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.


Problem 5.3a: $\displaystyle \frac{2x+5}{x-2}=\displaystyle \frac{x+7}{x+2}+\displaystyle \frac{x-6}{
x-1}$


Answer: $x=\displaystyle \frac{4}{3},$ -4



Solution:


Recall that you cannot divide by zero. Therefore, the first fraction is valid if , $\quad x\neq 2,$ the second fraction is valid if $
x\neq -2,\quad $and the third fraction is valid is $x\neq 1$. If $
\quad \pm 2$ $1\quad $ turn out to be the solutions, you must discard them as extraneous solutions.



Multiply both sides by the least common multiple $\left( x-2\right)
\left( x+2\right) \left( x-1\right) \qquad $(the smallest number that all the denominators will divide into evenly). This step will eliminate all the denominators.


\begin{eqnarray*}\displaystyle \frac{2x+5}{x-2} &=&\displaystyle \frac{x+7}{x+2}...
...isplaystyle \frac{x+7}{x+2}+\displaystyle \frac{x-6}{x-1}\right)
\end{eqnarray*}

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


which is equivalent to


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


which can be rewritten as


\begin{eqnarray*}\frac{\left( x-2\right) \left( x+2\right) \left( x-1\right) \le...
...right) \left( x-1\right) \left(
x-6\right) }{\left( x-1\right) }
\end{eqnarray*}


which can be rewritten as


\begin{eqnarray*}\frac{\left( x-2\right) }{\left( x-2\right) }\cdot \frac{\left(...
...\frac{\left(
x-2\right) \left( x+2\right) \left( x-6\right) }{1}
\end{eqnarray*}


which can be simplified to

\begin{eqnarray*}1\cdot \frac{\left( x+2\right) \left( x-1\right) \left( 2x+5\ri...
...+7\right) +\left( x-2\right) \left( x+2\right)
\left( x-6\right)
\end{eqnarray*}

\begin{eqnarray*}\left( x+2\right) \left[ \left( x-1\right) \left( 2x+5\right) \...
...[
x^{2}+6x-7\right] +\left( x-2\right) \left[ x^{2}-4x-12\right]
\end{eqnarray*}

\begin{eqnarray*}x\left[ 2x^{2}+3x-5\right] +2\left[ 2x^{2}+3x-5\right] &=&x\lef...
...&& \\
&&+x\left[ x^{2}-4x-12\right] -2\left[ x^{2}-4x-12\right]
\end{eqnarray*}

\begin{eqnarray*}2x^{3}+3x^{2}-5x+4x^{2}+6x-10 &=&x^{3}+6x^{2}-7x-2x^{2}-12x+14 \\
&& \\
&&+x^{3}-4x^{2}-12x-2x^{2}+8x+24
\end{eqnarray*}

\begin{eqnarray*}2x^{3}+7x^{2}+x-10 &=&2x^{3}-2x^{2}-23x+38 \\
&& \\
&& \\
9x^{2}+24x-48 &=&0 \\
&& \\
&& \\
3x^{2}+8x-16 &=&0
\end{eqnarray*}


Solve for x using the quadratic formula $x=\displaystyle \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}
$


$a=3,\quad b=8,\quad c=-16$

\begin{eqnarray*}3x^{2}+8x-16 &=&0 \\
&& \\
&& \\
x &=&\displaystyle \frac{-\...
... \\
&& \\
&& \\
x &=&\displaystyle \frac{-8\pm \sqrt{256}}{6}
\end{eqnarray*}

\begin{eqnarray*}x &=&\displaystyle \frac{-8\pm 16}{6} \\
&& \\
&& \\
x &=&\d...
...& \\
&& \\
x &=&\displaystyle \frac{-8-16}{6}=\frac{-24}{6}=-4
\end{eqnarray*}


The answers are $x=\displaystyle \frac{4}{3}$ and $-4.\medskip\medskip $


Check the solution $x=\displaystyle \frac{4}{3}$ by substituting 1.333333 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 left side of the equation does not quite equal the right side of the equation because we rounded our answer. However, they are close enough for us to conclude that $x=\displaystyle \frac{4}{3}$ is a solution.



Check the solution x=-4 by substituting -4 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.



Since the left side of the original equation is equal to the right side of the original equation after we substitute the value -4 for x, then x=4is a solution.


You can also check your answer by graphing $\quad f(x)=\displaystyle \frac{2x+5}{x-2}-
\displaystyle \frac{x+7}{x+2}-\displaystyle \frac{x-6}{x-1}\mathbf{\bigskip\bigskip }.\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; that will be the real solution. Note that the graph crosses the x-axis at two places: $
x=1.333333\approx \displaystyle \frac{4}{3}$ and -4.


This verifies our solution graphically.


We have verified our solution both algebraically and graphically.




If you would like to review the solution to problem 5.3b, click on Problem.


If you would like to go back to the equation table of contents, click on Contents


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Author: Nancy Marcus

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