EQUATIONS INVOLVING FRACTIONS (RATIONAL EQUATIONS)


Note:




If you would like an in-depth review of fractions, click on Fractions.



Solve for x in the following equation.


Example 5: $\displaystyle \frac{4x+5}{9x+10}=\displaystyle \frac{3x-1}{3x-4}-\displaystyle \frac{2x-1}{x+6}$



Recall that you cannot divide by zero. Therefore, the first fraction is valid if , $\quad x\neq -\displaystyle \frac{10}{9},$ the second fraction is valid if $x\neq \displaystyle \frac{4}{3},\quad $and the third fraction is valid $
x\neq -6.$ If $-\displaystyle \frac{10}{9},$
$\displaystyle \frac{4}{3},$ or -6 turn out to be the solutions, you must discard them as extraneous solutions.


Multiply both sides by the least common multiple $\left( x+6\right)
\left( 9x+10\right) \left( 3x-4\right) \qquad $(the smallest number that all the denominators will divide into evenly). This step will eliminate all the denominators in the equation. The resulting equation may be equivalent (same solutions as the original equation) or it may not be equivalent (extraneous solutions),


\begin{eqnarray*}\displaystyle \frac{4x+5}{9x+10} &=&\displaystyle \frac{3x-1}{3...
...laystyle \frac{3x-1}{3x-4}-\displaystyle \frac{2x-1}{x+6}\right)
\end{eqnarray*}

\begin{eqnarray*}\left( x+6\right) \left( 9x+10\right) \left( 3x-4\right) \left(...
...\left( 3x-4\right) \left( \displaystyle \frac{
2x-1}{x+6}\right)
\end{eqnarray*}


which is equivalent to


\begin{eqnarray*}\frac{\left( x+6\right) \left( 9x+10\right) \left( 3x-4\right) ...
...t( 3x-4\right) }{1}\left(
\displaystyle \frac{2x-1}{x+6}\right)
\end{eqnarray*}


which can be rewritten as


\begin{eqnarray*}\frac{\left( x+6\right) \left( 9x+10\right) \left( 3x-4\right) ...
...ght) \left( 3x-4\right) \left(
2x-1\right) }{\left( x+6\right) }
\end{eqnarray*}


which can be rewritten again as


\begin{eqnarray*}\frac{\left( 9x+10\right) }{\left( 9x+10\right) }\cdot \frac{\l...
...c{\left(
9x+10\right) \left( 3x-4\right) \left( 2x-1\right) }{1}
\end{eqnarray*}


which can be rewritten yet again as


\begin{eqnarray*}1\cdot \frac{\left( x+6\right) \left( 3x-4\right) \left( 4x+5\r...
...eft( 9x+10\right) \left( 3x-4\right) \left( 2x-1\right) }{1}
\\
\end{eqnarray*}

\begin{eqnarray*}\left( x+6\right) \left( 3x-4\right) \left( 4x+5\right) &=&\lef...
...{2}+21x-10\right) -\left( 9x+10\right) \left( 6x^{2}-11x+4\right)\end{eqnarray*}

\begin{eqnarray*}12x^{3}+71x^{2}-26x-120 &=&27x^{3}+183x^{2}+116x-60-\left(
54x^...
...x^{2}-26x-120 &=&27x^{3}+183x^{2}+116x-60-54x^{3}+39x^{2}+74x-40
\end{eqnarray*}

\begin{eqnarray*}12x^{3}+71x^{2}-26x-120 &=&-27x^{3}+222x^{2}+190x-100 \\
&& \\
&& \\
39x^{3}-151x^{2}-216x-20 &=&0
\end{eqnarray*}



Solve for x by factoring:


\begin{eqnarray*}39x^{3}-151x^{2}-216x-20 &=&0 \\
&& \\
\left( x-5\right) \left( 39x^{2}+44x+4\right) &=&0
\end{eqnarray*}


The only way that a product can have a value of zero is if at least one of the factors is equal to zero. The factor 3 can never be zero. The factor $
\left( x-5\right) $ is zero when x=5, and the factor $\left(
39x^{2}+44x+4\right) $ is zero when


\begin{eqnarray*}x &=&\frac{-44\pm \sqrt{1312}}{78} \\
&& \\
&& \\
x &=&\frac...
...rt{328}}{78} \\
&& \\
&& \\
x &=&\frac{-22\pm \sqrt{328}}{39}
\end{eqnarray*}


The exact answers are $x=\displaystyle \frac{-22\pm \sqrt{328}}{39}$ and 5.


Check the answer in the original equation.


Check the solution $x=\displaystyle \frac{-22+\sqrt{328}}{39}\smallskip $ by substituting the approximate solution x=-0.09972383907 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.




The checks tells us that the answer $x=\displaystyle \frac{-22+\sqrt{328}}{39}\smallskip $is a solution to the original problem.


Check the solution $x=\displaystyle \frac{-22-\sqrt{328}}{39}\smallskip $ by substituting the approximate solution x=-1.02848128914 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.




The checks tells us that the answer $x=\displaystyle \frac{-22-\sqrt{328}}{39}$ is a solution to the original problem.


Check the solution x=5 by substituting the approximate solution x=5 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.




You can also check your answer by graphing $\quad f(x)=\displaystyle \frac{4x+5}{9x+10}-
\displaystyle \frac{3x-1}{3x-4}+\displaystyle \frac{2x-1}{x+6}\smallskip .\quad $(formed by subtracting the right side of the original equation from the left side). Look to see where the graph crosses the x-axis; the intercept(s) will be the real solution(s). Note that the graph crosses the x-axis in three places: -0.09972383907, -1.02848128914 and 5.




If you would like to work another example, click on Example


If you would like to test yourself by working some problems similar to this example, click on Problem


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Author: Nancy Marcus

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