\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus

Please login to create an exam or a quiz.

Algebra
Rational Functions and Equations
New Random

Solve \(\displaystyle \frac{x}{x - 1} - \frac{4}{x - 2}=- \frac{4}{x^{2} - 3 x + 2}\).

Factoring the denominator on the right hand side gives \(\displaystyle \left(x - 2\right) \left(x - 1\right)\). This gives the LCD as \(\displaystyle \left(x - 2\right) \left(x - 1\right)\). Multiplying by the LCD gives \(\displaystyle x \left(x - 2\right) - 4 x + 4 = -4\). Getting zero on one side gives \(\displaystyle x^{2} - 6 x + 8=0\). Factoring gives \(\displaystyle \left(x - 4\right) \left(x - 2\right)=0\). The two possible solutions are \(\displaystyle x = 4\) and \(\displaystyle x = 2\). Checking the possible solutions gives:
Since \(\displaystyle 2\) is zero of the denominator it is not in the domain and must be rejected as a solution. Since \(\displaystyle 4\) is not zero of the denominator it is a solution.

Download \(\LaTeX\) 

\begin{question}Solve $\frac{x}{x - 1} - \frac{4}{x - 2}=- \frac{4}{x^{2} - 3 x + 2}$. 
    \soln{9cm}{Factoring the denominator on the right hand side gives $\left(x - 2\right) \left(x - 1\right)$. This gives the LCD as $\left(x - 2\right) \left(x - 1\right)$.  Multiplying by the LCD gives $x \left(x - 2\right) - 4 x + 4 = -4$.  Getting zero on one side gives $x^{2} - 6 x + 8=0$. Factoring gives  $\left(x - 4\right) \left(x - 2\right)=0$. The two possible solutions are $x = 4$ and $x = 2$. Checking the possible solutions gives:\newline
Since $2$ is zero of the denominator it is not in the domain and must be rejected as a solution. Since $4$ is not zero of the denominator it is a solution. }

\end{question}

Download Question and Solution Environment\(\LaTeX\)
\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}
HTML for Canvas 
<p> <p>Solve  <img class="equation_image" title=" \displaystyle \frac{x}{x - 1} - \frac{4}{x - 2}=- \frac{4}{x^{2} - 3 x + 2} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bx%7D%7Bx%20-%201%7D%20-%20%5Cfrac%7B4%7D%7Bx%20-%202%7D%3D-%20%5Cfrac%7B4%7D%7Bx%5E%7B2%7D%20-%203%20x%20%2B%202%7D%20" alt="LaTeX:  \displaystyle \frac{x}{x - 1} - \frac{4}{x - 2}=- \frac{4}{x^{2} - 3 x + 2} " data-equation-content=" \displaystyle \frac{x}{x - 1} - \frac{4}{x - 2}=- \frac{4}{x^{2} - 3 x + 2} " /> . </p> </p>
HTML for Canvas
<p> <p>Factoring the denominator on the right hand side gives  <img class="equation_image" title=" \displaystyle \left(x - 2\right) \left(x - 1\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20-%202%5Cright%29%20%5Cleft%28x%20-%201%5Cright%29%20" alt="LaTeX:  \displaystyle \left(x - 2\right) \left(x - 1\right) " data-equation-content=" \displaystyle \left(x - 2\right) \left(x - 1\right) " /> . This gives the LCD as  <img class="equation_image" title=" \displaystyle \left(x - 2\right) \left(x - 1\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20-%202%5Cright%29%20%5Cleft%28x%20-%201%5Cright%29%20" alt="LaTeX:  \displaystyle \left(x - 2\right) \left(x - 1\right) " data-equation-content=" \displaystyle \left(x - 2\right) \left(x - 1\right) " /> .  Multiplying by the LCD gives  <img class="equation_image" title=" \displaystyle x \left(x - 2\right) - 4 x + 4 = -4 " src="/equation_images/%20%5Cdisplaystyle%20x%20%5Cleft%28x%20-%202%5Cright%29%20-%204%20x%20%2B%204%20%3D%20-4%20" alt="LaTeX:  \displaystyle x \left(x - 2\right) - 4 x + 4 = -4 " data-equation-content=" \displaystyle x \left(x - 2\right) - 4 x + 4 = -4 " /> .  Getting zero on one side gives  <img class="equation_image" title=" \displaystyle x^{2} - 6 x + 8=0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20-%206%20x%20%2B%208%3D0%20" alt="LaTeX:  \displaystyle x^{2} - 6 x + 8=0 " data-equation-content=" \displaystyle x^{2} - 6 x + 8=0 " /> . Factoring gives   <img class="equation_image" title=" \displaystyle \left(x - 4\right) \left(x - 2\right)=0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20-%204%5Cright%29%20%5Cleft%28x%20-%202%5Cright%29%3D0%20" alt="LaTeX:  \displaystyle \left(x - 4\right) \left(x - 2\right)=0 " data-equation-content=" \displaystyle \left(x - 4\right) \left(x - 2\right)=0 " /> . The two possible solutions are  <img class="equation_image" title=" \displaystyle x = 4 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%204%20" alt="LaTeX:  \displaystyle x = 4 " data-equation-content=" \displaystyle x = 4 " />  and  <img class="equation_image" title=" \displaystyle x = 2 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%202%20" alt="LaTeX:  \displaystyle x = 2 " data-equation-content=" \displaystyle x = 2 " /> . Checking the possible solutions gives:<br>
Since  <img class="equation_image" title=" \displaystyle 2 " src="/equation_images/%20%5Cdisplaystyle%202%20" alt="LaTeX:  \displaystyle 2 " data-equation-content=" \displaystyle 2 " />  is zero of the denominator it is not in the domain and must be rejected as a solution. Since  <img class="equation_image" title=" \displaystyle 4 " src="/equation_images/%20%5Cdisplaystyle%204%20" alt="LaTeX:  \displaystyle 4 " data-equation-content=" \displaystyle 4 " />  is not zero of the denominator it is a solution. </p> </p>