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Solve \(\displaystyle \frac{x}{x + 6} - \frac{4}{x + 8}=\frac{8}{x^{2} + 14 x + 48}\).

Factoring the denominator on the right hand side gives \(\displaystyle \left(x + 6\right) \left(x + 8\right)\). This gives the LCD as \(\displaystyle \left(x + 6\right) \left(x + 8\right)\). Multiplying by the LCD gives \(\displaystyle x \left(x + 8\right) - 4 x - 24 = 8\). Getting zero on one side gives \(\displaystyle x^{2} + 4 x - 32=0\). Factoring gives \(\displaystyle \left(x - 4\right) \left(x + 8\right)=0\). The two possible solutions are \(\displaystyle x = 4\) and \(\displaystyle x = -8\). Checking the possible solutions gives:
Since \(\displaystyle -8\) is zero of the denominator it is not in the domain and must be rejected as a solution. Since \(\displaystyle 4\) is not zero of the denominator it is a solution.

Download \(\LaTeX\) 

\begin{question}Solve $\frac{x}{x + 6} - \frac{4}{x + 8}=\frac{8}{x^{2} + 14 x + 48}$. 
    \soln{9cm}{Factoring the denominator on the right hand side gives $\left(x + 6\right) \left(x + 8\right)$. This gives the LCD as $\left(x + 6\right) \left(x + 8\right)$.  Multiplying by the LCD gives $x \left(x + 8\right) - 4 x - 24 = 8$.  Getting zero on one side gives $x^{2} + 4 x - 32=0$. Factoring gives  $\left(x - 4\right) \left(x + 8\right)=0$. The two possible solutions are $x = 4$ and $x = -8$. Checking the possible solutions gives:\newline
Since $-8$ is zero of the denominator it is not in the domain and must be rejected as a solution. Since $4$ 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 + 6} - \frac{4}{x + 8}=\frac{8}{x^{2} + 14 x + 48} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bx%7D%7Bx%20%2B%206%7D%20-%20%5Cfrac%7B4%7D%7Bx%20%2B%208%7D%3D%5Cfrac%7B8%7D%7Bx%5E%7B2%7D%20%2B%2014%20x%20%2B%2048%7D%20" alt="LaTeX:  \displaystyle \frac{x}{x + 6} - \frac{4}{x + 8}=\frac{8}{x^{2} + 14 x + 48} " data-equation-content=" \displaystyle \frac{x}{x + 6} - \frac{4}{x + 8}=\frac{8}{x^{2} + 14 x + 48} " /> . </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 + 8\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%206%5Cright%29%20%5Cleft%28x%20%2B%208%5Cright%29%20" alt="LaTeX:  \displaystyle \left(x + 6\right) \left(x + 8\right) " data-equation-content=" \displaystyle \left(x + 6\right) \left(x + 8\right) " /> . This gives the LCD as  <img class="equation_image" title=" \displaystyle \left(x + 6\right) \left(x + 8\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%206%5Cright%29%20%5Cleft%28x%20%2B%208%5Cright%29%20" alt="LaTeX:  \displaystyle \left(x + 6\right) \left(x + 8\right) " data-equation-content=" \displaystyle \left(x + 6\right) \left(x + 8\right) " /> .  Multiplying by the LCD gives  <img class="equation_image" title=" \displaystyle x \left(x + 8\right) - 4 x - 24 = 8 " src="/equation_images/%20%5Cdisplaystyle%20x%20%5Cleft%28x%20%2B%208%5Cright%29%20-%204%20x%20-%2024%20%3D%208%20" alt="LaTeX:  \displaystyle x \left(x + 8\right) - 4 x - 24 = 8 " data-equation-content=" \displaystyle x \left(x + 8\right) - 4 x - 24 = 8 " /> .  Getting zero on one side gives  <img class="equation_image" title=" \displaystyle x^{2} + 4 x - 32=0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%204%20x%20-%2032%3D0%20" alt="LaTeX:  \displaystyle x^{2} + 4 x - 32=0 " data-equation-content=" \displaystyle x^{2} + 4 x - 32=0 " /> . Factoring gives   <img class="equation_image" title=" \displaystyle \left(x - 4\right) \left(x + 8\right)=0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20-%204%5Cright%29%20%5Cleft%28x%20%2B%208%5Cright%29%3D0%20" alt="LaTeX:  \displaystyle \left(x - 4\right) \left(x + 8\right)=0 " data-equation-content=" \displaystyle \left(x - 4\right) \left(x + 8\right)=0 " /> . The two possible solutions are  <img class="equation_image" title=" \displaystyle x = 4 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%204%20" alt="LaTeX:  \displaystyle x = 4 " data-equation-content=" \displaystyle x = 4 " />  and  <img class="equation_image" title=" \displaystyle x = -8 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-8%20" alt="LaTeX:  \displaystyle x = -8 " data-equation-content=" \displaystyle x = -8 " /> . Checking the possible solutions gives:<br>
Since  <img class="equation_image" title=" \displaystyle -8 " src="/equation_images/%20%5Cdisplaystyle%20-8%20" alt="LaTeX:  \displaystyle -8 " data-equation-content=" \displaystyle -8 " />  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 4 " src="/equation_images/%20%5Cdisplaystyle%204%20" alt="LaTeX:  \displaystyle 4 " data-equation-content=" \displaystyle 4 " />  is not zero of the denominator it is a solution. </p> </p>