% \renewcommand{\familydefault}{\sfdefault} % \renewcommand{\sfdefault}{phv} \documentclass[11pt]{report} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{amssymb} \usepackage{amsfonts} \usepackage{amsmath} \usepackage{multicol} \usepackage{hyperref} \usepackage{appendix} \usepackage{boxedminipage} \usepackage{color} \usepackage{fancyhdr} \usepackage{geometry} \usepackage{graphicx} \usepackage{helvet} \usepackage{times} \usepackage{fancybox} \usepackage{mathptmx} \usepackage{sectsty} \usepackage{makeidx} \usepackage[none]{hyphenat} \usepackage{setspace} \setcounter{MaxMatrixCols}{10} %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Version=5.00.0.2570} %TCIDATA{} %TCIDATA{Created=Thursday, December 29, 2016 04:53:04} %TCIDATA{LastRevised=Saturday, September 25, 2021 13:36:45} %TCIDATA{} %TCIDATA{} %TCIDATA{CSTFile=40 LaTeX Report.cst} \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} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.5in,right=0.5in,top=0.6in,bottom=0.2in} \pagestyle{fancy} \lhead{\color{blue} Lecture Notes} \chead{\large One- and Two-Step Equations} \rhead{\small page \thepage} \cfoot{} \lfoot{\footnotesize\copyright \; Hidegkuti, 2017} \rfoot{\footnotesize Last revised: July 21, 2018} \setlength{\headheight}{20pt} \renewcommand{\headrulewidth}{0.8pt} \setlength{\textwidth}{7.6in} \setlength{\textheight}{9.8in} \setlength{\parindent}{0in} \setlength{\parskip}{0.12in} \linespread{1.12} \begin{document} Equations are a fundamental concept and tool in mathematics. %TCIMACRO{% %\TeXButton{def start}{\begin{boxedminipage}{7.5in} %\begin{flushleft} %\setlength{\fboxrule}{15pt} %\setlength{\leftskip}{10pt} %\setlength{\rightskip}{10pt} %\setlength{\hangindent}{58pt} %\color{white} %. %\color{black}}}% %BeginExpansion \begin{boxedminipage}{7.5in} \begin{flushleft} \setlength{\fboxrule}{15pt} \setlength{\leftskip}{10pt} \setlength{\rightskip}{10pt} \setlength{\hangindent}{58pt} \color{white} . \color{black}% %EndExpansion \textbf{Definition}: \ $\ $An \textbf{equation }is a statement in which two expressions (algebraic or numeric) are connected with an equal sign.% %TCIMACRO{% %\TeXButton{def end}{\end{flushleft} %\end{boxedminipage}}}% %BeginExpansion \end{flushleft} \end{boxedminipage}% %EndExpansion For example, $3x^{2}-x=4x+28$ is an equation. \ So is $x^{2}+5y=-y^{2}+x+2$. %TCIMACRO{% %\TeXButton{def start}{\begin{boxedminipage}{7.5in} %\begin{flushleft} %\setlength{\fboxrule}{15pt} %\setlength{\leftskip}{10pt} %\setlength{\rightskip}{10pt} %\setlength{\hangindent}{58pt} %\color{white} %. %\color{black}}}% %BeginExpansion \begin{boxedminipage}{7.5in} \begin{flushleft} \setlength{\fboxrule}{15pt} \setlength{\leftskip}{10pt} \setlength{\rightskip}{10pt} \setlength{\hangindent}{58pt} \color{white} . \color{black}% %EndExpansion \textbf{Definition}: \ $\ $A \textbf{solution} of an equation is a number (or an ordered set of numbers) that, when substituted into the\ variable(s) in the equation, makes the statement of equality true. \ %TCIMACRO{% %\TeXButton{def end}{\end{flushleft} %\end{boxedminipage}}}% %BeginExpansion \end{flushleft} \end{boxedminipage}% %EndExpansion %TCIMACRO{% %\TeXButton{\setlength{\parskip}{0.05in}}{\setlength{\parskip}{0.05in}}}% %BeginExpansion \setlength{\parskip}{0.05in}% %EndExpansion \vspace{0.1in} %TCIMACRO{% %\TeXButton{par, hang}{\setlength{\parindent}{0pt} %\setlength{\hangindent}{61pt}}}% %BeginExpansion \setlength{\parindent}{0pt} \setlength{\hangindent}{61pt}% %EndExpansion \textbf{Example 1}. \ \ a) \ Verify that $-2$ is not a solution of the equation $3x^{2}-x=4x+28$.\vspace{0.05in}\newline b) \ Verify that $4$ is a solution of the equation $3x^{2}-x=4x+28$.\vspace{% 0.05in}\newline c) \ Verify that the pair of numbers $x=3$ and $y=-4$ is a solution of the equation $x^{2}+5y=-y^{2}+x+2$.\vspace{0.05in}\newline d) \ Verify that the pair of numbers $x=-4$ and $y=3$ is not a solution of the equation $x^{2}+5y=-y^{2}+x+2$.\vspace{0.1in} %TCIMACRO{% %\TeXButton{par, hang}{\setlength{\parindent}{0pt} %\setlength{\hangindent}{61pt}}}% %BeginExpansion \setlength{\parindent}{0pt} \setlength{\hangindent}{61pt}% %EndExpansion \textbf{Solution:} \ a) \ Consider the equation $3x^{2}-x=4x+28$ with $x=-2$% . We substitute $x=-2$ into both sides of the equation and evaluate the expressions.