\hfill \thepage} %} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.5in,right=0.4in,top=0.5in,bottom=0.4in} \pagestyle{fancy} \lhead{\large \color{blue}Lecture Notes} \lfoot{} \cfoot{} \chead{\Large Solving One- and Two-Step Linear Equations} \rhead{ page \thepage} \textwidth 7.7in \textheight 9.6in \setlength{\headheight}{27pt} \lfoot{\small \copyright \; Hidegkuti, Powell, 2008} \rfoot{\small Last revised: January 15, 2017} \setlength{\parindent}{0in} \setlength{\parskip}{0.05in} \begin{document} Equations are a fundamental concept and tool in mathematics. %TCIMACRO{% %\TeXButton{box start - 7.5in}{\begin{boxedminipage}{7.5in} %\setlength{\fboxrule}{15pt} %\setlength{\fboxsep}{15pt}}}% %BeginExpansion \begin{boxedminipage}{7.5in} \setlength{\fboxrule}{15pt} \setlength{\fboxsep}{15pt}% %EndExpansion %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \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{end of box}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion For example, $3x^{2}-x=4x+28$ is an equation. \ So is $x^{2}+5y=-y^{2}+x+2$. %TCIMACRO{% %\TeXButton{box start - 7.5in}{\begin{boxedminipage}{7.5in} %\setlength{\fboxrule}{15pt} %\setlength{\fboxsep}{15pt}}}% %BeginExpansion \begin{boxedminipage}{7.5in} \setlength{\fboxrule}{15pt} \setlength{\fboxsep}{15pt}% %EndExpansion %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \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 \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ variable(s) in the equation, makes the statement of equality true. \ %TCIMACRO{\TeXButton{end of box}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion %TCIMACRO{% %\TeXButton{\setlength{\parskip}{0.05in}}{\setlength{\parskip}{0.05in}}}% %BeginExpansion \setlength{\parskip}{0.05in}% %EndExpansion \vspace{0.1in} \textbf{Example 1}: \ \ a) \ Verify that $-2$ is not a solution of the equation $3x^{2}-x=4x+28$. %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ \ b) \ Verify that $4$ is a solution of the equation $% 3x^{2}-x=4x+28$. %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ \ c) \ Verify that the pair of numbers $x=3$ and $% y=-4 $ is a solution of the equation $x^{2}+5y=-y^{2}+x+2$. %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \qquad \qquad\ \ \ \ 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} Solution: \ a) \ Consider the equation $3x^{2}-x=4x+28$ with $x=-2$.\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ If $x=-2$, the left-hand side of the equation is \ LHS $=3x^{2}-x=3\left( -2\right) ^{2}-\left( -2\right) =3\cdot 4+2=12+2=14$ \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ and the right-hand side is RHS$% =4x+28=4\left( -2\right) +28=-8+28=20$\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Since the two sides are not equal, $14\not=20$, the number $-2$ is not a solution of this equation.\vspace{% 0.15in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ b) \ Consider the equation $3x^{2}-x=4x+28$ with $x=4$.\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ If $x=4,$ the left-hand side of the equation is \ LHS $=3x^{2}-x=3\cdot 4^{2}-4=3\cdot 16-4=48-4=44$ and \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ the right-hand side is RHS $% =4x+28=4\cdot 4+28=16+28=44$\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Since the two sides are equal, $% x=4 $ is a solution of this equation.\vspace{0.15in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ c) \ Consider the equation $% x^{2}+5y=-y^{2}+x+2$ with $x=3$ and $y=-4$.\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ If $x=3$ and $y=-4$, the left-hand side of the equation is \ LHS $=x^{2}+5y=3^{2}+5\left( -4\right) =9-20=-11$ \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ and the right-hand side is RHS $% =-y^{2}+x+2=-\left( -4\right) ^{2}+3+2=-16+3+2=-11$\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Since the two sides are equal, $% x=3 $ and $y=-4$ is a solution of this equation.\vspace{0.15in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ d) \ Consider the equation $% x^{2}+5y=-y^{2}+x+2$ with $x=-4$ and $y=3$.\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \thinspace If $x=-4$ and $y=3$, the left-hand side of the equation is \ LHS $=x^{2}+5y=\left( -4\right) ^{2}+5\cdot 3=16+15=41$ \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ and the right-hand side is RHS $% =-y^{2}+x+2=-3^{2}+\left( -4\right) +2=-9-4+2=-11$\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Since $41\not=-11$, \ the two sides are not equal, and so $x=-4$ and $y=3$ is not a solution of this equation.