\,\text{ factor} \text{ equal}\,\text{ to}\,\text{ zero}\\ 3 x - 2 = 0 \text{ or}\,2 x + 1 = 0 & & \text{ Solve}\,\text{ each} \text{ equation}\\ \underline{+ 2 \quad + 2}\, \underline{- 1 - 1}\,& & \\ 3 x = 2 \hspace{3em}\,2 x = - 1 & & \\ x = \frac{2}{3}\,\text{ or}\,x = - \frac{1}{2}\,& & \text{ Check} \text{ solutions}, \text{ LCD}\,\text{ can}' t \text{ be}\,\text{ zero}\\ 2 \left( \frac{3}{2}\,\right) =2 \left( - \frac{1}{2}\,\right) = - 1 & & \text{ Neither}\,\text{ make}\,\text{ LCD}\,\text{ zero}, \text{ both}\,\text{ are} \text{ solutions}\\ x = \frac{2}{3}\,\text{ or}\,x = - \frac{1}{2}\,& & \text{ Our} \text{ Solution} \end{eqnarray*}[/latex]
Answer
[latex-display]\displaystyle x = \frac{2}{3}\,\text{ or}\,x = - \frac{1}{2}\,[/latex-display]Example
Solve the equation [latex]\displaystyle \frac{8}{x+1}=\frac{4}{3}[/latex].Answer: Clear the denominators by multiplying each side by the common denominator. The common denominator is [latex]3\left(x+1\right)[/latex] since [latex]3\text{ and }x+1[/latex] don't have any common factors.
[latex]\begin{array}{c}3\left(x+1\right)\left(\frac{8}{x+1}\right)=3\left(x+1\right)\left(\frac{4}{3}\right)\end{array}[/latex]
Simplify common factors.
[latex]\begin{array}{c}3\cancel{\left(x+1\right)}\left(\frac{8}{\cancel{x+1}}\right)=\cancel{3}\left(x+1\right)\left(\frac{4}{\cancel{3}}\right)\\24=4\left(x+1\right)\\24=4x+4\end{array}[/latex]
Now this looks like a linear equation, and we can use the addition and multiplication properties of equality to solve it.
[latex]\begin{array}{c}24=4x+4\\\underline{-4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-4}\\20=4x\,\,\,\,\,\,\,\,\\\\x=5\,\,\,\,\,\,\,\,\,\end{array}[/latex]
Check the solution in the original equation.[latex]\begin{array}{r}\,\,\,\,\,\frac{8}{\left(x+1\right)}=\frac{4}{3}\\\\\frac{8}{\left(5+1\right)}=\frac{4}{3}\\\\\frac{8}{6}=\frac{4}{3}\end{array}[/latex]
Reduce the fraction [latex]\frac{8}{6}[/latex] by simplifying the common factor of 2:
[latex]\large\frac{\cancel{2}\cdot4}{\cancel{2}\cdot3}\normalsize=\large\frac{4}{3}[/latex]
Answer
[latex-display]x=5[/latex-display]Example
Solve the equation: [latex]\displaystyle \frac{5 x + 5}{x + 2}\,+ 3 x = \frac{x^2}{x + 2}[/latex]Answer: [latex] \begin{eqnarray*}\frac{5 x + 5}{x + 2}\,+ 3 x = \frac{x^2}{x + 2}\,& & \text{ Multiply} \text{ each}\,\text{ term}\,\text{ by}\,\text{ LCD}, (x + 2)\\ & & \\ \frac{(5 x + 5) (x + 2)}{x + 2}\,+ 3 x (x + 2) = \frac{x^2 (x + 2)}{x + 2} & & \text{ Reduce}\,\text{ fractions}\\ & & \\ 5 x + 5 + 3 x (x + 2) = x^2 & & \text{ Distribute}\\ 5 x + 5 + 3 x^2 + 6 x = x^2 & & \text{ Combine}\,\text{ like}\,\text{ terms}\\ 3 x^2 + 11 x + 5 = x^2 & & \text{ Make}\,\text{ equation}\,\text{ equal} \text{ zero}\\ \underline{- x^2 - x^2}\,& & \text{ Subtract}\,x^2 \text{ from}\,\text{ both} \text{ sides}\\ 2 x^2 + 11 x + 5 = 0 & & \text{ Factor}\\ (2 x + 1) (x + 5) = 0 & & \text{ Set}\,\text{ each}\,\text{ factor} \text{ equal}\,\text{ to}\,\text{ zero}\\ 2 x + 1 = 0 \text{ or}\,x + 5 = 0 & & \text{ Solve}\,\text{ each} \text{ equation}\\ 2 x = - 1 \text{ or}\,x = - 5 & & \\ x = - \frac{1}{2}\,\text{ or}\,- 5 & & \text{ Check}\,\text{ solutions}, \text{ LCD}\,\text{ can}' t \text{ be}\,\text{ zero}\\ - \frac{1}{2}\,+ 2 = \frac{3}{2}\,- 5 + 2 = - 3 & & \text{ Neither} \text{ make}\,\text{ LCD}\,\text{ zero}, \text{ both}\,\text{ are} \text{ solutions}\\ x = - \frac{1}{2}\,\text{ or}\,- 5 & & \text{ Our}\,\text{ Solution} \end{eqnarray*}[/latex]
Answer [latex-display]x = - \frac{1}{2}\,\text{ or}\,x = - 5[/latex-display]Excluded Values and Extraneous Solutions
Some rational expressions have a variable in the denominator. When this is the case, there is an extra step in solving them. Since division by 0 is undefined, you must exclude values of the variable that would result in a denominator of 0. These values are called excluded values. Let’s look at an example.Example
Solve the equation [latex]\displaystyle \frac{2x-5}{x-5}=\frac{15}{x-5}[/latex].Answer: Determine any values for x that would make the denominator 0.
