Solving Rational Equations
Learning Outcomes
- Solve rational equations by clearing denominators
- Identify extraneous solutions in a rational equation
[latex]\begin{array}{c}\frac{1}{2}x-3=2-\frac{3}{4}x\\ 4\left(\frac{1}{2}x-3\right)=4\left(2-\frac{3}{4}x\right)\\\text{}\\\,\,\,\,4\left(\frac{1}{2}x\right)-4\left(3\right)=4\left(2\right)+4\left(-\frac{3}{4}x\right)\\2x-12=8-3x\\\underline{+3x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{+3x}\\5x-12=8\,\,\,\,\,\,\,\,\\\,\,\,\,\,\,\,\underline{+12}\,\,\,\,\underline{+12} \\5x=20\\x=4\end{array}[/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 term that has a polynomial in the numerator.Example
Solve the equation [latex] \frac{x+5}{8}=\frac{7}{4}[/latex].Answer: Find the least common denominator of [latex]4[/latex] and [latex]8[/latex]. Remember, to find the LCD, identify the greatest number of times each factor appears in each factorization. Here, [latex]2[/latex] appears [latex]3[/latex] times, so [latex]2\cdot2\cdot2[/latex], or [latex]8[/latex], will be the LCD. Multiply both sides of the equation by the common denominator, [latex]8[/latex], to keep the equation balanced and to eliminate the denominators.
[latex]\begin{array}{r}8\cdot \frac{x+5}{8}=\frac{7}{4}\cdot 8\,\,\,\,\,\,\,\\\\\frac{8(x+5)}{8}=\frac{7(8)}{4}\,\,\,\,\,\,\\\\\frac{8}{8}\cdot (x+5)=\frac{7(4\cdot 2)}{4}\\\\\frac{8}{8}\cdot (x+5)=7\cdot 2\cdot \frac{4}{4}\\\\1\cdot (x+5)=14\cdot 1\,\,\,\end{array}[/latex]
Simplify and solve for x.[latex]\begin{array}{r}x+5=14\\x=9\,\,\,\end{array}[/latex]
Check the solution by substituting [latex]9[/latex] for x in the original equation.[latex]\begin{array}{r}\frac{x+5}{8}=\frac{7}{4}\\\\\frac{9+5}{8}=\frac{7}{4}\\\\\frac{14}{8}=\frac{7}{4}\\\\\frac{7}{4}=\frac{7}{4}\end{array}[/latex]
Therefore, [latex]x=9[/latex].Example
Solve the equation [latex] \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] do not 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}\,\,\,\,\,\dfrac{8}{\left(x+1\right)}=\dfrac{4}{3}\\\\\dfrac{8}{\left(5+1\right)}=\dfrac{4}{3}\\\\\dfrac{8}{6}=\dfrac{4}{3}\\\dfrac{4}{3}=\dfrac{4}{3}\end{array}[/latex]
Therefore, [latex]x=5[/latex].Example
Solve the equation [latex]\frac{x}{3}+1=\frac{4}{3}[/latex].Answer: Both fractions in the equation have a denominator of [latex]3[/latex]. Multiply every term on both sides of the equation (not just the fractions!) by [latex]3[/latex].
[latex] 3\left( \frac{x}{3}+1 \right)=3\left( \frac{4}{3} \right)[/latex]
Apply the distributive property and multiply [latex]3[/latex] by each term within the parentheses. Then simplify and solve for x.[latex]\begin{array}{r}3\left( \frac{x}{3} \right)+3\left( 1 \right)=3\left( \frac{4}{3} \right)\\\\\cancel{3}\left( \frac{x}{\cancel{3}} \right)+3\left( 1 \right)=\cancel{3}\left( \frac{4}{\cancel{3}} \right)\\\\ x+3=4\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-3}\,\,\,\,\,\underline{-3}\\\\x=1\end{array}[/latex]
Therefore, [latex]x=1[/latex].Try it
[ohm_question]189370[/ohm_question]Excluded Values and Extraneous Solutions
As you have seen, 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 [latex]0[/latex] is undefined, you must exclude values of the variable that would result in a denominator of [latex]0[/latex] in the original equation. These values are called excluded values. Let us look at an example.Example
Solve the equation [latex] \frac{2x-5}{x-5}=\frac{15}{x-5}[/latex].Answer: Determine any values for x that would make the denominator [latex]0[/latex].
[latex] \frac{2x-5}{x-5}=\frac{15}{x-5}[/latex]
[latex]5[/latex] is an excluded value because it makes the denominator [latex]x-5[/latex] equal to [latex]0[/latex]. 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]
Therefore, [latex]x=10[/latex].Example
Solve the equation [latex] \frac{16}{m+4}=\frac{{{m}^{2}}}{m+4}[/latex].Answer: Determine any values for m that would make the denominator [latex]0[/latex]. [latex]−4[/latex] is an excluded value because it makes [latex]m+4[/latex] equal to [latex]0[/latex]. 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 [latex]0[/latex], 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 [latex]0[/latex]. Notice, however, that [latex]m=4[/latex] is a solution that results in a true statement:[latex]\begin{array}{c}\frac{16}{4+4}=\frac{{{(4)}^{2}}}{4+4}\\\\\frac{16}{8}=\frac{16}{8}\end{array}[/latex]
Therefore, [latex]m=4[/latex].try it
[ohm_question]4029[/ohm_question]Summary
You can solve rational equations by finding a common denominator. By rewriting the equation so that all terms have the common denominator, you can solve for the variable using just the numerators. Or, you can multiply both sides of the equation by the least common denominator of all fractions so that all terms become polynomials instead of rational expressions. An important step in solving rational equations is to reject any extraneous solutions from the final answer. Extraneous solutions are solutions that do not satisfy the original form of the equation because they produce untrue statements or are excluded values that cause a denominator to equal [latex]0[/latex].Contribute!
Licenses & Attributions
CC licensed content, Original
- Revision and Adaptation. Provided by: Lumen Learning 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.
- Solve Basic Rational Equations. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
CC licensed content, Shared previously
- Unit 15: Rational Expressions, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education License: CC BY: Attribution.
- Ex 2: Solve a Literal Equation for a Variable. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.