\vspace{0.15in} %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \thinspace If the equation is in more than one variable, like in parts c) and d) before, it is important to identify \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \thinspace which number is to be substituted into which variable. \ After all, $x=3$ and $y=-4$ \ (or, as an \textit{ordered pair}, \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \thinspace $\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}, \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \thinspace $\left( -4,3\right) $) is not.\vspace{0.12in} %TCIMACRO{% %\TeXButton{box start - 7.5in}{\begin{boxedminipage}{7.5in} %\setlength{\fboxrule}{15pt} %\setlength{\fboxsep}{15pt}}}% %BeginExpansion \begin{boxedminipage}{7.5in} \setlength{\fboxrule}{15pt} \setlength{\fboxsep}{15pt}% %EndExpansion %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \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{end of box}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion \vspace{0.1in} Caution! \ Finding one solution for an equation is not the same as solving it. \ For example, we found that $4$ is a solution of $3x^{2}-x=4x+28$. \ As it turns out, $4$ is not the only solution. \ We leave to the reader to verify that $-\dfrac{7}{3}$ is also a solution of the equation. \ We will have to deploy systematic methods to find all solutions. \ The methods we will use usually depends on the type of 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. \pagebreak \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) \ $x-\dfrac{1}{6}=\dfrac{2}{3}$ \ \ \ \ \ \ \ \ \ \ \ \ 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.) 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 add $\dfrac{1}{6}$ to both sides.% \begin{eqnarray*} x-\dfrac{1}{6} &=&\dfrac{2}{3}\text{ \ \ \ \ \ \ \ \ add }\dfrac{1}{6}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ margin work:\ \ }\dfrac{2}{3}+\dfrac{1% }{6}=\dfrac{4}{6}+\dfrac{1}{6}=\dfrac{5}{6} \\ x &=&\dfrac{5}{6} \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 $\dfrac{5}{6} $. \ We check; if $x=\dfrac{5}{6}$, then\ \begin{equation*} \text{LHS}=x-\dfrac{1}{6}=\dfrac{5}{6}-\dfrac{1}{6}=\dfrac{4}{6}=\dfrac{2}{3}% =\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{$x=\dfrac{5}{6}$} 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} Note: If the reader is interested in applications of one-step equations, basic percent problems and basic motion problems \newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}} }% %BeginExpansion \color{black} %EndExpansion \ \qquad can be easily handled by setting up and solving one-step equations. \ \FRAME{itbpF}{0.7048in}{0.7117in}{0.2006in}{}{}{question.bmp}{\special% {language "Scientific Word";type "GRAPHIC";display "USEDEF";valid_file "F";width 0.7048in;height 0.7117in;depth 0.2006in;original-width 0.4869in;original-height 0.6797in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'aaaaa_book143/question.bmp';file-properties "NPEU";}}\textbf{Discussion}: \ Solve each of the following equations. \ How are these unusual?\newline %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \ \ \ \ \ \ \ \qquad \qquad\ \ \ \ \ \ \ \ a) \ \ $5x=5$ \ \ \ \ \ \ \ b) \ $% 5x=0$\ \ \ \ \ \ \ \ c) \ $x-4=-4$ \ \ \ \ \ \ \ d) \ $\dfrac{x}{3}=0$ \ \ \ \ \ \ e) \ $\dfrac{1}{3}x=0$ \textbf{Example 3}: \ Solve each of the given equations. \ Make sure to check your solutions. \qquad \qquad \qquad\ \ \ \ a) \ $10=3x-11$ \ \ \ \ \ \ \ \ \ \ b) \ $\dfrac{% 1}{2}x+1=-7$ \ \ \ \ \ \ \ \ \ \ \ c) \ $\dfrac{t-7}{2}=-8$ \ \ \ \ \ \ \ \ \ \ \ \ d) \ $\dfrac{x}{-3}+4=15$\vspace{0.12in} 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 it was done to the unknown. Solution: \ a) \ This equation 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 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} %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 left-hand side 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 }$\dfrac{1}{2}$\textit{\ and addition of }$1$. \ 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 $1$ from both sides and then divide both sides by $\dfrac{1}{2}$.