\hfill \thepage} %} \input{tcilatex} \geometry{left=0.6in,right=0.7in,top=0.7in,bottom=0.9in} \pagestyle{fancy} \lhead{ \Large \color{blue}Lecture Notes } \chead{\LARGE Integers} \rhead{\large page \ \thepage} \lfoot{\small \copyright $\;$ copyright Hidegkuti, Powell, 2009} \cfoot{} \textwidth 7.2in \textheight 9.2in \setlength{\headheight}{35pt} \setlength{\parindent}{0in} \begin{document} \begin{center} {\LARGE Sample Problems}\bigskip \end{center} \begin{enumerate} \item Perform each of the indicated operations. \begin{enumerate} \item $21-2\left( -5\right) =$ \item $\sqrt{60\div 4\div 5+1}=$ \item $20-\left\vert 32\div \left( -2\right) \right\vert =$ \item $\dfrac{-3^{2}+30\div \left( -2\right) }{-4}=$ \item $\left( 9-6\right) ^{2}=$ \item $9^{2}-6^{2}=$ \item $16-\sqrt{2^{3}-2^{2}}-5=$ \item $\sqrt{25}\cdot \dfrac{-3^{5}+\left( -1\right) ^{3}}{\left\vert \left( -5\right) ^{3}-\left\vert 6+\left( -5\right) ^{3}\right\vert \right\vert }=$ \item $\dfrac{3+\left( 6^{2}-3^{2}\right) +\left( 3^{3}-2^{3}\right) -9}{% 4^{2}+\left( 5-3\right) ^{2}}=$ \item $\sqrt{\sqrt{3^{6}}-2\left( 3^{2}+3^{3}\right) +2\left( 4\left( 17-2^{3}\right) -3^{2}\right) }=$ \item $2\cdot 3^{2}-\left( 6-4+\left( 6\cdot 4-3\left( 2^{4}-12\right) \right) \right) =$ \end{enumerate} \item Let $a=-4$, $\ b=2$, and $x=-3$. Evaluate each of the following expressions. \begin{enumerate} \item $a^{2}-b^{2}=$ \item $\left( a-b\right) ^{2}=$ \item $a^{b}-2bx+x-\left\vert 2x\right\vert =$ \item $\dfrac{-x^{2}+\left( x+2\right) ^{2}}{\left( x-1\right) }=$ \item $\dfrac{x-1}{x+3}=$ \end{enumerate} \item Consider the equation $x^{2}-10x+x^{3}-4=4\left( x+5\right) $. \ In each case, determine whether the number given is a solution of the equation or not. \begin{enumerate} \item $x=0$ \item $x=-2$ \item $x=-3$\pagebreak \end{enumerate} \end{enumerate} \begin{center} {\LARGE Practice Problems}\bigskip \end{center} \begin{enumerate} \item Simplify each of the following expressions. \begin{enumerate} \item $\left\vert \left\vert -20-3\left( -4\right) \right\vert -10\right\vert =$ \item $15-3\left( -2\right) =$ \item $\left( \left( \left( 1-2\right) ^{2}-2\right) ^{2}-2\right) ^{2}-2=$ \item $\sqrt{-7^{2}+15\left( 2\right) ^{2}-18\div 3\div 3}=$ \item $\dfrac{-6^{2}\div 2\cdot 3+\left\vert 2^{2}\cdot 12-11\right\vert +2\left( -1\right) ^{3}-\left( -1\right) ^{2}}{\left( -2\right) ^{2}+\left( -2\right) ^{3}}=$ \item $\sqrt{36}-\sqrt{81}+2\sqrt{25}=$ \end{enumerate} \item Evaluate $\dfrac{8x+x^{2}-33}{x+11}$ \ \ ~if \ \ \ \ \ \ \ \ \ \begin{enumerate} \item $x=0$ \item $x=-11$ \end{enumerate} \item Evaluate $\dfrac{x-2}{2-x}$ \ if \begin{enumerate} \item $x=0$ \item $x=2$ \end{enumerate} \item Consider the equation $-x+x^{3}+6=3\left( -x^{2}+x+2\right) $. \ In each case, determine whether the number given is a solution of the equation or not. \begin{enumerate} \item $x=0$ \item $x=1$ \item $x=-1$ \item $x=2$ \item $x=-2$ \item $x=3$ \item $x=-3$ \item $x=4$ \item $x=-4$\pagebreak \end{enumerate} \end{enumerate} \begin{center} {\LARGE Sample Problems - Solutions}\bigskip \end{center} \begin{enumerate} \item Perform each of the indicated operations. \begin{enumerate} \item $21-2\left( -5\right) =~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 31$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: \ \begin{eqnarray*} 21-2\left( -5\right) &=&\text{ \ \ \ \ \ \ \ \ \ multiplication first: \ }% 2\left( -5\right) =-10 \\ 21-\left( -10\right) &=&\text{ \ \ \ \ \ \ \ \ \ to subtract is to add the opposite} \\ 21+10 &=&31 \end{eqnarray*} \item $\sqrt{60\div 4\div 5+1}=~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 2$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: The square root sign stretching across other operations is a case of an "invisible parentheses." \ We will work out everything else and then take the square root of the result. \ We need to be careful; the two divisions have to be executed left to right. \begin{eqnarray*} \sqrt{60\div 4\div 5+1} &=&\text{ \ \ \ division first, left to right!