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\begin{center}\huge MATH 1011 - EXAM 4 - FALL 2016\\
SOLUTION
\end{center}
\begin{flushleft}\vspace{0.5in}
Thursday 27 October 2016\\
Instructor: Tom Cuchta
\end{flushleft}
\textbf{Instructions:}
\begin{itemize}
	\item  Show all work, clearly and in order, if you want to get full
	credit. If you claim something is true \textbf{you must show work backing up your claim}. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct).
	\item Justify your answers algebraically whenever possible to ensure full credit. 	
	\item  Circle or otherwise indicate your final answers.
	\item  Please keep your written answers brief; be clear and to the point.
	\item  Good luck!
\end{itemize}
\vspace*{10pt}
\end{coverpages}

\begin{questions}
\question[15] For which values of $x$ are the following expressions defined?
\begin{parts}
\part[5] $\dfrac{2x^2-9x+3}{x+1}$ \\
\textit{Solution:} The expression is valid when
$$x+1 \neq 0,$$
or
$$x \neq -1.$$
\part[5] $\dfrac{x^2+7}{(x+3)(x-6)}$ \\
\textit{Solution:} The expression is valid when
$$(x+3)(x-6) \neq 0,$$
or
$$x \neq -3, 6.$$
\part[5] $\dfrac{6x^3+9x+2}{x^2+9x+14}$ \\
\textit{Solution:} The expression is valid when
$$x^2+9x+14 \neq 0,$$
or
$$(x+7)(x+2) \neq 0,$$
or
$$x \neq -7, -2.$$
\end{parts}
\question[15] Find the least common multiple of \ldots 
\begin{parts}
\part[5] $2$, $6$, and $7$ \\
\textit{Solution:} Since $6 = 3 \cdot 2$ we see that 
$$\mathrm{lcm}(2,6,7) = \rm{lcm}(2,3\cdot 2, 7)= 2 \cdot 3 \cdot 7 = 42.$$
\part[5] $(x+2)(x+6)^2$, $(x+2)^2(x+6)$, and $(x+2)(x+6)$ \\
\textit{Solution:} 
$$\mathrm{lcm}\left( (x+2)(x+6)^2, (x+2)^2(x+6), (x+2)(x+6) \right) = (x+2)^2(x+6)^2$$
\part[5] $x^2+5x+6$, $x^2+3x+2$, and $x^2+7x+12$ \\
\textit{Solution:} Since
$$x^2+5x+6=(x+3)(x+2),$$
$$x^2+3x+2=(x+1)(x+2),$$
and
$$x^2+7x+12=(x+3)(x+4),$$
we see that
\begin{eqnarray*}
\lefteqn{\mathrm{lcm}(x^2+5x+6, x^2+3x+2, x^2+7x+12)} \\
& &= \mathrm{lcm}((x+3)(x+2), (x+2)(x+1), (x+4)(x+3)) \\
& &= (x+4)(x+3)(x+2)(x+1)
\end{eqnarray*}
\end{parts}

