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Solve \(\displaystyle \frac{x}{x + 9} + \frac{3}{x + 6}=\frac{9}{x^{2} + 15 x + 54}\).

Factoring the denominator on the right hand side gives \(\displaystyle \left(x + 6\right) \left(x + 9\right)\). This gives the LCD as \(\displaystyle \left(x + 6\right) \left(x + 9\right)\). Multiplying by the LCD gives \(\displaystyle x \left(x + 6\right) + 3 x + 27 = 9\). Getting zero on one side gives \(\displaystyle x^{2} + 9 x + 18=0\). Factoring gives \(\displaystyle \left(x + 3\right) \left(x + 6\right)=0\). The two possible solutions are \(\displaystyle x = -3\) and \(\displaystyle x = -6\). Checking the possible solutions gives:
Since \(\displaystyle -6\) 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.

Download \(\LaTeX\) 

\begin{question}Solve $\frac{x}{x + 9} + \frac{3}{x + 6}=\frac{9}{x^{2} + 15 x + 54}$. 
    \soln{9cm}{Factoring the denominator on the right hand side gives $\left(x + 6\right) \left(x + 9\right)$. This gives the LCD as $\left(x + 6\right) \left(x + 9\right)$.  Multiplying by the LCD gives $x \left(x + 6\right) + 3 x + 27 = 9$.  Getting zero on one side gives $x^{2} + 9 x + 18=0$. Factoring gives  $\left(x + 3\right) \left(x + 6\right)=0$. The two possible solutions are $x = -3$ and $x = -6$. Checking the possible solutions gives:\newline
Since $-6$ 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}

Download Question and Solution Environment\(\LaTeX\)
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\begin{document}\begin{question}(10pts) The question goes here!
    \soln{9cm}{The solution goes here.}

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HTML for Canvas 
<p> <p>Solve  <img class="equation_image" title=" \displaystyle \frac{x}{x + 9} + \frac{3}{x + 6}=\frac{9}{x^{2} + 15 x + 54} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bx%7D%7Bx%20%2B%209%7D%20%2B%20%5Cfrac%7B3%7D%7Bx%20%2B%206%7D%3D%5Cfrac%7B9%7D%7Bx%5E%7B2%7D%20%2B%2015%20x%20%2B%2054%7D%20" alt="LaTeX:  \displaystyle \frac{x}{x + 9} + \frac{3}{x + 6}=\frac{9}{x^{2} + 15 x + 54} " data-equation-content=" \displaystyle \frac{x}{x + 9} + \frac{3}{x + 6}=\frac{9}{x^{2} + 15 x + 54} " /> . </p> </p>
HTML for Canvas
<p> <p>Factoring the denominator on the right hand side gives  <img class="equation_image" title=" \displaystyle \left(x + 6\right) \left(x + 9\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%206%5Cright%29%20%5Cleft%28x%20%2B%209%5Cright%29%20" alt="LaTeX:  \displaystyle \left(x + 6\right) \left(x + 9\right) " data-equation-content=" \displaystyle \left(x + 6\right) \left(x + 9\right) " /> . This gives the LCD as  <img class="equation_image" title=" \displaystyle \left(x + 6\right) \left(x + 9\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%206%5Cright%29%20%5Cleft%28x%20%2B%209%5Cright%29%20" alt="LaTeX:  \displaystyle \left(x + 6\right) \left(x + 9\right) " data-equation-content=" \displaystyle \left(x + 6\right) \left(x + 9\right) " /> .  Multiplying by the LCD gives  <img class="equation_image" title=" \displaystyle x \left(x + 6\right) + 3 x + 27 = 9 " src="/equation_images/%20%5Cdisplaystyle%20x%20%5Cleft%28x%20%2B%206%5Cright%29%20%2B%203%20x%20%2B%2027%20%3D%209%20" alt="LaTeX:  \displaystyle x \left(x + 6\right) + 3 x + 27 = 9 " data-equation-content=" \displaystyle x \left(x + 6\right) + 3 x + 27 = 9 " /> .  Getting zero on one side gives  <img class="equation_image" title=" \displaystyle x^{2} + 9 x + 18=0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%209%20x%20%2B%2018%3D0%20" alt="LaTeX:  \displaystyle x^{2} + 9 x + 18=0 " data-equation-content=" \displaystyle x^{2} + 9 x + 18=0 " /> . Factoring gives   <img class="equation_image" title=" \displaystyle \left(x + 3\right) \left(x + 6\right)=0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%203%5Cright%29%20%5Cleft%28x%20%2B%206%5Cright%29%3D0%20" alt="LaTeX:  \displaystyle \left(x + 3\right) \left(x + 6\right)=0 " data-equation-content=" \displaystyle \left(x + 3\right) \left(x + 6\right)=0 " /> . The two possible solutions are  <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 " />  and  <img class="equation_image" title=" \displaystyle x = -6 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-6%20" alt="LaTeX:  \displaystyle x = -6 " data-equation-content=" \displaystyle x = -6 " /> . Checking the possible solutions gives:<br>
Since  <img class="equation_image" title=" \displaystyle -6 " src="/equation_images/%20%5Cdisplaystyle%20-6%20" alt="LaTeX:  \displaystyle -6 " data-equation-content=" \displaystyle -6 " />  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>