\vspace{0.05in} %TCIMACRO{% %\TeXButton{push in}{\begin{minipage}{68pt} %\color{white} %. %\color{black} %\end{minipage}}}% %BeginExpansion \begin{minipage}{68pt} \color{white} . \color{black} \end{minipage}% %EndExpansion %TCIMACRO{\TeXButton{minipage begin}{\begin{boxedminipage}{3.2in}}}% %BeginExpansion \begin{boxedminipage}{3.2in}% %EndExpansion If $x=-2$, the left-hand side of the equation is \ \vspace{0.05in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ LHS $=3x^{2}-x$\vspace{0.05in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\ \ \ \ \ \ \ \ =3\left( -2\right) ^{2}-\left( -2\right) \vspace{0.05in}$\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ $\ \ \ \ \ \ \ \ \ \ \ \ \ =3\cdot 4+2=12+2=14\vspace{% 0.05in}$ %TCIMACRO{\TeXButton{minipage end}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion %TCIMACRO{\TeXButton{minipage begin}{\begin{boxedminipage}{3.2in}}}% %BeginExpansion \begin{boxedminipage}{3.2in}% %EndExpansion If $x=-2$, the right-hand side of the equation is \vspace{0.05in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ RHS$\,=4x+28\vspace{0.05in}$\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\ \ \ \ \ \ \ \ \,=4\left( -2\right) +28% \vspace{0.05in}\newline $% %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\ \ \ \ \ \ \ \ \,=-8+28=20\vspace{0.05in}$% %TCIMACRO{\TeXButton{minipage end}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion %TCIMACRO{% %\TeXButton{par, hang}{\setlength{\parindent}{61pt} %\setlength{\hangindent}{61pt}}}% %BeginExpansion \setlength{\parindent}{61pt} \setlength{\hangindent}{61pt}% %EndExpansion Since the two sides are not equal, $14\not=20$, the number $-2$ is not a solution of this equation.\vspace{0.1in} %TCIMACRO{% %\TeXButton{par, hang}{\setlength{\parindent}{48pt} %\setlength{\hangindent}{62pt}}}% %BeginExpansion \setlength{\parindent}{48pt} \setlength{\hangindent}{62pt}% %EndExpansion b) \ Consider the equation $3x^{2}-x=4x+28$ with $x=4$. \ We evaluate both sides of the equation after substituting $4$ into $x$.\vspace{0.05in} %TCIMACRO{% %\TeXButton{par, hang}{\setlength{\parindent}{0pt} %\setlength{\hangindent}{0pt}}}% %BeginExpansion \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt}% %EndExpansion %TCIMACRO{% %\TeXButton{push in}{\begin{minipage}{68pt} %\color{white} %. %\color{black} %\end{minipage}}}% %BeginExpansion \begin{minipage}{68pt} \color{white} . \color{black} \end{minipage}% %EndExpansion %TCIMACRO{\TeXButton{minipage begin}{\begin{boxedminipage}{3.2in}}}% %BeginExpansion \begin{boxedminipage}{3.2in}% %EndExpansion If $x=4,$ $\vspace{0.05in}$the left-hand side of the equation is \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ LHS $=3x^{2}-x\vspace{0.05in}$\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\ \ \ \ \ \ \ \ =3\cdot 4^{2}-4\vspace{0.05in} $\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\ \ \ \ \ \ \ \ =3\cdot 16-4=48-4=44\vspace{% 0.05in}$ %TCIMACRO{\TeXButton{minipage end}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion %TCIMACRO{\TeXButton{minipage begin}{\begin{boxedminipage}{3.2in}}}% %BeginExpansion \begin{boxedminipage}{3.2in}% %EndExpansion If $x=4$, the right-hand side of the equation is \vspace{0.05in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ RHS $=4x+28\vspace{0.05in}$\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\ \ \ \ \ \ \ =4\cdot 4+28\vspace{0.05in}$% \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\ \ \ \ \ \ \ =16+28=44\vspace{0.05in}$% %TCIMACRO{\TeXButton{minipage end}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion %TCIMACRO{% %\TeXButton{par, hang}{\setlength{\parindent}{61pt} %\setlength{\hangindent}{61pt}}}% %BeginExpansion \setlength{\parindent}{61pt} \setlength{\hangindent}{61pt}% %EndExpansion Since the two sides are equal, $x=4$ is a solution of this equation.\vspace{% 0.1in} %TCIMACRO{% %\TeXButton{par, hang}{\setlength{\parindent}{48pt} %\setlength{\hangindent}{62pt}}}% %BeginExpansion \setlength{\parindent}{48pt} \setlength{\hangindent}{62pt}% %EndExpansion c) \ Consider the equation $x^{2}+5y=-y^{2}+x+2$ with $x=3$ and $y=-4$. \ This is an equation in two variables, but the idea is the same. \ We substitute $x=3$ and $y=-4$ into both sides of the equation, and evaluate them to see if they are equal to each other with the given values for $x$ and $y$.