} \\ \sqrt{15\div 5+1} &=&\text{ \ \ \ division} \\ \sqrt{3+1} &=&\text{ \ \ \ addition} \\ \sqrt{4} &=&\text{ \ \ \ square root} \\ &=&2 \end{eqnarray*} \item $20-\left\vert 32\div \left( -2\right) \right\vert =~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 4$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: We start with the absolute value sign since it also functions as parentheses. \ We work out what's inside and then take the absolute value of the result. This absolute value is what we subtract from $20$.% \begin{eqnarray*} 20-\left\vert 32\div \left( -2\right) \right\vert &=&\text{ \ \ \ \ \ \ division within absolute value sign} \\ 20-\left\vert -16\right\vert &=&\text{ \ \ \ \ \ \ \ absolute value of }-16% \text{ is }16 \\ 20-16 &=&\text{ \ \ \ \ \ \ \ \ subtraction} \\ &=&4 \end{eqnarray*} \item $\dfrac{-3^{2}+30\div \left( -2\right) }{-4}=~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 6$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: The division bar stretching over an expression is a case of an "invisible parentheses." \ We completely work out the top and the bottom, and finally divide.% \begin{eqnarray*} \dfrac{-3^{2}+30\div \left( -2\right) }{-4} &=&\text{ \ \ \ \ \ \ \ \ \ exponents} \\ \dfrac{-9+30\div \left( -2\right) }{-4} &=&\text{ \ \ \ \ \ \ \ \ \ division upstrairs} \\ \dfrac{-9+\left( -15\right) }{-4} &=&\text{ \ \ \ \ \ \ \ \ \ addition upstairs} \\ \dfrac{-24}{-4} &=&\text{ \ \ \ \ \ \ \ \ \ division} \\ &=&6 \end{eqnarray*} \item $\left( 9-6\right) ^{2}=~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 9$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: We start with the parentheses. \begin{eqnarray*} \left( 9-6\right) ^{2} &=&\text{ \ \ \ \ \ subtraction within parentheses} \\ \left( 3\right) ^{2} &=&\text{ \ \ \ \ \ drop parentheses} \\ 3^{2} &=&\text{ \ \ \ \ \ exponents} \\ &=&9 \end{eqnarray*} \item $9^{2}-6^{2}=~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 45$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: This and the previous problem is here to remind you that $% a^{2}-b^{2}$ \ and $\left( a-b\right) ^{2}$ are different! In this case there is no parentheses, and so we start with the exponents.% \begin{eqnarray*} 9^{2}-6^{2} &=&\text{ exponents} \\ 81-36 &=&\text{ subtraction} \\ &=&45 \end{eqnarray*} \item $16-\sqrt{2^{3}-2^{2}}-5=~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 9$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: The square root sign stretching across other operations is a case of an "invisible parentheses." We start there. \ \begin{eqnarray*} 16-\sqrt{2^{3}-2^{2}}-5 &=&\text{ \ \ \ \ exponents under square root} \\ 16-\sqrt{8-4}-5 &=&\text{ \ \ subtraction under square root} \\ 16-\sqrt{4}-5 &=&\text{ \ \ square root} \\ 16-2-5 &=&\text{ \ \ subtractions, left to right} \\ 14-5 &=& \\ &=&9 \end{eqnarray*} \item $\sqrt{25}\cdot \dfrac{-3^{5}+\left( -1\right) ^{3}}{\left\vert \left( -5\right) ^{3}-\left\vert 6+\left( -5\right) ^{3}\right\vert \right\vert }=~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion -5% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $\newline Solution: \begin{eqnarray*} \sqrt{25}\cdot \dfrac{-3^{5}+\left( -1\right) ^{3}}{\left\vert \left( -5\right) ^{3}-\left\vert 6+\left( -5\right) ^{3}\right\vert \right\vert } &=&\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ exponents upstairs} \\ \sqrt{25}\cdot \dfrac{-243+\left( -1\right) ^{3}}{\left\vert \left( -5\right) ^{3}-\left\vert 6+\left( -5\right) ^{3}\right\vert \right\vert } &=& \\ \sqrt{25}\cdot \dfrac{-244}{\left\vert \left( -5\right) ^{3}-\left\vert 6+\left( -5\right) ^{3}\right\vert \right\vert } &=&\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ addition upstairs} \end{eqnarray*}% \pagebreak We now start working downstairs, starting in the innermost parentheses. \ The absolute value signs are also parentheses! \ We start with exponents.% \begin{eqnarray*} \sqrt{25}\cdot \dfrac{-244}{\left\vert \left( -5\right) ^{3}-\left\vert 6+\left( -5\right) ^{3}\right\vert \right\vert } &=& \\ \sqrt{25}\cdot \dfrac{-244}{\left\vert \left( -5\right) ^{3}-\left\vert 6+\left( -125\right) \right\vert \right\vert } &=&\text{ \ \ \ \ addition} \\ \sqrt{25}\cdot \dfrac{-244}{\left\vert \left( -5\right) ^{3}-\left\vert -119\right\vert \right\vert } &=&\text{ \ \ \ \ the absolute value of }-119% \text{ \ is }119 \\ \sqrt{25}\cdot \dfrac{-244}{\left\vert \left( -5\right) ^{3}-119\right\vert } &=&\text{ \ \ \ \ \ exponents within absolute value sign} \\ \sqrt{25}\cdot \dfrac{-244}{\left\vert -125-119\right\vert } &=&\text{ \ \ \ subtraction within absolute value sign} \\ \sqrt{25}\cdot \dfrac{-244}{\left\vert -244\right\vert } &=&\text{ \ \ \ \ \ the absolute value of }-244\text{ \ is }244 \\ \sqrt{25}\cdot \dfrac{-244}{244} &=&\text{ \ \ \ \ \ \ \ square root} \\ 5\cdot \dfrac{-244}{244} &=&\text{ \ \ \ \ \ \ division} \\ 5\cdot \left( -1\right) &=&-5 \end{eqnarray*} \item $\dfrac{3+\left( 6^{2}-3^{2}\right) +\left( 3^{3}-2^{3}\right) -9}{% 4^{2}+\left( 5-3\right) ^{2}}=~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 2$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $\newline $Solution: $\allowbreak $The division bar stretching over an expression is a case of an "invisible parentheses" around the top and the bottom expressions. \ We completely work out the top and the bottom, and finally divide. \ We start within the three parentheses:% \begin{eqnarray*} \dfrac{3+\left( 6^{2}-3^{2}\right) +\left( 3^{3}-2^{3}\right) -9}{% 4^{2}+\left( 5-3\right) ^{2}} &=&\text{ \ \ \ \ first parentheses on the top} \\ \dfrac{3+\left( 36-9\right) +\left( 27-8\right) -9}{4^{2}+\left( 5-3\right) ^{2}} &=&\dfrac{3+27+19-9}{4^{2}+\left( 5-3\right) ^{2}}=\dfrac{30+19-9}{% 4^{2}+\left( 5-3\right) ^{2}} \\ &=&\dfrac{49-9}{4^{2}+\left( 5-3\right) ^{2}}=\dfrac{40}{4^{2}+\left( 5-3\right) ^{2}} \\ &=&\dfrac{40}{4^{2}+2^{2}}=\dfrac{40}{16+4}=\dfrac{40}{20}=2 \end{eqnarray*}% \pagebreak \item $\sqrt{\sqrt{3^{6}}-2\left( 3^{2}+3^{3}\right) +2\left( 4\left( 17-2^{3}\right) -3^{2}\right) }=~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 3$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $\newline $Solution: The square root sign stretching across other operations is a case of an "invisible parentheses." We will work out everything else and then take the square root of the result. \ We start with the innermost parentheses.