\question[20] Add or subtract and simplify
\begin{parts}
\part[5] $\dfrac{1}{3} + \dfrac{1}{7}$ \\
\textit{Solution}: The common denominator is $21$, so
$$\dfrac{1}{3} + \dfrac{1}{7} = \dfrac{7}{21} + \dfrac{3}{21} = \dfrac{10}{21}.$$
\part[5] $\dfrac{1}{x-1} + \dfrac{1}{x+3}$ \\
\textit{Solution:} The common denominator is $(x-1)(x+3)$, so
$$\dfrac{1}{x-1} + \dfrac{1}{x+3} = \dfrac{x+3}{(x-1)(x+3)} + \dfrac{x-1}{(x-1)(x+3)} = \dfrac{2x+2}{(x-1)(x+3)}.$$
\part[5] $\dfrac{5}{(x+2)(x+3)} - \dfrac{x}{(x+1)(x+2)}$ \\
\textit{Solution:} The common denominator is $(x+1)(x+2)(x+3)$, so
$$\begin{array}{ll}
\dfrac{5}{(x+2)(x+3)} - \dfrac{x}{(x+1)(x+2)} &= \dfrac{5(x+1)}{(x+1)(x+2)(x+3)} - \dfrac{x(x+3)}{(x+1)(x+2)(x+3)}\\
&= \dfrac{5x+5 - x^2 - 3x}{(x+1)(x+2)(x+3)} \\
&= \dfrac{-x^2 +2x+5}{(x+1)(x+2)(x+3)}.
\end{array}$$
\part[5] $\dfrac{1}{x^2+2x+1} + \dfrac{1}{x^2+6x+5}$ \\
\textit{Solution:} Since
$$x^2+2x+1 = (x+1)(x+1)$$
and
$$x^2+6x+5=(x+3)(x+2),$$
we see that the common denominator is $(x+1)^2(x+2)(x+3)$. So,
$$\begin{array}{ll}
\dfrac{1}{x^2+2x+1} + \dfrac{1}{x^2+6x+5} &= \dfrac{1}{(x+1)(x+1)} + \dfrac{1}{(x+3)(x+2)} \\
&= \dfrac{(x+3)(x+2)}{(x+1)^2(x+2)(x+3)} + \dfrac{(x+1)^2}{(x+1)^2(x+2)(x+3)} \\
&= \dfrac{x^2+5x+6}{(x+1)^2(x+2)(x+3)} + \dfrac{x^2+2x+1}{(x+1)^2(x+2)(x+3)} \\
&= \dfrac{2x^2+7x+7}{(x+1)^2(x+2)(x+3)}.
\end{array}$$
\end{parts}

\question[15] Simplify.
\begin{parts}
\part[7] $\dfrac{\frac{1}{x} + 2}{\frac{1}{x}+3}$ \\
\textit{Solution:} Get a common denominator on top and bottom then simplify the fraction of fractions:
$$\begin{array}{ll}
\dfrac{\frac{1}{x}+2}{\frac{1}{x}+3} &= \dfrac{\frac{1+2x}{x}}{\frac{1+3x}{x}} \\
&= \left( \dfrac{1+2x}{x} \right) \left( \dfrac{x}{1+3x} \right) \\
&= \dfrac{1+2x}{1+3x}.
\end{array}$$
\part[8] $\dfrac{x-3+\frac{1}{x}}{\frac{1}{x+8}+x}$ \\
\textit{Solution:} Get a common denominator on top and bottom and then simplify the fraction of fractions:
$$\begin{array}{ll}
\dfrac{x-3+\frac{1}{x}}{\frac{1}{x+8}+x} &= \dfrac{\frac{x^2-3x+1}{x}}{\frac{1+x^2+8x}{x}} \\
&= \left( \dfrac{x^2-3x+1}{x} \right) \left( \dfrac{x}{x^2+8x+1} \right) \\
&= \dfrac{x^2-3x+1}{x^2+8x+1}.
\end{array}$$
Note that neither the top nor the bottom factor, so this is as far simplified as it can go.
\end{parts}