\vspace{0.05in} %TCIMACRO{% %\TeXButton{par, hang}{\setlength{\parindent}{0pt} %\setlength{\hangindent}{0pt}}}% %BeginExpansion \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt}% %EndExpansion %TCIMACRO{% %\TeXButton{push in}{\begin{minipage}{68pt} %\color{white} %. %\color{black} %\end{minipage}}}% %BeginExpansion \begin{minipage}{68pt} \color{white} . \color{black} \end{minipage}% %EndExpansion %TCIMACRO{\TeXButton{minipage begin}{\begin{boxedminipage}{3.2in}}}% %BeginExpansion \begin{boxedminipage}{3.2in}% %EndExpansion If $x=3$ and $y=-4$, the left-hand side is$\vspace{0.05in}$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ LHS $=x^{2}+5y\vspace{0.05in}$\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\,=3^{2}+5\left( -4\right) \vspace{0.05in}$\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\,\ \ =9-20=-11\vspace{0.05in}$ %TCIMACRO{\TeXButton{minipage end}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion %TCIMACRO{\TeXButton{minipage begin}{\begin{boxedminipage}{3.2in}}}% %BeginExpansion \begin{boxedminipage}{3.2in}% %EndExpansion If $x=3$ and $y=-4$,$\vspace{0.05in}$\ the right-hand side is \ \ \ \ \ \ \ \ \ \ RHS $=-y^{2}+x+2\vspace{0.05in}$\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $=-\left( -4\right) ^{2}+3+2\vspace{0.05in} $\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $=-16+3+2=-11\vspace{0.05in}$% %TCIMACRO{\TeXButton{minipage end}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion $\vspace{0.05in}\vspace{0.05in}$ %TCIMACRO{% %\TeXButton{par, hang}{\setlength{\parindent}{61pt} %\setlength{\hangindent}{61pt}}}% %BeginExpansion \setlength{\parindent}{61pt} \setlength{\hangindent}{61pt}% %EndExpansion Since the two sides are equal, $x=3$ and $y=-4$ is a solution of this equation. \pagebreak %TCIMACRO{% %\TeXButton{par, hang}{\setlength{\parindent}{48pt} %\setlength{\hangindent}{62pt}}}% %BeginExpansion \setlength{\parindent}{48pt} \setlength{\hangindent}{62pt}% %EndExpansion d) \ Consider the equation $x^{2}+5y=-y^{2}+x+2$ with $x=-4$ and $y=3$.% \vspace{0.05in} %TCIMACRO{% %\TeXButton{par, hang}{\setlength{\parindent}{0pt} %\setlength{\hangindent}{0pt}}}% %BeginExpansion \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt}% %EndExpansion %TCIMACRO{% %\TeXButton{push in}{\begin{minipage}{68pt} %\color{white} %. %\color{black} %\end{minipage}}}% %BeginExpansion \begin{minipage}{68pt} \color{white} . \color{black} \end{minipage}% %EndExpansion %TCIMACRO{\TeXButton{minipage begin}{\begin{boxedminipage}{3.2in}}}% %BeginExpansion \begin{boxedminipage}{3.2in}% %EndExpansion If $x=-4$ and $y=3$, the left-hand side is$\vspace{0.05in}$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ LHS $=x^{2}+5y\vspace{0.05in}\newline $% %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,=\left( -4\right) ^{2}+5\cdot 3\vspace{0.05in}\newline $% %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,=16+15=41$ $\vspace{0.05in}$ %TCIMACRO{\TeXButton{minipage end}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion %TCIMACRO{\TeXButton{minipage begin}{\begin{boxedminipage}{3.2in}}}% %BeginExpansion \begin{boxedminipage}{3.2in}% %EndExpansion If $x=-4$ and $y=3$, $\vspace{0.05in}$ the right-hand side is \ \ \ \ \ \ \ \ \ \ RHS $=-y^{2}+x+2\vspace{0.05in}\newline $% %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,=-3^{2}+\left( -4\right) +2% \vspace{0.05in}\newline $% %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,=-9-4+2=-11$ $\vspace{0.05in}$% %TCIMACRO{\TeXButton{minipage end}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion %TCIMACRO{% %\TeXButton{par, hang}{\setlength{\parindent}{61pt} %\setlength{\hangindent}{61pt}}}% %BeginExpansion \setlength{\parindent}{61pt} \setlength{\hangindent}{61pt}% %EndExpansion Since $41\not=-11$, \ the two sides are not equal, $x=-4$ and $y=3$ is not a solution of this equation.\vspace{0.1in} %TCIMACRO{% %\TeXButton{par, hang}{\setlength{\parindent}{0pt} %\setlength{\hangindent}{0pt}}}% %BeginExpansion \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt}% %EndExpansion If the equation is in more than one variable, like in parts c) and d) before, it is important to identify which number is to be substituted into which variable. \ After all, $x=3$ and $y=-4$ \ (or, as an \textit{ordered pair}, $\ \left( 3,-4\right) $) is a solution of the equation $% x^{2}+5y=-y^{2}+x+2$, but $x=-4$ and $y=3$, \ (or, as an \textit{ordered pair% }, \thinspace $\left( -4,3\right) $) is not.