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 5.3d: $\displaystyle \frac{x+5}{x-5}=\displaystyle \frac{3x-1}{x+5}+\displaystyle \frac{3x+2
}{3x}$



Answer: x=-1, $-\displaystyle \frac{5}{9},$ 10




Solution:



Recall that you cannot divide by zero. Therefore, the first fraction is valid if , $\quad x\neq 5,$ the second fraction is valid if $x\neq -5,\quad $and the third fraction is valid is $x\neq 0$.If $\quad 5,$ -5, or $0\quad $ turn out to be the solutions, you must discard them as extraneous solutions.




Multiply both sides by the least common multiple $\left( x-5\right)
\left( x+5\right) \left( 3x\right) \qquad $(the smallest number that all the denominators will divide into evenly). This step will eliminate all the denominators.




\begin{eqnarray*}\displaystyle \frac{x+5}{x-5} &=&\displaystyle \frac{3x-1}{x+5}...
...ht) \left(
3x\right) \left[ \displaystyle \frac{3x+2}{3x}\right]
\end{eqnarray*}


which is equivalent to

\begin{eqnarray*}\displaystyle \frac{\left( x-5\right) \left( x+5\right) \left( ...
...left( 3x\right) }{1}\left[
\displaystyle \frac{3x+2}{3x}\right]
\end{eqnarray*}


which can be rewritten as

\begin{eqnarray*}\displaystyle \frac{\left( x-5\right) \left( x+5\right) \left( ...
...\right) \left( 3x\right) \left(
3x+2\right) }{\left( 3x\right) }
\end{eqnarray*}


which can be rewritten as

\begin{eqnarray*}\displaystyle \frac{\left( x-5\right) }{\left( x-5\right) }\cdo...
...frac{\left( x-5\right)
\left( x+5\right) \left( 3x+2\right) }{1}
\end{eqnarray*}


which can be simplified to

\begin{eqnarray*}1\cdot \displaystyle \frac{\left( x+5\right) \left( 3x\right) \...
...1\right) +\left( x-5\right) \left( x+5\right)
\left( 3x+2\right)
\end{eqnarray*}



\begin{eqnarray*}\left( 3x\right) \left[ \left( x+5\right) \left( x+5\right) \ri...
...^{2}-16x+5\right] +\left( x-5\right) \left[ 3x^{2}+17x+10\right]
\end{eqnarray*}



\begin{eqnarray*}3x^{3}+30x^{2}+75x &=&9x^{3}-48x^{2}+15x+3x^{3}+17x^{2}+10x-15x...
...0 \\
&& \\
&& \\
3x^{3}+30x^{2}+75x &=&12x^{3}-46x^{2}-60x-50
\end{eqnarray*}



\begin{eqnarray*}0 &=&9x^{3}-76x^{2}-135x-50 \\
&& \\
&& \\
0 &=&\left( x-10\right) \left( 9x^{2}+14x+5\right)
\end{eqnarray*}


The only way that the product of two numbers equals zero is if at least one of the numbers equations zero. Therefore set both expressions to zero and solve.

\begin{eqnarray*}x-10 &=&0 \\
&& \\
x &=&10
\end{eqnarray*}


Set 9x2+14x+5=0 and solve for x using the quadratic formula $x=\displaystyle \frac{
-b\pm \sqrt{b^{2}-4ac}}{2a}$ $a=9,\quad b=14,\quad c=5.$

\begin{eqnarray*}9x^{2}+14x+5 &=&0 \\
&& \\
x &=&\displaystyle \frac{-\left( 1...
...6}}{18} \\
&& \\
&& \\
x &=&\displaystyle \frac{-14\pm 4}{18}
\end{eqnarray*}



\begin{eqnarray*}x &=&\displaystyle \frac{-14+4}{18}=-\displaystyle \frac{10}{18...
...=&\displaystyle \frac{-14-4}{18}=-\displaystyle \frac{18}{18}=-1
\end{eqnarray*}



The answers are x=10, -1, and $-\displaystyle \frac{5}{9}.\medskip\bigskip\bigskip
\medskip $

Check the solution x=10 by substituting 10 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 equation equals the right side of the equation after the substitution, we have verified that x=10 is a solution.




Check the solution x=-1 by substituting -1 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 equation equals the right side of the equation after the substitution, we have verified that x=-1 is a solution.




Check the solution $x=-\displaystyle \frac{5}{9}$ by substituting $-\displaystyle \frac{5}{9}\approx
-0.555556$ 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 equation equals the right side of the equation after the substitution, we have verified that $x=-\displaystyle \frac{5}{9}$ is a solution.




You can also check your answer by graphing $\quad f(x)=\displaystyle \frac{x+5}{x-5}-
\displaystyle \frac{3x-1}{x+5}-\displaystyle \frac{3x+2}{3x}\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 three places: x=-1,10 and $-0.555556\approx -\displaystyle \frac{5}{9}$.



This verifies our solution graphically.



We have verified our solution both algebraically and graphically.




If you would like to review the solution to problem 5.3e, 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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