% \begin{eqnarray*} \sqrt{\sqrt{3^{6}}-2\left( 3^{2}+3^{3}\right) +2\left( 4\left( 17-2^{3}\right) -3^{2}\right) \allowbreak } &=&\text{ \ \ exponent in first parentheses from left} \\ \sqrt{\sqrt{3^{6}}-2\left( 9+3^{3}\right) +2\left( 4\left( 17-2^{3}\right) -3^{2}\right) \allowbreak } &=&\text{ \ \ exponent in first parentheses from left} \\ \sqrt{\sqrt{3^{6}}-2\left( 9+27\right) +2\left( 4\left( 17-2^{3}\right) -3^{2}\right) \allowbreak } &=&\text{ \ \ addition in first parentheses from left} \\ \sqrt{\sqrt{3^{6}}-2\cdot 36+2\left( 4\left( 17-2^{3}\right) -3^{2}\right) \allowbreak } &=&\text{ \ \ exponent in innermost parentheses} \\ \sqrt{\sqrt{3^{6}}-2\cdot 36+2\left( 4\left( 17-8\right) -3^{2}\right) \allowbreak } &=&\text{ \ \ subtraction in innermost parentheses} \\ \sqrt{\sqrt{3^{6}}-2\cdot 36+2\left( 4\cdot 9-3^{2}\right) \allowbreak } &=&% \text{ \ \ exponent in parentheses} \\ \sqrt{\sqrt{3^{6}}-2\cdot 36+2\left( 4\cdot 9-9\right) \allowbreak } &=&% \text{ \ \ multiplication in parentheses} \\ \sqrt{\sqrt{3^{6}}-2\cdot 36+2\left( 36-9\right) \allowbreak } &=&\text{ \ \ subtraction in parentheses} \\ \sqrt{\sqrt{3^{6}}-2\cdot 36+2\cdot 27\allowbreak } &=&\text{ \ \ exponent within square root sign} \\ \sqrt{\sqrt{729}-2\cdot 36+2\cdot 27\allowbreak } &=&\text{ \ \ square root} \\ \sqrt{27-2\cdot 36+2\cdot 27\allowbreak } &=&\text{ \ multiplications, left to right} \\ \sqrt{27-72+54} &=&\text{ \ \ subtraction} \\ \sqrt{-45+54\allowbreak } &=&\text{ \ \ addition} \\ \sqrt{9} &=&\text{ \ \ square root} \\ &=&3 \end{eqnarray*} \item $2\cdot 3^{2}-\left( 6-4+\left( 6\cdot 4-3\left( 2^{4}-12\right) \right) \right) =~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 4$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $\newline $Solution: $\ \allowbreak $We start with the innermost parentheses. \ Remember, every step we write down is a new, easier problem we have to solve now.% \begin{eqnarray*} 2\cdot 3^{2}-\left( 6-4+\left( 6\cdot 4-3\left( 2^{4}-12\right) \right) \right) &=&\text{ \ \ \ \ exponents in innermost parentheses} \\ 2\cdot 3^{2}-\left( 6-4+\left( 6\cdot 4-3\left( 16-12\right) \right) \right) &=&\text{ \ \ \ subtraction in innermost parentheses} \\ 2\cdot 3^{2}-\left( 6-4+\left( 6\cdot 4-3\left( 4\right) \right) \right) &=&% \text{ }\ \ \ \ \ \ \text{drop parentheses} \\ 2\cdot 3^{2}-\left( 6-4+\left( 6\cdot 4-3\cdot 4\right) \right) &=&\text{ \ \ \ \ \ multiplication in innermost parentheses, left to right} \\ 2\cdot 3^{2}-\left( 6-4+\left( 24-12\right) \right) &=&\text{ \ \ \ \ \ subtraction in innermost parentheses} \\ 2\cdot 3^{2}-\left( 6-4+12\right) &=&\text{ \ \ \ addition and subtraction in parentheses, left to right} \\ 2\cdot 3^{2}-\left( 2+12\right) &=& \\ 2\cdot 3^{2}-14 &=&\text{ \ \ \ exponent} \\ 2\cdot 9-14 &=&\text{ \ \ \ multiplication} \\ 18-14 &=&4 \end{eqnarray*}% \pagebreak \end{enumerate} \item Let $a=-4$, $\ b=2$, and $x=-3$. Evaluate each of the following expressions. \begin{enumerate} \item $a^{2}-b^{2}=~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 12$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $\newline $Solution: $\allowbreak $First we re-write the expression with one change, we write little pairs of parentheses instead of the letters.% \begin{equation*} a^{2}-b^{2}=\left( ~~\right) ^{2}-\left( ~~\right) ^{2} \end{equation*}% We now write the values inside the parentheses. \ From here on this is an order of operations problem. \begin{eqnarray*} a^{2}-b^{2} &=&\left( -4\right) ^{2}-\left( 2\right) ^{2}\text{ \ \ \ \ \ \ \ \ \ drop extra parentheses} \\ &=&\left( -4\right) ^{2}-2^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ exponents} \\ &=&16-4\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtraction} \\ &=&12 \end{eqnarray*} \item $\left( a-b\right) ^{2}=~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 36$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $\newline $Solution: $\allowbreak $First we re-write the expression with one change, we write little pairs of parentheses instead of the letters.