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Solve \(\displaystyle \frac{x}{x + 4} + \frac{1}{x + 2}=\frac{2}{x^{2} + 6 x + 8}\).

Factoring the denominator on the right hand side gives \(\displaystyle \left(x + 2\right) \left(x + 4\right)\). This gives the LCD as \(\displaystyle \left(x + 2\right) \left(x + 4\right)\). Multiplying by the LCD gives \(\displaystyle x \left(x + 2\right) + x + 4 = 2\). Getting zero on one side gives \(\displaystyle x^{2} + 3 x + 2=0\). Factoring gives \(\displaystyle \left(x + 1\right) \left(x + 2\right)=0\). The two possible solutions are \(\displaystyle x = -1\) and \(\displaystyle x = -2\). Checking the possible solutions gives:
Since \(\displaystyle -1\) is not zero of the denominator it is a solution. Since \(\displaystyle -2\) is zero of the denominator it is not in the domain and must be rejected as a solution.

Download \(\LaTeX\) 

\begin{question}Solve $\frac{x}{x + 4} + \frac{1}{x + 2}=\frac{2}{x^{2} + 6 x + 8}$. 
    \soln{9cm}{Factoring the denominator on the right hand side gives $\left(x + 2\right) \left(x + 4\right)$. This gives the LCD as $\left(x + 2\right) \left(x + 4\right)$.  Multiplying by the LCD gives $x \left(x + 2\right) + x + 4 = 2$.  Getting zero on one side gives $x^{2} + 3 x + 2=0$. Factoring gives  $\left(x + 1\right) \left(x + 2\right)=0$. The two possible solutions are $x = -1$ and $x = -2$. Checking the possible solutions gives:\newline
Since $-1$ is not zero of the denominator it is a solution. Since $-2$ is zero of the denominator it is not in the domain and must be rejected as 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 + 4} + \frac{1}{x + 2}=\frac{2}{x^{2} + 6 x + 8} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bx%7D%7Bx%20%2B%204%7D%20%2B%20%5Cfrac%7B1%7D%7Bx%20%2B%202%7D%3D%5Cfrac%7B2%7D%7Bx%5E%7B2%7D%20%2B%206%20x%20%2B%208%7D%20" alt="LaTeX:  \displaystyle \frac{x}{x + 4} + \frac{1}{x + 2}=\frac{2}{x^{2} + 6 x + 8} " data-equation-content=" \displaystyle \frac{x}{x + 4} + \frac{1}{x + 2}=\frac{2}{x^{2} + 6 x + 8} " /> . </p> </p>
HTML for Canvas
<p> <p>Factoring the denominator on the right hand side gives  <img class="equation_image" title=" \displaystyle \left(x + 2\right) \left(x + 4\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%202%5Cright%29%20%5Cleft%28x%20%2B%204%5Cright%29%20" alt="LaTeX:  \displaystyle \left(x + 2\right) \left(x + 4\right) " data-equation-content=" \displaystyle \left(x + 2\right) \left(x + 4\right) " /> . This gives the LCD as  <img class="equation_image" title=" \displaystyle \left(x + 2\right) \left(x + 4\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%202%5Cright%29%20%5Cleft%28x%20%2B%204%5Cright%29%20" alt="LaTeX:  \displaystyle \left(x + 2\right) \left(x + 4\right) " data-equation-content=" \displaystyle \left(x + 2\right) \left(x + 4\right) " /> .  Multiplying by the LCD gives  <img class="equation_image" title=" \displaystyle x \left(x + 2\right) + x + 4 = 2 " src="/equation_images/%20%5Cdisplaystyle%20x%20%5Cleft%28x%20%2B%202%5Cright%29%20%2B%20x%20%2B%204%20%3D%202%20" alt="LaTeX:  \displaystyle x \left(x + 2\right) + x + 4 = 2 " data-equation-content=" \displaystyle x \left(x + 2\right) + x + 4 = 2 " /> .  Getting zero on one side gives  <img class="equation_image" title=" \displaystyle x^{2} + 3 x + 2=0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%203%20x%20%2B%202%3D0%20" alt="LaTeX:  \displaystyle x^{2} + 3 x + 2=0 " data-equation-content=" \displaystyle x^{2} + 3 x + 2=0 " /> . Factoring gives   <img class="equation_image" title=" \displaystyle \left(x + 1\right) \left(x + 2\right)=0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%201%5Cright%29%20%5Cleft%28x%20%2B%202%5Cright%29%3D0%20" alt="LaTeX:  \displaystyle \left(x + 1\right) \left(x + 2\right)=0 " data-equation-content=" \displaystyle \left(x + 1\right) \left(x + 2\right)=0 " /> . The two possible solutions are  <img class="equation_image" title=" \displaystyle x = -1 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-1%20" alt="LaTeX:  \displaystyle x = -1 " data-equation-content=" \displaystyle x = -1 " />  and  <img class="equation_image" title=" \displaystyle x = -2 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-2%20" alt="LaTeX:  \displaystyle x = -2 " data-equation-content=" \displaystyle x = -2 " /> . Checking the possible solutions gives:<br>
Since  <img class="equation_image" title=" \displaystyle -1 " src="/equation_images/%20%5Cdisplaystyle%20-1%20" alt="LaTeX:  \displaystyle -1 " data-equation-content=" \displaystyle -1 " />  is not zero of the denominator it is a solution. Since  <img class="equation_image" title=" \displaystyle -2 " src="/equation_images/%20%5Cdisplaystyle%20-2%20" alt="LaTeX:  \displaystyle -2 " data-equation-content=" \displaystyle -2 " />  is zero of the denominator it is not in the domain and must be rejected as a solution. </p> </p>