[latex] \frac{2x-5}{x-5}=\frac{15}{x-5}[/latex]
5 is an excluded value because it makes the denominator [latex]x-5[/latex] equal to 0. Since the denominator of each expression in the equation is the same, the numerators must be equal. Set the numerators equal to one another and solve for x.[latex]\begin{array}{r}2x-5=15\\2x=20\\x=10\end{array}[/latex]
Check the solution in the original equation.[latex]\begin{array}{r}\frac{2x-5}{x-5}=\frac{15}{x-5}\,\,\\\\\frac{2(10)-5}{10-5}=\frac{15}{10-5}\\\\\frac{20-5}{10-5}=\frac{15}{10-5}\\\\\frac{15}{5}=\frac{15}{5}\,\,\,\,\,\,\,\,\,\end{array}[/latex]
Answer
[latex-display]x=10[/latex-display]Example
Solve the equation [latex]\displaystyle \frac{16}{m+4}=\frac{{{m}^{2}}}{m+4}[/latex].Answer: Determine any values for m that would make the denominator 0. [latex]−4[/latex] is an excluded value because it makes [latex]m+4[/latex] equal to 0. Since the denominator of each expression in the equation is the same, the numerators must be equal. Set the numerators equal to one another and solve for m.
[latex]\begin{array}{l}16=m^{2}\\\,\,\,0={{m}^{2}}-16\\\,\,\,0=\left( m+4 \right)\left( m-4 \right)\end{array}[/latex]
[latex]\begin{array}{c}0=m+4\,\,\,\,\,\,\text{or}\,\,\,\,\,\,0=m-4\\m=-4\,\,\,\,\,\,\text{or}\,\,\,\,\,\,m=4\\m=4,-4\end{array}[/latex]
Check the solutions in the original equation. Since [latex]m=−4[/latex] leads to division by 0, it is an extraneous solution.[latex]\begin{array}{c}\frac{16}{m+4}=\frac{{{m}^{2}}}{m+4}\\\\\frac{16}{-4+4}=\frac{{{(-4)}^{2}}}{-4+4}\\\\\frac{16}{0}=\frac{16}{0}\end{array}[/latex]
[latex]-4[/latex] is excluded because it leads to division by 0.[latex]\begin{array}{c}\frac{16}{4+4}=\frac{{{(4)}^{2}}}{4+4}\\\\\frac{16}{8}=\frac{16}{8}\end{array}[/latex]
Answer
[latex-display]m=4[/latex-display]More Rational Equations
Sometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out that there is no solution.Example: Solving a Rational Equation Leading to a Quadratic
Solve the following rational equation: [latex]\displaystyle \frac{-4x}{x - 1}+\frac{4}{x+1}=\frac{-8}{{x}^{2}-1}[/latex].Answer: We want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, [latex]{x}^{2}-1=\left(x+1\right)\left(x - 1\right)[/latex]. Then, the LCD is [latex]\left(x+1\right)\left(x - 1\right)[/latex]. Next, we multiply the whole equation by the LCD.
Example
Solve [latex]\displaystyle \frac{3x+2}{x - 2}+\frac{1}{x}=\frac{-2}{{x}^{2}-2x}[/latex].Answer: [latex]x=-1[/latex], ([latex]x=0[/latex] is not a solution).
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- Unit 15: Rational Expressions, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education License: CC BY: Attribution.
- Solve Basic Rational Equations. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Solve Rational Equations with Like Denominators. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- College Algebra. Authored by: Abramson, Jay et al.. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: Public Domain: No Known Copyright. License terms: Download for free at : http://cnx.org/contents/[email protected]:1/Preface.
- Question ID#44861. Authored by: Kulinsky,Carla. License: CC BY: Attribution.
- Question ID 3496. Authored by: Shawn Triplett. License: CC BY: Attribution.
- Rational Expressions - Equations examples. Authored by: Tyler Wallace. Located at: http://www.wallace.ccfaculty.org/book/book.html. License: CC BY: Attribution.