% \begin{equation*} \left( a-b\right) ^{2}=\left( \left( ~~\right) -\left( ~~\right) \right) ^{2} \end{equation*}% We now write the values inside the parentheses. \ From here on this is an order of operations problem. \begin{eqnarray*} \left( a-b\right) ^{2} &=&\left( \left( -4\right) -\left( 2\right) \right) ^{2}\text{ \ \ \ \ \ \ \ drop extra parentheses} \\ &=&\left( -4-2\right) ^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ subtraction in parentheses} \\ &=&\left( -6\right) ^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ exponent} \\ &=&36 \end{eqnarray*}% This and the previous problem is here to remind you that $\left( a-b\right) ^{2}$ and $a^{2}-b^{2}$ are two different expressions. \item $a^{b}-2bx+x-\left\vert 2x\right\vert =~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 19$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $\newline $Solution: $\allowbreak $ First we re-write the expression with one modification only: we write little pairs of parentheses instead of the letters.% \begin{equation*} a^{b}-2bx+x-\left\vert 2x\right\vert =\left( ~~~\right) ^{\left( ~~\right) }-2\left( ~~\right) \left( ~~\right) +\left( ~~\right) -\left\vert 2\left( ~~\right) \right\vert \end{equation*}% We now write the values inside the parentheses. \ From here on this is an order of operations problem.\newline $a^{b}-2bx+x-\left\vert 2x\right\vert =$% \begin{eqnarray*} &=&\left( ~~\right) ^{\left( ~~\right) }-2\left( ~~\right) \left( ~~\right) +\left( ~~\right) -\left\vert 2\left( ~~\right) \right\vert \\ &=&\left( -4\right) ^{\left( 2\right) }-2\left( 2\right) \left( -3\right) +\left( -3\right) -\left\vert 2\left( -3\right) \right\vert \text{ \ \ \ \ \ \ drop extra parentheses} \\ &=&\left( -4\right) ^{2}-2\left( 2\right) \left( -3\right) +\left( -3\right) -\left\vert 2\left( -3\right) \right\vert \text{ \ \ \ \ \ \ \ \ multiplication within absolute value sign } \\ &=&\left( -4\right) ^{2}-2\left( 2\right) \left( -3\right) +\left( -3\right) -\left\vert -6\right\vert \text{ \ \ \ \ \ \ \ \ \ \ \ \ the absolute value of }-6\text{ \ is }6 \\ &=&\left( -4\right) ^{2}-2\left( 2\right) \left( -3\right) +\left( -3\right) -6\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ exponent} \end{eqnarray*}% \begin{eqnarray*} &=&16-2\left( 2\right) \left( -3\right) +\left( -3\right) -6\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ multiplication, left to right} \\ &=&16-4\left( -3\right) +\left( -3\right) -6\text{ \ \ \ } \\ &=&16-\left( -12\right) +\left( -3\right) -6\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ additions, subtractions, left to right} \\ &=&16+12+\left( -3\right) -6 \\ &=&28+\left( -3\right) -6 \\ &=&25-6 \\ &=&19 \end{eqnarray*} \item $\dfrac{-x^{2}+\left( x+2\right) ^{2}}{\left( x-1\right) }=~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 2$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $\newline $Solution: First we re-write the expression with only one modification: we write little pairs of parentheses instead of the letters.% \begin{equation*} \dfrac{-x^{2}+\left( x+2\right) ^{2}}{\left( x-1\right) }=\dfrac{-\left( ~~\right) ^{2}+\left( \left( ~~\right) +2\right) ^{2}}{\left( \left( ~~\right) -1\right) } \end{equation*}% We now write the values inside the parentheses. \ From here on this is an order of operations problem.