\vspace{0.12in} %TCIMACRO{% %\TeXButton{def start}{\begin{boxedminipage}{7.5in} %\begin{flushleft} %\setlength{\fboxrule}{15pt} %\setlength{\leftskip}{10pt} %\setlength{\rightskip}{10pt} %\setlength{\hangindent}{58pt} %\color{white} %. %\color{black}}}% %BeginExpansion \begin{boxedminipage}{7.5in} \begin{flushleft} \setlength{\fboxrule}{15pt} \setlength{\leftskip}{10pt} \setlength{\rightskip}{10pt} \setlength{\hangindent}{58pt} \color{white} . \color{black}% %EndExpansion \textbf{Definition}: \ $\ $To \textbf{solve an equation }is to find \textit{% all} solutions of it. \ The set of all solutions is also called the solution set.% %TCIMACRO{% %\TeXButton{def end}{\end{flushleft} %\end{boxedminipage}}}% %BeginExpansion \end{flushleft} \end{boxedminipage}% %EndExpansion \vspace{0.1in} Caution! \ Finding one solution for an equation is not the same as solving it. \ For example, the number $2$ is a solution of the equation $x^{3}=4x$. \ However, $-2$ is also a solution of this equation. \ We will have to deploy systematic methods to find \textit{all} solutions. \ The method we will use to solve an equation usually depends on the type of the equation. \ We will start with the simplest equations, linear equations.\vspace{0.1in} Linear equations are a fundamental concept and tool in mathematics. \ To solve a linear equation, we isolate the unknown by applying the same operation(s) to both sides. $\vspace{0.05in}$ \textbf{Example 2}. \ Solve each of the given equations. \ Make sure to check your solutions. \qquad \qquad \qquad\ \ \ \ a) \ $x-8=10$ \ \ \ \ \ \ \ \ \ \ b) \ $3y=-12$ \ \ \ \ \ \ \ \ \ \ \ c) \ $\dfrac{x}{-3}=8$ \ \ \ \ \ \ \ \ \ \ \ \ d) \ $% m+10=-5$ Equations like these are called \textbf{one-step equations} because they can be solved in only one step. \ We need to\ isolate the unknown on one side. \ In order to do that, we perform the inverse operation. \ (The inverse operation of additon is subtraction and vica versa. \ The inverse operation of multiplication is division and vica versa.) %TCIMACRO{% %\TeXButton{par, hang}{\setlength{\parindent}{0pt} %\setlength{\hangindent}{61pt}}}% %BeginExpansion \setlength{\parindent}{0pt} \setlength{\hangindent}{61pt}% %EndExpansion \textbf{Solution:} \ a) \ In order to isolate the unknown, we add $8$ to both sides.\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $x-8=10$ \ \ \ \ \ \ \ \ add $8$\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $\ \ ~~\ \ x=18$\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ So the only solution of this equation is $18.$ \ We can also say that the solution set is $\left\{ 18\right\} $. \ We should check;\ \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ if $x=18,$ the left-hand side is\vspace{0.03in}% \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad LHS $=x-8=18-8=10=$% RHS\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ So our solution, $\,$\fbox{$x=18$} is correct.\vspace{% 0.03in}\vspace{0.03in} %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \qquad \qquad b) \ In order to isolate the unknown, we divide both sides by $% 3$.\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $3y=-12$ \ \ \ \ \ \ \ \ divide by $3$\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $~~y=-4$% \vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ So the only solution of this equation is $-4.$ \ We check; if $y=-4$, then\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $~~$\ LHS $% =3y=3\left( -4\right) =-12=$ RHS\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ So our solution, $\,$\fbox{$y=-4$} is correct.\vspace{% 0.03in} %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \qquad \qquad c) \ In order to isolate the unknown, we multiply both sides by $-3$.% \begin{eqnarray*} \dfrac{x}{-3} &=&8\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ multiply by }-3 \\ x &=&-24 \end{eqnarray*}% %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ So the only solution of this equation is $-24$. \ We check; if $x=-24$, then\ \begin{equation*} \text{LHS}=\dfrac{-24}{-3}=8=\text{RHS} \end{equation*}% %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ So our solution, $\,$\fbox{$x=-24$} is correct. %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \qquad \qquad d) \ In order to isolate the unknown, we we subtract $10$ from both sides.\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $m+10=-5$ \ \ \ \ \ \ \ \ subtract $10$\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $~~\ \ \ \ \ \,\,m=-15$\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ So the only solution of this equation is $-15.