\question[15] Multiply and simplify.
\begin{parts}
\part[5] $\dfrac{x+1}{x+2} \cdot \dfrac{(x+2)(x+3)}{x+1}$ \\
\textit{Solution:} Cancel the $x+2$'s and $x+1$'s to get
$$\left( \dfrac{x+1}{x+2} \right) \left( \dfrac{(x+2)(x+3)}{x+1} \right) = x+3.$$
\part[5] $\dfrac{(t+2)(4t+9)}{3t-2} \cdot \dfrac{6t-4}{(t-2)(4t+9)}$ \\
\textit{Solution:} Notice that $6t-4=2(3t-2)$. Cancel the factors of $3t-2$ and the factors of $4t+9$ to get
$$\begin{array}{ll}
\left( \dfrac{(t+2)(4t+9)}{3t-2} \right) \left( \dfrac{6t-4}{(t-2)(4t+9)} \right) &= \left( \dfrac{t+2}{3t-2} \right) \left( \dfrac{2(3t-2)}{t-2} \right) \\
&= \dfrac{2t+4}{t-2}.
\end{array}$$
\part[5] $\dfrac{x^2-16}{x^2+7x+6} \cdot \dfrac{x+6}{x+4}$ \\
\textit{Solution:} Notice that
$$x^2-16=(x-4)(x+4)$$
and
$$x^2+7x+6=(x+6)(x+1).$$
Substitute in these factorizations and cancel appropriately to get
$$\begin{array}{ll}
\left( \dfrac{x^2-16}{x^2+7x+6} \right) \left( \dfrac{x+6}{x+4} \right) &= \left( \dfrac{(x-4)(x+4)}{(x+6)(x+1)} \right) \left( \dfrac{x+6}{x+4} \right) \\
&= \dfrac{x-4}{x+1}.
\end{array}$$
\end{parts}

\question[10] Solve.
\begin{parts}
\part[5] $\dfrac{4}{5} - \dfrac{1}{3} = \dfrac{x}{2}$ \\
\textit{Solution:} The common denominator on the left is $15$ yielding
$$\dfrac{7}{15} = \dfrac{x}{2}.$$
Now multiply both sides by $2$ to get
$$\dfrac{14}{15} = x.$$
\part[5] $\dfrac{4}{x+1} + \dfrac{3}{x-5} = \dfrac{7}{x^2-4x-5}$ \\
\textit{Solution:} Notice that
$$x^2-4x-5 = (x-5)(x+1).$$
On the left the common denominator is $(x-5)(x+1)$ so write it in terms of that:
$$\dfrac{4(x-5)}{(x+1)(x-5)} + \dfrac{3(x+1)}{(x+1)(x-5)} = \dfrac{7}{(x+1)(x-5)}.$$
Simplify the left hand side to get
$$\dfrac{7x-17}{(x+1)(x-5)} = \dfrac{7}{(x+1)(x-5)}.$$
Now multiply both sides by $(x+1)(x-5)$ to get
$$7x-17=7.$$
Add $17$ to get
$$7x=7,$$
and finally divide by $7$ to get $x=1$. 
\end{parts}


\question[10] Do the following problems. 
\begin{parts}
\part[5] Bob mows a lawn in $7$ hours. It takes Mary $4$ hours to mow the same lawn. How many hours does it take them working together? \\
\textit{Solution:} The rate at which Bob mows the lawn is $\dfrac{1}{7} \dfrac{\rm{lawn}}{\rm{hour}}$ and the rate at which Mary mows the lawn is $\dfrac{1}{4} \dfrac{\rm{lawn}}{\rm{hour}}$. Therefore together they mow at a rate of 
$$\left( \dfrac{1}{7} + \dfrac{1}{4} \right) \dfrac{\rm{lawn}}{\rm{hour}}= \dfrac{11}{28} \dfrac{\rm{lawn}}{\rm{hour}}.$$
Therefore if they work together it should take $\dfrac{28}{11}$ hours to mow the one lawn.
\part[5] The following triangles are similar triangles. Find $x$. \\
\begin{center}
\begin{tikzpicture}[scale=2]
\draw (0,0)--(1,0)--(0.5,1)--cycle;
\draw (2,0)--(3.5,0)--(2.75,1.5)--cycle;
\draw (0.5,-0.1) node {$2$};
\draw (0.9,0.55) node {$7$};
\draw (2.75,-0.1) node {$x$};
\draw (3.3,0.7) node {$5$};
\end{tikzpicture}
\end{center} 
\textit{Solution:} Since the triangles are similar, we may write
$$\dfrac{2}{x} = \dfrac{7}{5}.$$
Therefore multiply by $x$ to get
$$2 = \dfrac{7x}{5}.$$
Now multiply by $5$ to get
$$10 = 7x,$$
and finally divide by $7$ to get
$$x = \dfrac{10}{7}.$$
\end{parts}
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\end{questions}

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