% \begin{eqnarray*} \dfrac{-x^{2}+\left( x+2\right) ^{2}}{\left( x-1\right) } &=&\dfrac{-\left( -3\right) ^{2}+\left( \left( -3\right) +2\right) ^{2}}{\left( \left( -3\right) -1\right) }\text{ \ \ drop parentheses} \\ &=&\dfrac{-\left( -3\right) ^{2}+\left( -3+2\right) ^{2}}{\left( -3-1\right) }\text{ \ \ \ \ addition in\ parentheses upstairs} \\ &=&\dfrac{-\left( -3\right) ^{2}+\left( -1\right) ^{2}}{\left( -3-1\right) }% \text{ \ \ \ \ \ \ \ \ \ subtraction downstairs in parentheses} \\ &=&\dfrac{-\left( -3\right) ^{2}+\left( -1\right) ^{2}}{\left( -4\right) }% \text{ \ \ \ \ \ \ \ \ \ drop parentheses} \\ &=&\dfrac{-\left( -3\right) ^{2}+\left( -1\right) ^{2}}{-4}\text{ \ \ \ \ \ \ \ \ exponents upstairs} \\ &=&\dfrac{-9+1}{-4}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ addition% } \\ &=&\dfrac{-8}{-4}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ division} \\ &=&2 \end{eqnarray*} \item $\dfrac{x-1}{x+3}=~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion \func{undefined}$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $\newline $Solution: First we re-write the expression with only one modification: we write little pairs of parentheses instead of the letters.% \begin{equation*} \dfrac{x-1}{x+3}=\dfrac{\left( ~~\right) -1}{\left( ~~\right) +3} \end{equation*}% We write the values inside the parentheses and evaluate the expression.% \begin{equation*} \dfrac{x-1}{x+3}=\dfrac{\left( -3\right) -1}{\left( -3\right) +3}=\dfrac{-4}{% 0}=\func{undefined} \end{equation*}% \pagebreak \end{enumerate} \item Consider the equation $x^{2}-10x+x^{3}-4=4\left( x+5\right) $. \ In each case, determine whether the number given is a solution of the equation or not. \begin{enumerate} \item $x=0~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion -4\not=20~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion no% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: \ We simply evaluate both sides of the equation when $x=0$.% \begin{eqnarray*} \text{LHS} &=&\left( 0\right) ^{2}-10\left( 0\right) +\left( 0\right) ^{3}-4 \\ &=&0^{2}-10\cdot 0+0^{3}-4 \\ &=&0-10\cdot 0+0-4=0-0+0-4=-4 \end{eqnarray*}% \begin{equation*} \text{RHS}=4\left( \left( 0\right) +5\right) =4\left( 0+5\right) =4\cdot 5=20 \end{equation*}% Since $-4\not=20$, \ $x=0$ \ is not a solution of this equation. \item $x=-2~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 12=12~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion yes% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: \ We simply evaluate both sides of the equation when $x=-2$.% \begin{eqnarray*} \text{LHS} &=&\left( -2\right) ^{2}-10\left( -2\right) +\left( -2\right) ^{3}-4 \\ &=&4-10\left( -2\right) +\left( -8\right) -4 \\ &=&4-\left( -20\right) +\left( -8\right) -4 \\ &=&24+\left( -8\right) -4 \\ &=&16-4 \\ &=&12 \end{eqnarray*}% \begin{equation*} \text{RHS}=4\left( \left( -2\right) +5\right) =4\cdot 3=12 \end{equation*}% Since $12=12$, \ $x=-2$ \ is a solution of this equation. \item $x=-3~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 8=8~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion yes% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: \ We simply evaluate both sides of the equation when $x=-3$.% \begin{eqnarray*} \text{LHS} &=&\left( -3\right) ^{2}-10\left( -3\right) +\left( -3\right) ^{3}-4 \\ &=&9-10\left( -3\right) +\left( -27\right) -4 \\ &=&9-\left( -30\right) +\left( -27\right) -4 \\ &=&39+\left( -27\right) -4 \\ &=&12-4 \\ &=&8 \end{eqnarray*}% \begin{equation*} \text{RHS}=4\left( \left( -3\right) +5\right) =4\left( -3+5\right) =4\cdot 2=8 \end{equation*}% Since $8=8$, \ $x=-3$ \ is a solution of this equation.