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%TCIDATA{Created=Wednesday, July 12, 2006 00:27:03}
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%TCIDATA{<META NAME="Title" CONTENT="Sample Problems - Evaluating Algebraic Expressions - Solutions">}
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\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
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\lfoot{\small   \copyright $\;$ copyright  Hidegkuti,  Powell,  2007}
\cfoot{}
\chead{\Large Evaluating Algebraic Expressions\\ with Natural Numbers}
\rhead{page \thepage\\\color{red}\large Solutions}
\textwidth 7.0in
\textheight 9in
\setlength{\headheight}{35pt}

\begin{document}


{\Large Part 1. Sample Problems}

\bigskip

Evaluate each of the algebraic expressions when $p=7$ \ and \ $q=3$.

\begin{enumerate}
\item $15-p=~~%
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8%
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Solution: Step 1. We re-write the expression with one modification: we
replace each variable by an empty pair of parenthses.

Step 2. We insert the values into the parentheses. \ Now the problem becomes
an order of operations problem.

Step 3. We drop the unnecessary parentheses and work out the order of
operations problem. \ (It may appear awkward to create these parentheses but
they will later become extremely helpful.)%
\begin{eqnarray*}
\text{Step 1. \ \ \ \ \ \ \ \ \ \ }15-p &=&15-\left( ~~\right) \\
\text{Step 2. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } &=&15-\left( 7\right) \\
\text{Step 3. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } &=&15-7 \\
&=&8
\end{eqnarray*}

\item $pq=~~%
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21%
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Solution:%
\begin{eqnarray*}
\text{Step 1. \ \ \ \ \ \ \ \ \ \ \ \ \ \ }pq &=&\left( ~~\right) \left(
~~\right) \\
\text{Step 2. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } &=&\left( 7\right) \left(
3\right) \\
\text{Step 3. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } &=&21
\end{eqnarray*}

\item $4p-q=~~%
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25%
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Solution:%
\begin{eqnarray*}
\text{Step 1. \ \ \ \ \ \ \ \ \ \ \ \ \ \ }4p-q &=&4\left( ~~\right) -\left(
~~\right) \\
\text{Step 2. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } &=&4\left(
7\right) -\left( 3\right) \\
\text{Step 3.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } &=&4\cdot 7-3%
\text{ \ \ \ \ \ \ \ \ \ \ \ \ multiplication} \\
&=&28-3\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtraction} \\
&=&25
\end{eqnarray*}

\item $p-2q=~~%
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1%
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Solution:%
\begin{eqnarray*}
\text{Step 1. \ \ \ \ \ }p-2q &=&\left( ~~\right) -2\left( ~~\right)  \\
\text{Step 2.\ \ \ \ \ \ \ \ \ \ \ \ \ \ } &=&\left( 7\right) -2\left(
3\right)  \\
\text{Step 3.\ \ \ \ \ \ \ \ \ \ \ \ \ \ } &=&7-2\cdot 3\text{ \ \ \ \ \ \
multiplication} \\
&=&7-6\text{ \ \ \ \ \ \ \ \ \ \ \ subtraction} \\
&=&1
\end{eqnarray*}

\item $p^{2}-q^{2}=~~%
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40%
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Solution: 
\begin{eqnarray*}
p^{2}-q^{2} &=&\left( ~~\right) ^{2}-\left( ~~\right) ^{2} \\
&=&\left( 7\right) ^{2}-\left( 3\right) ^{2} \\
&=&7^{2}-3^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ exponents,} \\
&=&49-3^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ left to right} \\
&=&49-9\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtraction} \\
&=&40
\end{eqnarray*}

\item $\left( p-q\right) ^{2}=~~%
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16%
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Solution: 
\begin{eqnarray*}
\left( p-q\right) ^{2} &=&\left[ \left( ~~\right) -\left( ~~\right) \right]
^{2} \\
&=&\left[ \left( 7\right) -\left( 3\right) \right] ^{2} \\
&=&\left( 7-3\right) ^{2}\text{ \ \ \ \ \ \ \ \ \ subtraction in parentheses}
\\
&=&4^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ exponentiation} \\
&=&16
\end{eqnarray*}

