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Solve \(\displaystyle \frac{x}{x + 7} - \frac{1}{x + 5}=- \frac{2}{x^{2} + 12 x + 35}\).

Factoring the denominator on the right hand side gives \(\displaystyle \left(x + 5\right) \left(x + 7\right)\). This gives the LCD as \(\displaystyle \left(x + 5\right) \left(x + 7\right)\). Multiplying by the LCD gives \(\displaystyle x \left(x + 5\right) - x - 7 = -2\). Getting zero on one side gives \(\displaystyle x^{2} + 4 x - 5=0\). Factoring gives \(\displaystyle \left(x - 1\right) \left(x + 5\right)=0\). The two possible solutions are \(\displaystyle x = 1\) and \(\displaystyle x = -5\). Checking the possible solutions gives:
Since \(\displaystyle 1\) is not zero of the denominator it is a solution. Since \(\displaystyle -5\) is zero of the denominator it is not in the domain and must be rejected as a solution.

Download \(\LaTeX\) 

\begin{question}Solve $\frac{x}{x + 7} - \frac{1}{x + 5}=- \frac{2}{x^{2} + 12 x + 35}$. 
    \soln{9cm}{Factoring the denominator on the right hand side gives $\left(x + 5\right) \left(x + 7\right)$. This gives the LCD as $\left(x + 5\right) \left(x + 7\right)$.  Multiplying by the LCD gives $x \left(x + 5\right) - x - 7 = -2$.  Getting zero on one side gives $x^{2} + 4 x - 5=0$. Factoring gives  $\left(x - 1\right) \left(x + 5\right)=0$. The two possible solutions are $x = 1$ and $x = -5$. Checking the possible solutions gives:\newline
Since $1$ is not zero of the denominator it is a solution. Since $-5$ 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 + 7} - \frac{1}{x + 5}=- \frac{2}{x^{2} + 12 x + 35} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bx%7D%7Bx%20%2B%207%7D%20-%20%5Cfrac%7B1%7D%7Bx%20%2B%205%7D%3D-%20%5Cfrac%7B2%7D%7Bx%5E%7B2%7D%20%2B%2012%20x%20%2B%2035%7D%20" alt="LaTeX:  \displaystyle \frac{x}{x + 7} - \frac{1}{x + 5}=- \frac{2}{x^{2} + 12 x + 35} " data-equation-content=" \displaystyle \frac{x}{x + 7} - \frac{1}{x + 5}=- \frac{2}{x^{2} + 12 x + 35} " /> . </p> </p>
HTML for Canvas
<p> <p>Factoring the denominator on the right hand side gives  <img class="equation_image" title=" \displaystyle \left(x + 5\right) \left(x + 7\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%205%5Cright%29%20%5Cleft%28x%20%2B%207%5Cright%29%20" alt="LaTeX:  \displaystyle \left(x + 5\right) \left(x + 7\right) " data-equation-content=" \displaystyle \left(x + 5\right) \left(x + 7\right) " /> . This gives the LCD as  <img class="equation_image" title=" \displaystyle \left(x + 5\right) \left(x + 7\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%205%5Cright%29%20%5Cleft%28x%20%2B%207%5Cright%29%20" alt="LaTeX:  \displaystyle \left(x + 5\right) \left(x + 7\right) " data-equation-content=" \displaystyle \left(x + 5\right) \left(x + 7\right) " /> .  Multiplying by the LCD gives  <img class="equation_image" title=" \displaystyle x \left(x + 5\right) - x - 7 = -2 " src="/equation_images/%20%5Cdisplaystyle%20x%20%5Cleft%28x%20%2B%205%5Cright%29%20-%20x%20-%207%20%3D%20-2%20" alt="LaTeX:  \displaystyle x \left(x + 5\right) - x - 7 = -2 " data-equation-content=" \displaystyle x \left(x + 5\right) - x - 7 = -2 " /> .  Getting zero on one side gives  <img class="equation_image" title=" \displaystyle x^{2} + 4 x - 5=0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%204%20x%20-%205%3D0%20" alt="LaTeX:  \displaystyle x^{2} + 4 x - 5=0 " data-equation-content=" \displaystyle x^{2} + 4 x - 5=0 " /> . Factoring gives   <img class="equation_image" title=" \displaystyle \left(x - 1\right) \left(x + 5\right)=0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%28x%20%2B%205%5Cright%29%3D0%20" alt="LaTeX:  \displaystyle \left(x - 1\right) \left(x + 5\right)=0 " data-equation-content=" \displaystyle \left(x - 1\right) \left(x + 5\right)=0 " /> . The two possible solutions are  <img class="equation_image" title=" \displaystyle x = 1 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%201%20" alt="LaTeX:  \displaystyle x = 1 " data-equation-content=" \displaystyle x = 1 " />  and  <img class="equation_image" title=" \displaystyle x = -5 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-5%20" alt="LaTeX:  \displaystyle x = -5 " data-equation-content=" \displaystyle x = -5 " /> . Checking the possible solutions gives:<br>
Since  <img class="equation_image" title=" \displaystyle 1 " src="/equation_images/%20%5Cdisplaystyle%201%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 -5 " src="/equation_images/%20%5Cdisplaystyle%20-5%20" alt="LaTeX:  \displaystyle -5 " data-equation-content=" \displaystyle -5 " />  is zero of the denominator it is not in the domain and must be rejected as a solution. </p> </p>