\pagebreak \end{enumerate} \end{enumerate} \begin{center} {\LARGE Practice Problems - Answers}\bigskip \end{center} \begin{enumerate} \item Simplify each of the following expressions. \begin{enumerate} \item $\left\vert \left\vert -20-3\left( -4\right) \right\vert -10\right\vert =~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 2$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \item $15-3\left( -2\right) =~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 21$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \item $\left( \left( \left( 1-2\right) ^{2}-2\right) ^{2}-2\right) ^{2}-2=~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion -1$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \item $\sqrt{-7^{2}+15\left( 2\right) ^{2}-18\div 3\div 3}=~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 3$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \item $\dfrac{-6^{2}\div 2\cdot 3+\left\vert 2^{2}\cdot 12-11\right\vert +2\left( -1\right) ^{3}-\left( -1\right) ^{2}}{\left( -2\right) ^{2}+\left( -2\right) ^{3}}=~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 5$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \item $\sqrt{36}-\sqrt{81}+2\sqrt{25}=~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 7$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \end{enumerate} \item Evaluate $\dfrac{8x+x^{2}-33}{x+11}$ \ \ ~if \ \ \ \ \ \ \ \ \ \begin{enumerate} \item $x=0~~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion -3$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \item $x=-11~~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion \func{undefined}$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \end{enumerate} \item Evaluate $\dfrac{x-2}{2-x}$ \ if \begin{enumerate} \item $x=0~~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion -1$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \item $x=2~~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion \func{undefined}$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \end{enumerate} \item Consider the equation $-x+x^{3}+6=3\left( -x^{2}+x+2\right) $. \ In each case, determine whether the number given is a solution of the equation or not. \begin{enumerate} \item $x=0~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 6=6~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion yes% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \item $x=1~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 6=6~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion yes% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \item $x=-1~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 6\not=0~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion no% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \item $x=2~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 12\not=0~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion no% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \item $x=-2~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 0\not=-12~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion no% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \item $x=3~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 30\not=-12~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion no% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \item $x=-3~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion -18\not=-30~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion no% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \item $x=4~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 66\not=-30~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion no% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \item $x=-4~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion -54=-54~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion yes% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \end{enumerate} \end{enumerate} \end{document}