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Solve the inequality \(\displaystyle \frac{2}{x^{2} - 1}<\frac{8}{x^{2} + 4 x - 5}\)
Getting zero on one side and factoring gives \(\displaystyle - \frac{8}{\left(x - 1\right) \left(x + 5\right)} + \frac{2}{\left(x - 1\right) \left(x + 1\right)}< 0\). This gives the least common denominator as \(\displaystyle \left(x - 1\right) \left(x + 1\right) \left(x + 5\right)\). Building each fraction to get the common denominator gives \(\displaystyle \frac{2 x + 10 - (8 x + 8)}{\left(x - 1\right) \left(x + 1\right) \left(x + 5\right)} < 0\). Simplifying gives \(\displaystyle \frac{2 - 6 x}{\left(x - 1\right) \left(x + 1\right) \left(x + 5\right)}<0\). The inequality can change signs at the zeros of the numerator, \(\displaystyle \left\{\frac{1}{3}\right\}\), or the zeros of the denominator \(\displaystyle \left\{-5, -1, 1\right\}\). Making a sign chart gives: This gives the solution \(\displaystyle \left(-\infty, -5\right) \cup \left(-1, \frac{1}{3}\right) \cup \left(1, \infty\right)\)
\begin{question}Solve the inequality $\frac{2}{x^{2} - 1}<\frac{8}{x^{2} + 4 x - 5}$
\soln{9cm}{Getting zero on one side and factoring gives $- \frac{8}{\left(x - 1\right) \left(x + 5\right)} + \frac{2}{\left(x - 1\right) \left(x + 1\right)}< 0$. This gives the least common denominator as $\left(x - 1\right) \left(x + 1\right) \left(x + 5\right)$. Building each fraction to get the common denominator gives $\frac{2 x + 10 - (8 x + 8)}{\left(x - 1\right) \left(x + 1\right) \left(x + 5\right)} < 0$. Simplifying gives $\frac{2 - 6 x}{\left(x - 1\right) \left(x + 1\right) \left(x + 5\right)}<0$. The inequality can change signs at the zeros of the numerator, $\left\{\frac{1}{3}\right\}$, or the zeros of the denominator $\left\{-5, -1, 1\right\}$. Making a sign chart gives: \begin{tikzpicture}
\draw[latex-latex, thick] (-3, 0) -- (7, 0);
\draw(-1, 0.2) -- (-1, -0.2);
\draw (-1,-0.2) node[below]{$-5$};
\draw (-2,.2) node[above]{$-$};
\draw(1, 0.2) -- (1, -0.2);
\draw (1,-0.2) node[below]{$-1$};
\draw (0,.2) node[above]{$+$};
\draw(3, 0.2) -- (3, -0.2);
\draw (3,-0.2) node[below]{$1/3$};
\draw (2,.2) node[above]{$-$};
\draw(5, 0.2) -- (5, -0.2);
\draw (5,-0.2) node[below]{$1$};
\draw (6,.2) node[above]{$-$};
\draw (4,.2) node[above]{$+$};
\end{tikzpicture}
This gives the solution $\left(-\infty, -5\right) \cup \left(-1, \frac{1}{3}\right) \cup \left(1, \infty\right)$}
\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 the inequality <img class="equation_image" title=" \displaystyle \frac{2}{x^{2} - 1}<\frac{8}{x^{2} + 4 x - 5} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B2%7D%7Bx%5E%7B2%7D%20-%201%7D%3C%5Cfrac%7B8%7D%7Bx%5E%7B2%7D%20%2B%204%20x%20-%205%7D%20" alt="LaTeX: \displaystyle \frac{2}{x^{2} - 1}<\frac{8}{x^{2} + 4 x - 5} " data-equation-content=" \displaystyle \frac{2}{x^{2} - 1}<\frac{8}{x^{2} + 4 x - 5} " /> </p> </p><p> <p>Getting zero on one side and factoring gives <img class="equation_image" title=" \displaystyle - \frac{8}{\left(x - 1\right) \left(x + 5\right)} + \frac{2}{\left(x - 1\right) \left(x + 1\right)}< 0 " src="/equation_images/%20%5Cdisplaystyle%20-%20%5Cfrac%7B8%7D%7B%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%28x%20%2B%205%5Cright%29%7D%20%2B%20%5Cfrac%7B2%7D%7B%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%28x%20%2B%201%5Cright%29%7D%3C%200%20" alt="LaTeX: \displaystyle - \frac{8}{\left(x - 1\right) \left(x + 5\right)} + \frac{2}{\left(x - 1\right) \left(x + 1\right)}< 0 " data-equation-content=" \displaystyle - \frac{8}{\left(x - 1\right) \left(x + 5\right)} + \frac{2}{\left(x - 1\right) \left(x + 1\right)}< 0 " /> . This gives the least common denominator as <img class="equation_image" title=" \displaystyle \left(x - 1\right) \left(x + 1\right) \left(x + 5\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%28x%20%2B%201%5Cright%29%20%5Cleft%28x%20%2B%205%5Cright%29%20" alt="LaTeX: \displaystyle \left(x - 1\right) \left(x + 1\right) \left(x + 5\right) " data-equation-content=" \displaystyle \left(x - 1\right) \left(x + 1\right) \left(x + 5\right) " /> . Building each fraction to get the common denominator gives <img class="equation_image" title=" \displaystyle \frac{2 x + 10 - (8 x + 8)}{\left(x - 1\right) \left(x + 1\right) \left(x + 5\right)} < 0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B2%20x%20%2B%2010%20-%20%288%20x%20%2B%208%29%7D%7B%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%28x%20%2B%201%5Cright%29%20%5Cleft%28x%20%2B%205%5Cright%29%7D%20%3C%200%20" alt="LaTeX: \displaystyle \frac{2 x + 10 - (8 x + 8)}{\left(x - 1\right) \left(x + 1\right) \left(x + 5\right)} < 0 " data-equation-content=" \displaystyle \frac{2 x + 10 - (8 x + 8)}{\left(x - 1\right) \left(x + 1\right) \left(x + 5\right)} < 0 " /> . Simplifying gives <img class="equation_image" title=" \displaystyle \frac{2 - 6 x}{\left(x - 1\right) \left(x + 1\right) \left(x + 5\right)}<0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B2%20-%206%20x%7D%7B%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%28x%20%2B%201%5Cright%29%20%5Cleft%28x%20%2B%205%5Cright%29%7D%3C0%20" alt="LaTeX: \displaystyle \frac{2 - 6 x}{\left(x - 1\right) \left(x + 1\right) \left(x + 5\right)}<0 " data-equation-content=" \displaystyle \frac{2 - 6 x}{\left(x - 1\right) \left(x + 1\right) \left(x + 5\right)}<0 " /> . The inequality can change signs at the zeros of the numerator, <img class="equation_image" title=" \displaystyle \left\{\frac{1}{3}\right\} " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%5C%7B%5Cfrac%7B1%7D%7B3%7D%5Cright%5C%7D%20" alt="LaTeX: \displaystyle \left\{\frac{1}{3}\right\} " data-equation-content=" \displaystyle \left\{\frac{1}{3}\right\} " /> , or the zeros of the denominator <img class="equation_image" title=" \displaystyle \left\{-5, -1, 1\right\} " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%5C%7B-5%2C%20-1%2C%201%5Cright%5C%7D%20" alt="LaTeX: \displaystyle \left\{-5, -1, 1\right\} " data-equation-content=" \displaystyle \left\{-5, -1, 1\right\} " /> . Making a sign chart gives: <?xml version="1.0" encoding="UTF-8"?>
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This gives the solution <img class="equation_image" title=" \displaystyle \left(-\infty, -5\right) \cup \left(-1, \frac{1}{3}\right) \cup \left(1, \infty\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28-%5Cinfty%2C%20-5%5Cright%29%20%5Ccup%20%5Cleft%28-1%2C%20%5Cfrac%7B1%7D%7B3%7D%5Cright%29%20%5Ccup%20%5Cleft%281%2C%20%5Cinfty%5Cright%29%20" alt="LaTeX: \displaystyle \left(-\infty, -5\right) \cup \left(-1, \frac{1}{3}\right) \cup \left(1, \infty\right) " data-equation-content=" \displaystyle \left(-\infty, -5\right) \cup \left(-1, \frac{1}{3}\right) \cup \left(1, \infty\right) " /> </p> </p>