\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 $\dfrac{1}{2}% x+1=-7$ \ \ \ \ \ \ \ \ \ \ subtract $1$\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\ \ \ \ \ \ $\,% \dfrac{1}{2}x=-8$ \ \ \ \ \ \ \ \ \ divide by $\dfrac{1}{2}$ \ \ \ \ \ \ \ \ margin work: to divide is to multiply by \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 \qquad \qquad \qquad \qquad \qquad\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ the reciprocal: \ $% \dfrac{-8}{~~\dfrac{1}{2}~~}=-8\cdot 2=-16$\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\ \ \ \ $\ \ \ \ \ x=-16$\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 $-16$. \ We check: if $x=-16$, 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 $=\dfrac{1}{2}x+1=\dfrac{1}{2}% \left( -16\right) +1=-8+1=-7=$ 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{$x=-16$} 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. \pagebreak %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} \bigskip \bigskip \bigskip \bigskip \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";}}\ \ {\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 $\dfrac{1}{5}x-\dfrac{2}{3}=\dfrac{26}{15}\medskip $ \item $\dfrac{3}{8}x+\left( 1\dfrac{4}{5}\right) =\dfrac{3}{10}\medskip $ \end{enumerate} %TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \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";}}\ \ {\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+\dfrac{3}{8}}{2\dfrac{4}{5}}=\dfrac{5}{16}\medskip $ \item $\dfrac{2}{3}b+\dfrac{3}{5}=-\dfrac{1}{15}\medskip $ \item $\dfrac{1}{3}x+\dfrac{2}{5}=-\dfrac{34}{15}$ \end{enumerate} %TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \bigskip \pagebreak {\LARGE \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";}} \ \ \ Answers} {\large Discussion \ } a) \ $1$ \ \ \ \ b) \ $0$ \ \ \ \ c) \ $0$ \ \ \ d) \ $0$ \ \ \ e) \ $0$ 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.) \ $12$ \ \ \ \ 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.) \ $\dfrac{1}{2}$ \ \ \ \ 14.) \ $-1$ \ \ \ \ \ 15.) \ $-8$\vspace{0.1in}\vspace{0.15in} \begin{center} \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";}}{\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*}% We check: if $t=53$, then% \begin{equation*} \text{RHS}=\dfrac{53-5}{12}=\dfrac{48}{12}=4=\text{LHS} \end{equation*}% 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:% \begin{equation*} \text{LHS}=\dfrac{-14+8}{3}=\dfrac{-6}{3}=-2=\text{RHS} \end{equation*}% 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 $\dfrac{1}{5}x-\dfrac{2}{3}=\dfrac{26}{15}$\vspace{0.1in}\newline Solution:% \begin{eqnarray*} \dfrac{1}{5}x-\dfrac{2}{3} &=&\dfrac{26}{15}\text{ \ \ \ \ \ \ \ \ \ add \ }% \dfrac{2}{3}\text{ \ to both sides \ \ \ \ \ \ \ \ \ \ }\dfrac{26}{15}+% \dfrac{2}{3}=\dfrac{26}{15}+\dfrac{2\cdot 5}{3\cdot 5}= \\ \dfrac{1}{5}x &=&\dfrac{12}{5}\text{ \ \ \ \ \ \ \ \ \ \ divide by }\dfrac{1% }{5}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\dfrac{26}{15}+% \dfrac{10}{15}=\dfrac{36}{15}=\dfrac{\NEG{3}\cdot 12}{\NEG{3}\cdot 5}=\dfrac{% 12}{5} \\ x &=&12\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\dfrac{\dfrac{12}{5}}{\dfrac{1}{5}}=% \dfrac{12}{5}\cdot \dfrac{5}{1}=\dfrac{12\cdot \NEG{5}}{1\cdot \NEG{5}}=% \dfrac{12}{1}=12 \end{eqnarray*}% We check: if $x=12$, then% \begin{eqnarray*} \text{LHS} &=&\dfrac{1}{5}\cdot 12-\dfrac{2}{3}=\dfrac{1}{5}\cdot \dfrac{12}{% 1}-\dfrac{2}{3}=\dfrac{12}{5}-\dfrac{2}{3}=\dfrac{12\cdot 3}{5\cdot 3}-% \dfrac{2\cdot 5}{3\cdot 5}=\dfrac{36}{15}-\dfrac{10}{15}=\dfrac{26}{15} \\ \text{RHS} &=&\dfrac{26}{15} \end{eqnarray*}% Thus our solution, \fbox{$x=12$}\ is correct. \item $\dfrac{3}{8}x+\left( 1\dfrac{4}{5}\right) =\dfrac{3}{10}$\vspace{0.1in% }\newline Solution: this is a very simple equation, much like $2x+1=7$, only the numbers in it are fractions. However, the principles and operations regarding equations are the same.% \begin{eqnarray*} \dfrac{3}{8}x+\left( 1\dfrac{4}{5}\right) &=&\allowbreak \dfrac{3}{10}\text{ \ \ \ \ \ \ convert mixed number to improper fraction} \\ \dfrac{3}{8}x+\dfrac{9}{5} &=&\allowbreak \dfrac{3}{10}\text{ \ \ \ \ \ \ subtract }\dfrac{9}{5}\text{ from both sides; }\dfrac{3}{10}-\dfrac{9}{5}=% \dfrac{3}{10}-\dfrac{18}{10}=\dfrac{3-18}{10}=\dfrac{-15}{10}=\dfrac{-3}{2} \\ \dfrac{3}{8}x &=&\dfrac{-3}{2}\text{ \ \ \ \ \ \ divide \ both sides by }% \dfrac{3}{8} \\ x &=&-4\text{ \ \ \ \ \ \ \ \ \ \ \ \ }\left( \dfrac{-3}{2}\right) \div \left( \dfrac{3}{8}\right) =\dfrac{-3}{2}\cdot \dfrac{8}{3}=\dfrac{-24}{6}=-4% \text{\ \ \ \ \ \ \ } \end{eqnarray*}% We check:% \begin{equation*} \text{RHS}=\dfrac{3}{8}\left( -4\right) +\left( 1\dfrac{4}{5}\right) =\dfrac{% 3}{8}\cdot \dfrac{-4}{1}+\dfrac{9}{5}=\dfrac{-12}{8}+\dfrac{9}{5}=\dfrac{-3}{% 2}+\dfrac{9}{5}=\dfrac{-15}{10}+\dfrac{18}{10}=\dfrac{3}{10}=\text{LHS } \end{equation*}% Thus our solution, \fbox{$x=-4$} is correct. \end{enumerate} \vspace{5.3in} {\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}