$ \ We check; if $m=-15$, then\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $~~$\ LHS $% =m+10=-15+10=-5=$ RHS\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ So our solution, $\,$\fbox{$m=-15$} is correct.% \vspace{0.03in}\vspace{0.1in} %TCIMACRO{% %\TeXButton{par, hang}{\setlength{\parindent}{0pt} %\setlength{\hangindent}{27pt}}}% %BeginExpansion \setlength{\parindent}{0pt} \setlength{\hangindent}{27pt}% %EndExpansion Note: If the reader is interested in applications of one-step equations, basic percent problems and basic motion problems can be easily handled by setting up and solving one-step equations. \ \vspace{0.1in}\vspace{0.1in} %TCIMACRO{\TeXButton{begin minipage 1 in}{\begin{minipage}{1in}}}% %BeginExpansion \begin{minipage}{1in}% %EndExpansion %TCIMACRO{% %\FRAME{itbpF}{0.7048in}{0.7117in}{0in}{}{}{question.bmp}{% %\special{language "Scientific Word";type "GRAPHIC";display "USEDEF";valid_file "F";width 0.7048in;height 0.7117in;depth 0in;original-width 0.4869in;original-height 0.6797in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'question.bmp';file-properties "XNPEU";}}}% %BeginExpansion \includegraphics[ natheight=0.6797in, natwidth=0.4869in, height=0.7117in, width=0.7048in] {//Mac/Dropbox/lnotes/1Prealgebra/equations/twostepequations/graphics/question__1.pdf}% %EndExpansion %TCIMACRO{\TeXButton{end minipage}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{\TeXButton{begin minipage 6.6 in}{\begin{minipage}{6.6in}}}% %BeginExpansion \begin{minipage}{6.6in}% %EndExpansion \textbf{Discussion}: \ Solve each of the following equations. \ How are these unusual? a) \ \ $5x=5$ \ \ \ \ \ \ \ b) \ $5x=0$\ \ \ \ \ \ \ \ c) \ $x-4=-4$ \ \ \ \ \ \ \ d) \ $\dfrac{x}{3}=0$ %TCIMACRO{\TeXButton{end minipage}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{% %\TeXButton{par, hang}{\setlength{\parindent}{0pt} %\setlength{\hangindent}{61pt}}}% %BeginExpansion \setlength{\parindent}{0pt} \setlength{\hangindent}{61pt}% %EndExpansion \textbf{Example 3}. \ Solve each of the given equations. \ Make sure to check your solutions. %TCIMACRO{% %\TeXButton{par, hang}{\setlength{\parindent}{61pt} %\setlength{\hangindent}{61pt}}}% %BeginExpansion \setlength{\parindent}{61pt} \setlength{\hangindent}{61pt}% %EndExpansion a) \ $10=3x-11$ \ \ \ \ \ \ \ \ \ \ b) \ $3x+8=-7$ \ \ \ \ \ \ \ \ \ \ \ c) \ $\dfrac{t-7}{2}=-8$ \ \ \ \ \ \ \ \ \ \ \ \ d) \ $\dfrac{x}{-3}+4=15$% \vspace{0.12in} %TCIMACRO{% %\TeXButton{par, hang}{\setlength{\parindent}{0pt} %\setlength{\hangindent}{0pt}}}% %BeginExpansion \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt}% %EndExpansion Equations like these are called \textbf{two-step equations}. \ We need to\ isolate the unknown on one side. \ In order to do that, we perform the inverse operations, in the reverse order. \textbf{Solution:} \ a) \ The equation \ $10=3x-11$ looks unusual in the sense that two-step equations often contain the unknown on the \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ left-hand side. \ We are always allowed to swap two sides of an equation. \ If $A=B$, then clearly, also $B=A.$ \ \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ We will do this first. \ Notice that this is an optional step.\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $10=3x-11$ \ \ \ \ \ \ \ \ swap the two sides\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $3x-11=10$\vspace{% 0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ We now look at the side that contains $x$ and ask: \textit{What happened to the unknown}? \ The answer is: \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ \textit{Multiplication by }$3$\textit{\ and then subtraction of }$11$. \ We need to apply the inverse operations, in a reverse order. \ \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ In this case, this means that we will add $11$ to both sides and then divide both sides by $3$.\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $\,\,3x-11=10$ \ \ \ \ \ \ \ \ \ \ add $11$\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\ \ \ \ $\ \ \ \ \ \,3x=21$ \ \ \ \ \ \ \ \ \ divide by $3$\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\ \ \ \ $\ \ \ \ \ \ \ \,x=7$\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ So the only solution of this equation is $7$. \ We check: if $x=7$, then\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad\ LHS $=3x-11=3\cdot 7-11=21-11=10=$ RHS\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ So our solution, $\,$\fbox{$x=7$} is correct.