\hfill \hfill } %} \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}{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} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.3in,right=0.5in,top=0.5in,bottom=0.5in} \pagestyle{fancy} \lhead{\color{blue} \Large Lecture Notes} \chead{\color{black} \LARGE Equations} \rhead{\large page \ \thepage} \cfoot{} \lfoot{\small \copyright $\;$ copyright Hidegkuti, Powell, 2008} \textwidth 7.5in \textheight 9.5in \setlength{\headheight}{28pt} \setlength{\parindent}{0in} \begin{document} \begin{center} {\Large Sample Problems}\bigskip \end{center} \begin{enumerate} \item Consider the equation $2x^{2}+x+34=21x-8$. a) \ Is the number $1$ a solution of the equation? b) \ Is $3$ a solution of the equation? c) \ Is $x=4$ a solution of the equation? d) \ Is $7$ a solution of the equation? \item Consider the equation \ $3a-2b-1=\left( a-b\right) ^{2}+4$. a) \ Is the pair of numbers $a=8$ and $b=5$ a solution of the equation? b) \ Is the pair of numbers $a=10$ and $b=7$ a solution of the equation?\bigskip \bigskip \end{enumerate} \begin{center} {\Large Practice Problems}\bigskip \end{center} \begin{enumerate} \item Consider the equation $\dfrac{2x^{2}-11x-21}{2x+3}=\allowbreak 3x-\left( 2x+7\right) $. a) \ Is the number $8$ a solution of the equation? b) \ Is $13$ a solution of the equation? c) \ Is $x=10$ a solution of the equation? \item Consider the equation $y=\dfrac{5x-3}{2}$. a) \ Is the pair of numbers $x=1$ and $y=1$ a solution of the equation? b) \ Is the pair of numbers $x=9$ and $y=4$ a solution of the equation? c) \ Is the pair of numbers $x=3$ and $y=6$ a solution of the equation? d) \ Is the pair of numbers $x=17$ and $y=41$ a solution of the equation? \item Consider the equation \ $\left( p-q\right) ^{2}+\dfrac{3p-1}{6-q}% =4\left( p+1\right) $. a) \ Is the pair of numbers $p=8$ and $q=5$ a solution of the equation? b) \ Is the pair of numbers $p=7$ and $q=1$ a solution of the equation?\pagebreak \end{enumerate} \begin{center} {\Large Sample Problems - Answers}\bigskip \end{center} \begin{enumerate} \item a) \ no, since $37\not=13\qquad \ \ $b) \ yes, since $55=55\qquad \ \ $% c) \ no, since $70\not=76\qquad \ \ $ d) \ yes, since $139=139$ \item a) \ yes, since $13=13\qquad \qquad $b) \ no, since $15\not=13$% \bigskip \bigskip \bigskip \end{enumerate} \begin{center} {\Large Practice Problems - Answers}\bigskip \end{center} \begin{enumerate} \item a) \ yes, since $1=1\qquad \qquad $b) \ yes, since $6=6\qquad \qquad $% c)\ \ yes, since $3=3$ \item a) \ yes, since $1=1\qquad \qquad $b) \ no, since $21\not=4\qquad \qquad $c) \ yes, since $6=6$ d) \ yes, since $41=41$ \item a) \ no, since $32\not=36\qquad \qquad $b) \ yes, since $40=40$ \end{enumerate} \pagebreak \begin{center} {\Large Sample Problems - Solutions}\bigskip \end{center} \begin{enumerate} \item Consider the equation $2x^{2}+x+34=21x-8$. a) \ Is the number $1$ a solution of the equation?$~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion no, since $% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 37\not=13$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: \ We need to substitute $1$ for $x$ into both the left-hand side and right-hand side of the equation and evaluate those algebraic expressions to see whether the left-hand side equals to the right-hand side.\newline The left-hand side:% \begin{eqnarray*} \text{LHS} &=&2x^{2}+x+34=2\left( 1\right) ^{2}+\left( 1\right) +34=2\cdot 1+1+34 \\ &=&2+1+34=3+34=37 \end{eqnarray*}% The right-hand side:% \begin{equation*} \text{RHS}=21x-8=21\left( 1\right) -8=21-8=13 \end{equation*}% When $x=1$, the two expressions are not equal. \ Thus\ $1$ is NOT a solution of the equation. b) \ Is $3$ a solution of the equation?$~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion yes, since $% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 55=55$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: \ We need to substitute $3$ for $x$ into both the left-hand side and right-hand side of the equation and evaluate those algebraic expressions to see whether the left-hand side equals to the right-hand side.