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Solve the inequality \(\displaystyle \frac{3}{x^{2} - 1}<\frac{3}{x^{2} + 4 x + 3}\)
Getting zero on one side and factoring gives \(\displaystyle - \frac{3}{\left(x + 1\right) \left(x + 3\right)} + \frac{3}{\left(x - 1\right) \left(x + 1\right)}< 0\). This gives the least common denominator as \(\displaystyle \left(x - 1\right) \left(x + 1\right) \left(x + 3\right)\). Building each fraction to get the common denominator gives \(\displaystyle \frac{3 x + 9 - (3 x - 3)}{\left(x - 1\right) \left(x + 1\right) \left(x + 3\right)} < 0\). Simplifying gives \(\displaystyle \frac{12}{\left(x - 1\right) \left(x + 1\right) \left(x + 3\right)}<0\). The inequality can change signs at the zeros of the numerator, \(\displaystyle \emptyset\), or the zeros of the denominator \(\displaystyle \left\{-3, -1, 1\right\}\). Making a sign chart gives: This gives the solution \(\displaystyle \left(-\infty, -3\right) \cup \left(-1, 1\right)\)
\begin{question}Solve the inequality $\frac{3}{x^{2} - 1}<\frac{3}{x^{2} + 4 x + 3}$
\soln{9cm}{Getting zero on one side and factoring gives $- \frac{3}{\left(x + 1\right) \left(x + 3\right)} + \frac{3}{\left(x - 1\right) \left(x + 1\right)}< 0$. This gives the least common denominator as $\left(x - 1\right) \left(x + 1\right) \left(x + 3\right)$. Building each fraction to get the common denominator gives $\frac{3 x + 9 - (3 x - 3)}{\left(x - 1\right) \left(x + 1\right) \left(x + 3\right)} < 0$. Simplifying gives $\frac{12}{\left(x - 1\right) \left(x + 1\right) \left(x + 3\right)}<0$. The inequality can change signs at the zeros of the numerator, $\emptyset$, or the zeros of the denominator $\left\{-3, -1, 1\right\}$. 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]{$-3$};
\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]{$1$};
\draw (4,.2) node[above]{$+$};
\draw (2,.2) node[above]{$-$};
\end{tikzpicture}
This gives the solution $\left(-\infty, -3\right) \cup \left(-1, 1\right)$}
\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{3}{x^{2} - 1}<\frac{3}{x^{2} + 4 x + 3} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B3%7D%7Bx%5E%7B2%7D%20-%201%7D%3C%5Cfrac%7B3%7D%7Bx%5E%7B2%7D%20%2B%204%20x%20%2B%203%7D%20" alt="LaTeX: \displaystyle \frac{3}{x^{2} - 1}<\frac{3}{x^{2} + 4 x + 3} " data-equation-content=" \displaystyle \frac{3}{x^{2} - 1}<\frac{3}{x^{2} + 4 x + 3} " /> </p> </p><p> <p>Getting zero on one side and factoring gives <img class="equation_image" title=" \displaystyle - \frac{3}{\left(x + 1\right) \left(x + 3\right)} + \frac{3}{\left(x - 1\right) \left(x + 1\right)}< 0 " src="/equation_images/%20%5Cdisplaystyle%20-%20%5Cfrac%7B3%7D%7B%5Cleft%28x%20%2B%201%5Cright%29%20%5Cleft%28x%20%2B%203%5Cright%29%7D%20%2B%20%5Cfrac%7B3%7D%7B%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%28x%20%2B%201%5Cright%29%7D%3C%200%20" alt="LaTeX: \displaystyle - \frac{3}{\left(x + 1\right) \left(x + 3\right)} + \frac{3}{\left(x - 1\right) \left(x + 1\right)}< 0 " data-equation-content=" \displaystyle - \frac{3}{\left(x + 1\right) \left(x + 3\right)} + \frac{3}{\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 + 3\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%203%5Cright%29%20" alt="LaTeX: \displaystyle \left(x - 1\right) \left(x + 1\right) \left(x + 3\right) " data-equation-content=" \displaystyle \left(x - 1\right) \left(x + 1\right) \left(x + 3\right) " /> . Building each fraction to get the common denominator gives <img class="equation_image" title=" \displaystyle \frac{3 x + 9 - (3 x - 3)}{\left(x - 1\right) \left(x + 1\right) \left(x + 3\right)} < 0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B3%20x%20%2B%209%20-%20%283%20x%20-%203%29%7D%7B%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%28x%20%2B%201%5Cright%29%20%5Cleft%28x%20%2B%203%5Cright%29%7D%20%3C%200%20" alt="LaTeX: \displaystyle \frac{3 x + 9 - (3 x - 3)}{\left(x - 1\right) \left(x + 1\right) \left(x + 3\right)} < 0 " data-equation-content=" \displaystyle \frac{3 x + 9 - (3 x - 3)}{\left(x - 1\right) \left(x + 1\right) \left(x + 3\right)} < 0 " /> . Simplifying gives <img class="equation_image" title=" \displaystyle \frac{12}{\left(x - 1\right) \left(x + 1\right) \left(x + 3\right)}<0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B12%7D%7B%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%28x%20%2B%201%5Cright%29%20%5Cleft%28x%20%2B%203%5Cright%29%7D%3C0%20" alt="LaTeX: \displaystyle \frac{12}{\left(x - 1\right) \left(x + 1\right) \left(x + 3\right)}<0 " data-equation-content=" \displaystyle \frac{12}{\left(x - 1\right) \left(x + 1\right) \left(x + 3\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\{-3, -1, 1\right\} " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%5C%7B-3%2C%20-1%2C%201%5Cright%5C%7D%20" alt="LaTeX: \displaystyle \left\{-3, -1, 1\right\} " data-equation-content=" \displaystyle \left\{-3, -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, -3\right) \cup \left(-1, 1\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28-%5Cinfty%2C%20-3%5Cright%29%20%5Ccup%20%5Cleft%28-1%2C%201%5Cright%29%20" alt="LaTeX: \displaystyle \left(-\infty, -3\right) \cup \left(-1, 1\right) " data-equation-content=" \displaystyle \left(-\infty, -3\right) \cup \left(-1, 1\right) " /> </p> </p>