\vspace{% 0.12in} \pagebreak %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad b) \ As we look at the equation $3x+8=-7$ and ask: \textit{% What happened to the unknown}? \ The answer is: \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ \textit{Multiplication by }$3$\textit{\ and then addition of }$8$. \ We need to apply the inverse operations, in a reverse order. \ \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ In this case, this means that we will subtract $8$ from both sides and then divide both sides by $3$.\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\ $\ 3x+8=-7$ \ \ \ \ \ \ \ \ \ \ subtract $8$\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\ \ \ \ \ \ $\ \ \,3x=-15$ \ \ \ \ \ \ \ \ \ divide by $3$ \ \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\ \ \ \ $\ \ \ \ \ \ \ x=-5$\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ So the only solution of this equation is $-5$. \ We check: if $x=-5$, then\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad\ LHS $=3x+8=3\left( -5\right) +8=-15+8=-7=$ RHS\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ So our solution, $\,$\fbox{$x=-5$} is correct.\vspace{% 0.12in}\vspace{0.03in} %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \qquad \qquad c) \ What happened to the unknown? \ On the left-hand side, there was a subtraction of $7$ and then a division by $2$.\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ \ To reverse that, we will multiply both sides by $2$ and \ then add $7$ to both sides.\vspace{0.03in}\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $\dfrac{t-7}{2% }=-8$ \ \ \ \ \ \ \ \ \ \ \ multiply by $2$\vspace{0.03in}\vspace{0.03in}% \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\ \ $t-7=-16$ \ \ \ \ \ \ \ \ \ add \ $7$\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\ \ \ \ $\ \ \ \ t=-9$\vspace{0.03in}\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ So the only solution of this equation is $-9$. \ We check: if $t=-9$, then\vspace{0.03in}\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad\ LHS $=\dfrac{t-7}{2}=\dfrac{-9-7}{% 2}=\dfrac{-16}{2}=-8=$ RHS\vspace{0.03in}\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ So our solution, $\,$\fbox{$t=-9$} is correct.\vspace{% 0.1in} %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \qquad \qquad d) \ What happened to the unknown? \ On the left-hand side, there was a division by $-3$ and then an addition of $4$.\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ \ To reverse that, we will subtract $4$ from both sides by and then multiply both sides by $-3$.\vspace{0.03in}\vspace{0.03in}% \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $\dfrac{x}{-3}% +4=15$ \ \ \ \ \ \ \ \ \ \ \ subtract $4$\vspace{0.03in}\vspace{0.03in}% \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\ \ $\ \ \ \ \dfrac{x}{-3}=11$ \ \ \ \ \ \ \ \ \ \ \ \ multiply by \ $-3$\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\ \ \ \ $\ \ \ \ x=-33$\vspace{0.03in}\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ So the only solution of this equation is $-33$. \ We check: if $x=-33$, then\vspace{0.03in}\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \qquad \qquad \qquad \qquad \qquad \qquad\ LHS $=\dfrac{x}{-3}+4=\dfrac{-33}{% -3}+4=11+4=15=$ RHS\vspace{0.03in}\vspace{0.03in}\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ So our solution, $\,$\fbox{$x=-33$} is correct.% \vspace{0.03in} \pagebreak %TCIMACRO{% %\FRAME{itbpF}{0.8553in}{0.6962in}{0.2811in}{}{}{sample.jpg}{% %\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 0.8553in;height 0.6962in;depth 0.2811in;original-width 8.4267in;original-height 6.8441in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'sample.jpg';file-properties "XNPEU";}}}% %BeginExpansion \raisebox{-0.2811000000in}{ \includegraphics[ natheight=6.8441in, natwidth=8.4267in, height=0.6962in, width=0.8553in] {//Mac/Dropbox/lnotes/1Prealgebra/equations/twostepequations/graphics/sample__2.jpg}% }% %EndExpansion \vspace{0.1in}\ \ {\LARGE Sample Problems} Solve each of the following equations. Make sure to check your solutions. %TCIMACRO{\TeXButton{3 col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion \begin{enumerate} \item $2x-5=17\medskip $ \item $\dfrac{a-10}{5}=-3\medskip $ \item $\dfrac{t}{4}-10=-4\medskip $ \item $\dfrac{t-5}{12}=4\medskip $ \item $2x-7=-3\medskip $ \item $\dfrac{x+8}{3}=-2\medskip $ \item $\dfrac{x}{3}+8=-2\medskip $ \item $-2x+3=3\medskip $ \item $3\left( x+7\right) =36\medskip $ \item $3x-10=-10\medskip $ \item $-4x+6=-18\medskip $ \end{enumerate} %TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion %TCIMACRO{% %\FRAME{itbpF}{0.8951in}{0.9997in}{0.3528in}{}{}{work.jpg}{% %\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 0.8951in;height 0.9997in;depth 0.3528in;original-width 4.6977in;original-height 5.2607in;cropleft "0";croptop "1.003173";cropright "1";cropbottom "0.003173";filename 'work.jpg';file-properties "XNPEU";}}}% %BeginExpansion \raisebox{-0.3528000000in}{ \includegraphics[ natheight=5.2607in, natwidth=4.6977in, height=0.9997in, width=0.8951in] {//Mac/Dropbox/lnotes/1Prealgebra/equations/twostepequations/graphics/work__3.jpg}% }% %EndExpansion \ \ {\LARGE Practice Problems }\bigskip Solve each of the following equations. Make sure to check your solutions.\bigskip %TCIMACRO{\TeXButton{3 col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion \begin{enumerate} \item $2x-3=-11\medskip $ \item $-2x-3=7\medskip $ \item $5x-3=17\medskip $ \item $\dfrac{x-3}{7}=-2\medskip $ \item $\dfrac{x}{7}-3=-1\medskip $ \item $-4x-3=13\medskip $ \item $\dfrac{a+1}{4}=-9\medskip $ \item $5x-6=-6\medskip $ \item $\dfrac{x}{7}-1=-3\medskip $ \item $-x+5=-7\medskip $ \item $\dfrac{2x-1}{7}=-3\medskip $ \item $5\left( x-2\right) =-20\medskip $ \item $\dfrac{x-8}{7}=-2\medskip $ \item $3b+13=-5\medskip $ \item $\dfrac{x}{3}-7=7$ \end{enumerate} %TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \bigskip \pagebreak {\LARGE %TCIMACRO{% %\FRAME{itbpF}{0.9055in}{0.9055in}{0.1609in}{}{}{answers.jpg}{% %\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 0.9055in;height 0.9055in;depth 0.1609in;original-width 8.6455in;original-height 8.6455in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'answers.jpg';file-properties "XNPEU";}}}% %BeginExpansion \raisebox{-0.1609000000in}{ \includegraphics[ natheight=8.6455in, natwidth=8.6455in, height=0.9055in, width=0.9055in] {//Mac/Dropbox/lnotes/1Prealgebra/equations/twostepequations/graphics/answers__4.jpg}% }% %EndExpansion \ \ \ Answers} {\large Discussion \ } a) \ $1$ \ \ \ \ b) \ $0$ \ \ \ \ c) \ $0$ \ \ \ d) \ $0$ One thing that is unusual in this problem is the idea of cancellation. \ Cancellation results in $0$ or $1$, depending on the operation.\vspace{0.1in} {\large Sample Problems} 1. \ $11$ \ \ \ \ \ \ 2. \ $-5$ \ \ \ \ 3. \ $24$\ \ \ \ 4. \ $53$\ \ \ \ 5. \ $2$\ \ \ \ 6. \ $-14$\ \ \ \ \ 7. \ $-30$ \ \ \ \ 8. $0$\ \ \ \ \ \ \ 9. $5 $\ \ \ \ \ \ 10. \ $0$ \ \ \ \ 11. \ $-4$\vspace{0.1in} {\large Practice Problems} 1. \ $-4$ \ \ \ \ \ 2. \ $-5$ \ \ \ \ \ 3. \ $4$ \ \ \ \ \ 4. \ $-11$ \ \ \ \ \ 5. \ $14$ \ \ \ \ 6. \ $-4$ \ \ \ \ \ 7. \ $-37$ \ \ \ \ 8. \ $0$ \ \ \ \ \ \ 9. \ $-14$ \ \ \ \ 10. \ $12$\vspace{0.1in} 11. \ $-10$ \ \ \ 12. \ $-2$\ \ \ \ \ \ \ \ 13. \ $-6$ \ \ \ \ 14. \ $-6$ \ \ \ 15. \ $42$\vspace{0.1in}\vspace{0.15in} \begin{center} %TCIMACRO{% %\FRAME{itbpF}{1.1096in}{0.6512in}{0.1807in}{}{}{pencil.bmp}{% %\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.1096in;height 0.6512in;depth 0.1807in;original-width 1.9735in;original-height 1.1467in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'pencil.bmp';file-properties "XNPEU";}}}% %BeginExpansion \raisebox{-0.1807000000in}{ \includegraphics[ natheight=1.1467in, natwidth=1.9735in, height=0.6512in, width=1.1096in] {//Mac/Dropbox/lnotes/1Prealgebra/equations/twostepequations/graphics/pencil__5.pdf}% }% %EndExpansion {\LARGE Sample Problems - Solutions}\bigskip \end{center} Solve each of the following equations. \ Make sure to check your solutions. \begin{enumerate} \item $2x-5=17$\vspace{0.08in}\newline Solution:% \begin{eqnarray*} 2x-5 &=&17\text{ \ \ \ \ \ \ \ add }5\text{ to both sides} \\ 2x &=&22\text{ \ \ \ \ \ \ \ divide by }2 \\ x &=&11 \end{eqnarray*}% We check: if $x=11,$ then% \begin{equation*} \text{RHS}=2\left( 11\right) -5=22-5=17=\text{LHS} \end{equation*}% Thus our solution, \fbox{$x=11$} \ is correct. \item $\dfrac{a-10}{5}=-3$\vspace{0.1in}\newline Solution: \begin{eqnarray*} \dfrac{a-10}{5} &=&-3\text{ \ \ \ \ \ \ \ \ \ \ multiply both sides by }5 \\ a-10 &=&-15\text{ \ \ \ \ \ \ \ \ \ add \ }10\text{ \ to both sides} \\ a &=&-5 \end{eqnarray*}% We check: if $a=-5$, then% \begin{equation*} \text{LHS}=\dfrac{-5-10}{5}=\dfrac{-15}{5}=-3=\text{RHS} \end{equation*}% Thus our solution, \fbox{$a=-5$} is correct. \item $\dfrac{t}{4}-10=-4$\vspace{0.1in}\newline Solution: \begin{eqnarray*} \dfrac{t}{4}-10 &=&-4\text{ \ \ \ \ \ \ \ \ \ add \ }10\text{ \ to both sides% } \\ \dfrac{t}{4} &=&6\text{ \ \ \ \ \ \ \ \ \ \ \ multiply both sides by }4 \\ t &=&24 \end{eqnarray*}% We check: \ if \ $t=24$, \ then% \begin{equation*} \text{RHS}=\dfrac{t}{4}-10=\dfrac{24}{4}-10=6-10=-4=\text{LHS} \end{equation*}% Thus our solution, \fbox{$t=24$} \ is correct. \item $\dfrac{t-5}{12}=4$\vspace{0.1in}\newline Solution: \begin{eqnarray*} \dfrac{t-5}{12} &=&4\text{ \ \ \ \ \ \ \ multiply both sides by }12 \\ t-5 &=&48\text{ \ \ \ \ \ add }5\text{ to both sides} \\ t &=&53 \end{eqnarray*}% \vspace{0.1in}We check: if $t=53$, then \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ LHS $=\dfrac{53-5}{12}=\dfrac{48}{12}=4=$ RHS. Thus our solution, \fbox{$t=53$} \ is correct. \item $2x-7=-3$\vspace{0.08in}\newline Solution: \ We apply all operations to both sides.% \begin{eqnarray*} 2x-7 &=&-3\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ add \ }7 \\ 2x &=&4\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }2 \\ x &=&2 \end{eqnarray*}% We check: if $x=2,$ then% \begin{equation*} \text{LHS}=2\left( 2\right) -7=4-7=-3=\text{RHS} \end{equation*}% Thus our solution, \fbox{$x=2$} is correct. \item $\dfrac{x+8}{3}=-2$\vspace{0.1in}\newline Solution: \ We apply all operations to both sides.% \begin{eqnarray*} \dfrac{x+8}{3} &=&-2\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ multiply by \ }3 \\ x+8 &=&-6\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract \ }8 \\ x &=&-14 \end{eqnarray*}% We check: \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ LHS$=\dfrac{-14+8}{% 3}=\dfrac{-6}{3}=-2=$RHS\vspace{0.05in}\newline Thus our solution, \fbox{$x=-14$} is correct. \item $\dfrac{x}{3}+8=-2$\vspace{0.1in}\newline Solution: \ We apply all operations to both sides.% \begin{eqnarray*} \dfrac{x}{3}+8 &=&-2\text{ \ \ \ \ \ \ \ \ subtract \ }8\text{\ } \\ \dfrac{x}{3} &=&-10\text{ \ \ \ \ \ \ \ \ \ multiply by \ }3 \\ x &=&-30 \end{eqnarray*}% We check:% \begin{equation*} \text{LHS}=\dfrac{-30}{3}+8=-10+8=-2=\text{RHS} \end{equation*}% Thus our solution, \fbox{$x=-30$} is correct. \item $-2x+3=3$\vspace{0.08in}\newline Solution: \ We apply all operations to both sides.% \begin{eqnarray*} -2x+3 &=&3\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract \ }3\text{\ } \\ -2x &=&0\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by \ }-2 \\ x &=&0 \end{eqnarray*}% We check: if $x=0$, then \begin{equation*} \text{LHS}=-2\cdot 0+3=0+3=3=\text{RHS} \end{equation*}% Thus our solution, \fbox{$x=0$} is correct. \item $3\left( x+7\right) =36$\vspace{0.08in}\newline Solution: \ We apply all operation to both sides, \begin{eqnarray*} 3\left( x+7\right) &=&36\text{ \ \ \ \ \ \ \ \ \ \ \ \ divide by }3 \\ x+7 &=&12\text{ \ \ \ \ \ \ \ \ \ \ \ \ subtract }7 \\ x &=&5 \end{eqnarray*}% We check: if $x=5$, then \begin{equation*} \text{LHS}=3\left( 5+7\right) =3\cdot 12=36=\text{RHS} \end{equation*}% Thus our solution, \fbox{$x=5$} is correct. \item $3x-10=-10$\vspace{0.1in}\newline Solution:% \begin{eqnarray*} 3x-10 &=&-10\text{ \ \ \ \ \ \ \ \ \ add \ }10\text{ \ to both sides} \\ 3x &=&0\text{ \ \ \ \ \ \ \ \ \ \ \ \ divide by }3 \\ x &=&0 \end{eqnarray*}% We check: if $x=0$, then% \begin{equation*} \text{LHS}=3\cdot 0-10=0-10=-10=\text{RHS} \end{equation*}% Thus our solution, \fbox{$x=0$}\ is correct. \item $-4x+6=-18$\vspace{0.1in}\newline Solution: \begin{eqnarray*} -4x+6 &=&-18\text{ \ \ \ \ \ \ subtract }6 \\ -4x &=&-24\text{ \ \ \ \ \ \ divide by }-4 \\ x &=&6 \end{eqnarray*}% We check:% \begin{equation*} \text{RHS}=-4x+6=-4\cdot 6+6=-24+6=-18=\text{LHS } \end{equation*}% Thus our solution, \fbox{$x=6$} is correct. \end{enumerate} \vspace{0.59in}\vspace{6in}\bigskip \vspace{0.2in} %TCIMACRO{\TeXButton{small}{\small}}% %BeginExpansion \small% %EndExpansion {\Large %TCIMACRO{\TeXButton{\small}{\small}}% %BeginExpansion \small% %EndExpansion } \href{https://teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html}{% For more documents like this, visit our page at\ https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}