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Solve \(\displaystyle \frac{x}{x - 9} + \frac{3}{x - 11}=\frac{6}{x^{2} - 20 x + 99}\).
Factoring the denominator on the right hand side gives \(\displaystyle \left(x - 11\right) \left(x - 9\right)\). This gives the LCD as \(\displaystyle \left(x - 11\right) \left(x - 9\right)\). Multiplying by the LCD gives \(\displaystyle x \left(x - 11\right) + 3 x - 27 = 6\). Getting zero on one side gives \(\displaystyle x^{2} - 8 x - 33=0\). Factoring gives \(\displaystyle \left(x - 11\right) \left(x + 3\right)=0\). The two possible solutions are \(\displaystyle x = 11\) and \(\displaystyle x = -3\). Checking the possible solutions gives:
Since \(\displaystyle 11\) is zero of the denominator it is not in the domain and must be rejected as a solution. Since \(\displaystyle -3\) is not zero of the denominator it is a solution.
\begin{question}Solve $\frac{x}{x - 9} + \frac{3}{x - 11}=\frac{6}{x^{2} - 20 x + 99}$.
\soln{9cm}{Factoring the denominator on the right hand side gives $\left(x - 11\right) \left(x - 9\right)$. This gives the LCD as $\left(x - 11\right) \left(x - 9\right)$. Multiplying by the LCD gives $x \left(x - 11\right) + 3 x - 27 = 6$. Getting zero on one side gives $x^{2} - 8 x - 33=0$. Factoring gives $\left(x - 11\right) \left(x + 3\right)=0$. The two possible solutions are $x = 11$ and $x = -3$. Checking the possible solutions gives:\newline
Since $11$ is zero of the denominator it is not in the domain and must be rejected as a solution. Since $-3$ is not zero of the denominator it is a solution. }
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Solve <img class="equation_image" title=" \displaystyle \frac{x}{x - 9} + \frac{3}{x - 11}=\frac{6}{x^{2} - 20 x + 99} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bx%7D%7Bx%20-%209%7D%20%2B%20%5Cfrac%7B3%7D%7Bx%20-%2011%7D%3D%5Cfrac%7B6%7D%7Bx%5E%7B2%7D%20-%2020%20x%20%2B%2099%7D%20" alt="LaTeX: \displaystyle \frac{x}{x - 9} + \frac{3}{x - 11}=\frac{6}{x^{2} - 20 x + 99} " data-equation-content=" \displaystyle \frac{x}{x - 9} + \frac{3}{x - 11}=\frac{6}{x^{2} - 20 x + 99} " /> . </p> </p><p> <p>Factoring the denominator on the right hand side gives <img class="equation_image" title=" \displaystyle \left(x - 11\right) \left(x - 9\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20-%2011%5Cright%29%20%5Cleft%28x%20-%209%5Cright%29%20" alt="LaTeX: \displaystyle \left(x - 11\right) \left(x - 9\right) " data-equation-content=" \displaystyle \left(x - 11\right) \left(x - 9\right) " /> . This gives the LCD as <img class="equation_image" title=" \displaystyle \left(x - 11\right) \left(x - 9\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20-%2011%5Cright%29%20%5Cleft%28x%20-%209%5Cright%29%20" alt="LaTeX: \displaystyle \left(x - 11\right) \left(x - 9\right) " data-equation-content=" \displaystyle \left(x - 11\right) \left(x - 9\right) " /> . Multiplying by the LCD gives <img class="equation_image" title=" \displaystyle x \left(x - 11\right) + 3 x - 27 = 6 " src="/equation_images/%20%5Cdisplaystyle%20x%20%5Cleft%28x%20-%2011%5Cright%29%20%2B%203%20x%20-%2027%20%3D%206%20" alt="LaTeX: \displaystyle x \left(x - 11\right) + 3 x - 27 = 6 " data-equation-content=" \displaystyle x \left(x - 11\right) + 3 x - 27 = 6 " /> . Getting zero on one side gives <img class="equation_image" title=" \displaystyle x^{2} - 8 x - 33=0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20-%208%20x%20-%2033%3D0%20" alt="LaTeX: \displaystyle x^{2} - 8 x - 33=0 " data-equation-content=" \displaystyle x^{2} - 8 x - 33=0 " /> . Factoring gives <img class="equation_image" title=" \displaystyle \left(x - 11\right) \left(x + 3\right)=0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20-%2011%5Cright%29%20%5Cleft%28x%20%2B%203%5Cright%29%3D0%20" alt="LaTeX: \displaystyle \left(x - 11\right) \left(x + 3\right)=0 " data-equation-content=" \displaystyle \left(x - 11\right) \left(x + 3\right)=0 " /> . The two possible solutions are <img class="equation_image" title=" \displaystyle x = 11 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%2011%20" alt="LaTeX: \displaystyle x = 11 " data-equation-content=" \displaystyle x = 11 " /> and <img class="equation_image" title=" \displaystyle x = -3 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-3%20" alt="LaTeX: \displaystyle x = -3 " data-equation-content=" \displaystyle x = -3 " /> . Checking the possible solutions gives:<br>
Since <img class="equation_image" title=" \displaystyle 11 " src="/equation_images/%20%5Cdisplaystyle%2011%20" alt="LaTeX: \displaystyle 11 " data-equation-content=" \displaystyle 11 " /> is zero of the denominator it is not in the domain and must be rejected as a solution. Since <img class="equation_image" title=" \displaystyle -3 " src="/equation_images/%20%5Cdisplaystyle%20-3%20" alt="LaTeX: \displaystyle -3 " data-equation-content=" \displaystyle -3 " /> is not zero of the denominator it is a solution. </p> </p>