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Solve the inequality \(\displaystyle \frac{2}{x^{2} - 25}<\frac{2}{x^{2} + 4 x - 5}\)
Getting zero on one side and factoring gives \(\displaystyle - \frac{2}{\left(x - 1\right) \left(x + 5\right)} + \frac{2}{\left(x - 5\right) \left(x + 5\right)}< 0\). This gives the least common denominator as \(\displaystyle \left(x - 5\right) \left(x - 1\right) \left(x + 5\right)\). Building each fraction to get the common denominator gives \(\displaystyle \frac{2 x - 2 - (2 x - 10)}{\left(x - 5\right) \left(x - 1\right) \left(x + 5\right)} < 0\). Simplifying gives \(\displaystyle \frac{8}{\left(x - 5\right) \left(x - 1\right) \left(x + 5\right)}<0\). The inequality can change signs at the zeros of the numerator, \(\displaystyle \emptyset\), or the zeros of the denominator \(\displaystyle \left\{-5, 1, 5\right\}\). Making a sign chart gives: This gives the solution \(\displaystyle \left(-\infty, -5\right) \cup \left(1, 5\right)\)
\begin{question}Solve the inequality $\frac{2}{x^{2} - 25}<\frac{2}{x^{2} + 4 x - 5}$ \soln{9cm}{Getting zero on one side and factoring gives $- \frac{2}{\left(x - 1\right) \left(x + 5\right)} + \frac{2}{\left(x - 5\right) \left(x + 5\right)}< 0$. This gives the least common denominator as $\left(x - 5\right) \left(x - 1\right) \left(x + 5\right)$. Building each fraction to get the common denominator gives $\frac{2 x - 2 - (2 x - 10)}{\left(x - 5\right) \left(x - 1\right) \left(x + 5\right)} < 0$. Simplifying gives $\frac{8}{\left(x - 5\right) \left(x - 1\right) \left(x + 5\right)}<0$. The inequality can change signs at the zeros of the numerator, $\emptyset$, or the zeros of the denominator $\left\{-5, 1, 5\right\}$. Making a sign chart gives: \begin{tikzpicture} \draw[latex-latex, thick] (-3, 0) -- (5, 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]{$5$}; \draw (4,.2) node[above]{$+$}; \draw (2,.2) node[above]{$-$}; \end{tikzpicture} This gives the solution $\left(-\infty, -5\right) \cup \left(1, 5\right)$} \end{question}
\documentclass{article} \usepackage{tikz} \usepackage{amsmath} \usepackage[margin=2cm]{geometry} \usepackage{tcolorbox} \newcounter{ExamNumber} \newcounter{questioncount} \stepcounter{questioncount} \newenvironment{question}{{\noindent\bfseries Question \arabic{questioncount}.}}{\stepcounter{questioncount}} \renewcommand{\labelenumi}{{\bfseries (\alph{enumi})}} \newif\ifShowSolution \newcommand{\soln}[2]{% \ifShowSolution% \noindent\begin{tcolorbox}[colframe=blue,title=Solution]#2\end{tcolorbox}\else% \vspace{#1}% \fi% }% \newcommand{\hideifShowSolution}[1]{% \ifShowSolution% % \else% #1% \fi% }% \everymath{\displaystyle} \ShowSolutiontrue \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} - 25}<\frac{2}{x^{2} + 4 x - 5} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B2%7D%7Bx%5E%7B2%7D%20-%2025%7D%3C%5Cfrac%7B2%7D%7Bx%5E%7B2%7D%20%2B%204%20x%20-%205%7D%20" alt="LaTeX: \displaystyle \frac{2}{x^{2} - 25}<\frac{2}{x^{2} + 4 x - 5} " data-equation-content=" \displaystyle \frac{2}{x^{2} - 25}<\frac{2}{x^{2} + 4 x - 5} " /> </p> </p>
<p> <p>Getting zero on one side and factoring gives <img class="equation_image" title=" \displaystyle - \frac{2}{\left(x - 1\right) \left(x + 5\right)} + \frac{2}{\left(x - 5\right) \left(x + 5\right)}< 0 " src="/equation_images/%20%5Cdisplaystyle%20-%20%5Cfrac%7B2%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-%205%5Cright%29%20%5Cleft%28x%20%2B%205%5Cright%29%7D%3C%200%20" alt="LaTeX: \displaystyle - \frac{2}{\left(x - 1\right) \left(x + 5\right)} + \frac{2}{\left(x - 5\right) \left(x + 5\right)}< 0 " data-equation-content=" \displaystyle - \frac{2}{\left(x - 1\right) \left(x + 5\right)} + \frac{2}{\left(x - 5\right) \left(x + 5\right)}< 0 " /> . This gives the least common denominator as <img class="equation_image" title=" \displaystyle \left(x - 5\right) \left(x - 1\right) \left(x + 5\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20-%205%5Cright%29%20%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%28x%20%2B%205%5Cright%29%20" alt="LaTeX: \displaystyle \left(x - 5\right) \left(x - 1\right) \left(x + 5\right) " data-equation-content=" \displaystyle \left(x - 5\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 - 2 - (2 x - 10)}{\left(x - 5\right) \left(x - 1\right) \left(x + 5\right)} < 0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B2%20x%20-%202%20-%20%282%20x%20-%2010%29%7D%7B%5Cleft%28x%20-%205%5Cright%29%20%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%28x%20%2B%205%5Cright%29%7D%20%3C%200%20" alt="LaTeX: \displaystyle \frac{2 x - 2 - (2 x - 10)}{\left(x - 5\right) \left(x - 1\right) \left(x + 5\right)} < 0 " data-equation-content=" \displaystyle \frac{2 x - 2 - (2 x - 10)}{\left(x - 5\right) \left(x - 1\right) \left(x + 5\right)} < 0 " /> . Simplifying gives <img class="equation_image" title=" \displaystyle \frac{8}{\left(x - 5\right) \left(x - 1\right) \left(x + 5\right)}<0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B8%7D%7B%5Cleft%28x%20-%205%5Cright%29%20%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%28x%20%2B%205%5Cright%29%7D%3C0%20" alt="LaTeX: \displaystyle \frac{8}{\left(x - 5\right) \left(x - 1\right) \left(x + 5\right)}<0 " data-equation-content=" \displaystyle \frac{8}{\left(x - 5\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 \emptyset " src="/equation_images/%20%5Cdisplaystyle%20%5Cemptyset%20" alt="LaTeX: \displaystyle \emptyset " data-equation-content=" \displaystyle \emptyset " /> , or the zeros of the denominator <img class="equation_image" title=" \displaystyle \left\{-5, 1, 5\right\} " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%5C%7B-5%2C%201%2C%205%5Cright%5C%7D%20" alt="LaTeX: \displaystyle \left\{-5, 1, 5\right\} " data-equation-content=" \displaystyle \left\{-5, 1, 5\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, 5\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28-%5Cinfty%2C%20-5%5Cright%29%20%5Ccup%20%5Cleft%281%2C%205%5Cright%29%20" alt="LaTeX: \displaystyle \left(-\infty, -5\right) \cup \left(1, 5\right) " data-equation-content=" \displaystyle \left(-\infty, -5\right) \cup \left(1, 5\right) " /> </p> </p>