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 3:

$\displaystyle \frac{3x+1}{x^{2}-7x+10}+3=5-\displaystyle \frac{107}{4x+5}$



Recall that you cannot divide by zero. Therefore, we must eliminate any values of x that will cause the denominator to have a value of zero. We determine these values by setting the denominators equal to zero and solving for x.

\begin{eqnarray*}&&\\
If\ x^{2}-7x+10 &=&0 \\
&& \\
\left( x-2\right) \left( x-5\right) &=&0\\
&&\\
&&
\end{eqnarray*}


The only way a product equals zero is if at least one of the factors equals zero.

\begin{eqnarray*}&&\\
If\ x-2 &=&0,\ then\ x=2 \\
&& \\
If\ x-5 &=&0,\ then\ x=5 \\
&& \\
&\rightarrow &x\neq 2\ 5\\
&&\\
&&
\end{eqnarray*}
\begin{eqnarray*}&&\\
If\ 4x+5 &=&0 \\
&& \\
x &=&-\displaystyle \frac{5}{4} ...
...& \\
&\rightarrow &x\neq -\displaystyle \frac{5}{4}\\
&&\\
&&
\end{eqnarray*}


If any of the solutions turn out to be either 2, $5\ -\displaystyle \frac{5}{4}$, we will discard them as extraneous solutions.



Simplify the equation by subtracting 3 from both sides of the original equation.

\begin{eqnarray*}&&\\
\mbox{ Original Equation } &:&\displaystyle \frac{3x+1}{x...
...rac{3x+1}{x^{2}-7x+10} &=&2-\displaystyle \frac{107}{4x+5}\\
&&
\end{eqnarray*}




Simplify the equation by getting rid of the denominators. You do this by multiplying both sides by the product of the denominators: $\left(
x^{2}-7x+10\right) \left( 4x+5\right) .$

\begin{eqnarray*}&& \\
\displaystyle \frac{3x+1}{x^{2}-7x+10}=2-\displaystyle \...
...t) \left( 2-\displaystyle \frac{107}{
4x+5}\right) && \\
&& \\
\end{eqnarray*}
\begin{eqnarray*}&& \\
\displaystyle \frac{\left( x^{2}-7x+10\right) \left( 4x+...
...t( x^{2}-7x+10\right) \left(
4x+5\right) \left( 2\right) }{1} &&
\end{eqnarray*}



\begin{eqnarray*}&&\\
-\displaystyle \frac{\left( x^{2}-7x+10\right) \left( 4x+...
...( x^{2}-7x+10\right)
\left( 4x+5\right) \left( 2\right) }{1}
&&
\end{eqnarray*}



\begin{eqnarray*}&&\\
-\displaystyle \frac{\left( 4x+5\right) }{\left( 4x+5\rig...
...frac{\left( x^{2}-7x+10\right) \left( 107\right) }{1} && \\
&&
\end{eqnarray*}



\begin{eqnarray*}&& \\
\left( 12x^{2}+19x+5\right) =2\left( 4x^{3}-23x^{2}+5x+5...
...& \\
&& \\
&& \\
0=8x^{3}-165x^{2}+740x-975 && \\
&& \\
&&
\end{eqnarray*}


Rewrite the last equation in factored form.

\begin{eqnarray*}&& \\
8x^{3}-165x^{2}+740x-975 &=&0 \\
&& \\
&& \\
\left( x-15\right) \left( 8x^{2}-45x+65\right) &=&0 \\
&& \\
&&
\end{eqnarray*}


The only way a product can equal zero is if at least one of the factors equals zero.

\begin{eqnarray*}&& \\
If\ x-15 &=&0,\ then\ x=15 \\
&& \\
If\ 8x^{2}-45x+65 ...
... then\ x=\displaystyle \frac{45\pm \sqrt{-55}}{16} \\
&& \\
&&
\end{eqnarray*}


Since $x=\displaystyle \frac{45\pm \sqrt{-55}}{16}$ is not a real number, the only real solutions is x=15.



Check the solution x=15 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.        


  • Left Side: $\qquad \displaystyle \frac{3x+1}{x^{2}-7x+10}+3=\displaystyle \frac{3\left( 15\...
...15\right) ^{2}-7\left( 15\right) +10}+3=\displaystyle \frac{46}{130}+3=3.353846$



  • Right Side: $\qquad 5-\displaystyle \frac{107}{4x+5}=5-\displaystyle \frac{107}{4\left( 15\right) +5}
3.353846$



    Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 15 for x, then x=15is a solution.





    You can also check your answer by graphing $\quad $

    \begin{eqnarray*}&& \\
f(x) &=&\displaystyle \frac{3x+1}{x^{2}-7x+10}-2+\displaystyle \frac{107}{4x+5} \\
&&
\end{eqnarray*}.

    (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 one place, 15.



    This means that there is one real solution and the solution is x=15.







    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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