\item $2q^{2}=~~%
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Solution: 
\begin{eqnarray*}
2q^{2} &=&2\left( ~~\right) ^{2} \\
&=&2\left( 3\right) ^{2} \\
&=&2\cdot 3^{2}\text{ \ \ \ \ \ \ \ \ exponentiation} \\
&=&2\cdot 9\text{ \ \ \ \ \ \ \ \ \ \ multiplication} \\
&=&18
\end{eqnarray*}

\item $\left( 2q\right) ^{2}=~~%
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36%
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Solution: 
\begin{eqnarray*}
\left( 2q\right) ^{2} &=&\left[ 2\left( ~~\right) \right] ^{2} \\
&=&\left[ 2\left( 3\right) \right] ^{2} \\
&=&\left( 2\cdot 3\right) ^{2}\text{ \ \ \ \ \ \ \ \ \ \ multiplication in
parentheses} \\
&=&6^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ exponents} \\
&=&36
\end{eqnarray*}%
\pagebreak

\item $15-\dfrac{p+q}{5}=~~%
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Solution: \ From here on, we show computations \textbf{in the form they
should appear}. \ Once you wrote down the expression with little parentheses
instead of the letters, you can insert the values into it. \ (In this
problem, we skipped the line $15-\dfrac{\left( ~~\right) +\left( ~~\right) }{%
5}$) 
\begin{eqnarray*}
15-\dfrac{p+q}{5} &=&15-\dfrac{\left( 7\right) +\left( 3\right) }{5} \\
&=&15-\dfrac{7+3}{5}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ invisible parentheses!%
} \\
&=&15-\dfrac{10}{5}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ division} \\
&=&15-2\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtraction} \\
&=&13
\end{eqnarray*}

\item $\left( p+q\right) ^{2}-\left( 5q-2p\right) ^{4}=~~%
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Solution:%
\begin{equation*}
\left( p+q\right) ^{2}-\left( 5q-2p\right) ^{4}=\left[ \left( 7\right)
+\left( 3\right) \right] ^{2}-\left[ 5\left( 3\right) -2\left( 7\right) %
\right] ^{4}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }
\end{equation*}%
\begin{eqnarray*}
&=&\left( 7+3\right) ^{2}-\left( 5\cdot 3-2\cdot 7\right) ^{4}\text{ \ \ \ \
\ \ \ \ \ \ \ \ \ \ addition in parentheses} \\
&=&10^{2}-\left( 5\cdot 3-2\cdot 7\right) ^{4}\text{ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ multiplications in parentheses,} \\
&=&10^{2}-\left( 15-2\cdot 7\right) ^{4}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ left to right} \\
&=&10^{2}-\left( 15-14\right) ^{4}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ subtraction in parentheses} \\
&=&10^{2}-1^{4}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ exponents, left to right} \\
&=&100-1^{4}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
careful! \ }1^{4}\not=4 \\
&=&100-1 \\
&=&99
\end{eqnarray*}%
\pagebreak
\end{enumerate}

{\Large Part 2. Practice Problems}\bigskip

\begin{enumerate}
\item Evaluate each of the algebraic expressions when $x=6$ \ and \ $y=8$.

\begin{enumerate}
\item $19-y+x=~~%
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\item $19-\left( y+x\right) =~~%
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5$%
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\item $2x^{2}-5y+3=~~%
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\item $x^{2}+y^{2}=~~%
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100$%
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\item $\left( x+y\right) ^{2}=~~%
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\item $3y-x=~~%
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\item $3\left( y-x\right) =~~%
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6$%
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\item $\dfrac{5x-y}{2}=~~%
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\item $5x-\dfrac{y}{2}=~~%
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\item $\dfrac{x^{2}-5x+4}{y-3}=~~%
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\end{enumerate}

\item Consider the expression $\dfrac{6a-3b-ab+2a^{2}}{2a-b}-3$.

\begin{enumerate}
\item Evaluate the expression if \ $a=5$ \ and \ $b=1$.~$~~%
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5$%
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\item Evaluate the expression if \ $a=5$ \ and \ $b=2$.~$~~%
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\item Evaluate the expression if \ $a=5$ \ and \ $b=3$.~$~~%
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\item Evaluate the expression if \ $a=4$ \ and \ $b=1$.~$~~%
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\end{enumerate}

\end{document}
