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Questions: Algebra BusinessCalculus
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Solve \(\displaystyle | 13 x + 1 |=129\)
Splitting into two linear equations gives \(\displaystyle 13 x + 1=-129\) and \(\displaystyle 13 x + 1=129\). Solving gives the two solutions \(\displaystyle x=\frac{128}{13}\) and \(\displaystyle x=-10\)
\begin{question}Solve $| 13 x + 1 |=129$
\soln{9cm}{Splitting into two linear equations gives $13 x + 1=-129$ and $13 x + 1=129$. Solving gives the two solutions $x=\frac{128}{13}$ and $x=-10$}
\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 <img class="equation_image" title=" \displaystyle | 13 x + 1 |=129 " src="/equation_images/%20%5Cdisplaystyle%20%7C%2013%20x%20%2B%201%20%7C%3D129%20" alt="LaTeX: \displaystyle | 13 x + 1 |=129 " data-equation-content=" \displaystyle | 13 x + 1 |=129 " /> </p> </p>
<p> <p>Splitting into two linear equations gives <img class="equation_image" title=" \displaystyle 13 x + 1=-129 " src="/equation_images/%20%5Cdisplaystyle%2013%20x%20%2B%201%3D-129%20" alt="LaTeX: \displaystyle 13 x + 1=-129 " data-equation-content=" \displaystyle 13 x + 1=-129 " /> and <img class="equation_image" title=" \displaystyle 13 x + 1=129 " src="/equation_images/%20%5Cdisplaystyle%2013%20x%20%2B%201%3D129%20" alt="LaTeX: \displaystyle 13 x + 1=129 " data-equation-content=" \displaystyle 13 x + 1=129 " /> . Solving gives the two solutions <img class="equation_image" title=" \displaystyle x=\frac{128}{13} " src="/equation_images/%20%5Cdisplaystyle%20x%3D%5Cfrac%7B128%7D%7B13%7D%20" alt="LaTeX: \displaystyle x=\frac{128}{13} " data-equation-content=" \displaystyle x=\frac{128}{13} " /> and <img class="equation_image" title=" \displaystyle x=-10 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-10%20" alt="LaTeX: \displaystyle x=-10 " data-equation-content=" \displaystyle x=-10 " /> </p> </p>