\newline The left-hand side:% \begin{eqnarray*} \text{LHS} &=&2x^{2}+x+34=2\left( 3\right) ^{2}+\left( 3\right) +34 \\ &=&2\cdot 9+3+34=18+3+34=21+34=55 \end{eqnarray*}% The right-hand side:% \begin{equation*} \text{RHS}=21x-8=21\left( 3\right) -8=63-8=55 \end{equation*}% When $x=3$, the two expressions are equal. Thus \ $3$ IS a solution of the equation. c) \ Is $x=4$ a solution of the equation?$~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion no, since $% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 70\not=76$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: \ We need to substitute $4$ for $x$ into both the left-hand side and right-hand side of the equation and evaluate those algebraic expressions to see whether the left-hand side equals to the right-hand side.\newline The left-hand side:% \begin{eqnarray*} \text{LHS} &=&2x^{2}+x+34=2\left( 4\right) ^{2}+\left( 4\right) +34 \\ &=&2\cdot 16+4+34=32+4+34=36+34=70 \end{eqnarray*}% The right-hand side:% \begin{equation*} \text{RHS}=21x-8=21\left( 4\right) -8=84-8=76 \end{equation*}% Since the two expressions are not equal when $x=4$, \ $4$ is NOT a solution of the equation. d) \ Is $7$ a solution of the equation?$~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion yes, since $% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 139=139$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: \ We need to substitute $7$ for $x$ into both the left-hand side and right-hand side of the equation and evaluate those algebraic expressions to see whether the left-hand side equals to the right-hand side.\newline The left-hand side:% \begin{eqnarray*} \text{LHS} &=&2x^{2}+x+34=2\left( 7\right) ^{2}+\left( 7\right) +34 \\ &=&2\cdot 49+7+34=98+7+34=105+34=139 \end{eqnarray*}% The right-hand side:% \begin{equation*} \text{RHS}=21x-8=21\left( 7\right) -8=147-8=139 \end{equation*}% Since the two expressions are equal when $x=7$, \ \ $7$ \ IS a solution of the equation. \item Consider the equation \ $3a-2b-1=\left( a-b\right) ^{2}+4$. a) \ Is the pair of numbers $a=8$ and $b=5$ a solution of the equation?$~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion yes, since $% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 13=13$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: \ We need to substitute $a=8$ \ \ and $b=5$ into both the left-hand side and right-hand side of the equation and evaluate those algebraic expressions to see whether the left-hand side equals to the right-hand side.\newline The left-hand side:% \begin{eqnarray*} \text{LHS} &=&3a-2b-1=3\left( 8\right) -2\left( 5\right) -1=3\cdot 8-2\cdot 5-1 \\ &=&24-2\cdot 5-1=24-10-1=14-1=13 \end{eqnarray*}% The right-hand side:% \begin{equation*} \text{RHS}=\left( a-b\right) ^{2}+4=\left( \left( 8\right) -\left( 5\right) \right) ^{2}+4=\left( 8-5\right) ^{2}+4=3^{2}+4=9+4=13 \end{equation*}% Since the two expressions are equal when $a=8$ \ \ and $b=5$, \ this pair IS a solution of the equation. b) \ Is the pair of numbers $a=10$ and $b=7$ a solution of the equation?$~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion no, since $% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 15\not=13$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: \ We need to substitute $a=10$ \ \ and $b=7$ into both the left-hand side and right-hand side of the equation and evaluate those algebraic expressions to see whether the left-hand side equals to the right-hand side.\newline The left-hand side:% \begin{eqnarray*} \text{LHS} &=&3a-2b-1=3\left( 10\right) -2\left( 7\right) -1=3\cdot 10-2\cdot 7-1 \\ &=&30-2\cdot 7-1=30-14-1=16-1=15 \end{eqnarray*}% The right-hand side:% \begin{equation*} \text{RHS}=\left( a-b\right) ^{2}+4=\left( \left( 10\right) -\left( 7\right) \right) ^{2}+4=\left( 10-7\right) ^{2}+4=3^{2}+4=9+4=13 \end{equation*}% Since the two expressions are not equal when $a=8$ \ \ and $b=5$, \ this pair is NOT a solution of the equation. \end{enumerate} \bigskip \vspace{2in} \href{http://www.teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html% }{For more documents like this, visit our page at\ http://www.teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}