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\lhead{\large \color{blue}Lecture Notes}
\lfoot{}
\cfoot{}
\chead{\Large Solving One- and Two-Step Linear Equations}
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\lfoot{\small \copyright \; Hidegkuti, Powell, 2008}
\rfoot{\small Last revised: August 18, 2018}
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\begin{document}


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Equations are a fundamental concept and tool in mathematics.

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\textbf{Definition}: \ $\ $An \textbf{equation }is a statement in which two
expressions (algebraic or numeric) are connected with an equal sign.

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For example, $3x^{2}-x=4x+28$ is an equation. \ So is $x^{2}+5y=-y^{2}+x+2$.

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\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. \ 

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\vspace{0.1in}

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\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}\vspace{0.1in}

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\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}

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If $x=-2$, the left-hand side of the equation is \ \vspace{0.05in}

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ LHS $=3x^{2}-x$\vspace{0.05in}\newline
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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\ \ \ \ \ \ \ \ =3\left( -2\right)
^{2}-\left( -2\right) \vspace{0.05in}$\newline
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\ \ \ \ \ \ \ \ \ \ $\ \ \ \ \ \ \ \ \ \ \ \ \ =3\cdot 4+2=12+2=14\vspace{%
0.05in}$

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If $x=-2$, the right-hand side of the equation is \vspace{0.05in}

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ RHS$\,=4x+28\vspace{0.05in}$\newline
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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\ \ \ \ \ \ \ \ \,=4\left( -2\right) +28%
\vspace{0.05in}\newline
$%
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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\ \ \ \ \ \ \ \ \,=-8+28=20\vspace{0.05in}$%
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Since the two sides are not equal, $14\not=20$, the number $-2$ is not a
solution of this equation.\vspace{0.1in}

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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}

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If $x=4,$ $\vspace{0.05in}$the left-hand side of the equation is

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ LHS $=3x^{2}-x\vspace{0.05in}$\newline
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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\ \ \ \ \ \ \ \ =3\cdot 4^{2}-4\vspace{0.05in}
$\newline
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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\ \ \ \ \ \ \ \ =3\cdot 16-4=48-4=44\vspace{%
0.05in}$

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If $x=4$, the right-hand side of the equation is \vspace{0.05in}

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ RHS $=4x+28\vspace{0.05in}$\newline
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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\ \ \ \ \ \ \ =4\cdot 4+28\vspace{0.05in}$%
\newline
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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\ \ \ \ \ \ \ =16+28=44\vspace{0.05in}$%
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Since the two sides are equal, $x=4$ is a solution of this equation.\vspace{%
0.1in}

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\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.

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\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. \ 

If we think about it a little, trial and error is never a legitimate method
because there is no way for us to gurantee that there are no other solutions
are there. \ It is impossible for us to try all real numbers because there
are infinitely many of them, and we have finite lives. \ 

So we will need to develop systematic methods to solve equations. \ 

\pagebreak

We will start with the easiest group of equations, linear equations. \ There
are several types of linear equations, and we will start with the easiest
type that is called one-step equations. \ 

To solve a linear equation, we isolate the unknown by applying the same
operation(s) to both sides. Consider, for example, Ann and Dewitt who has
the same monthly salary. \ This month they both get a $40$ dollar raise. \
Who is making ore money now? \ It is clear that if we start with two equal
quantities and we add the same amount to them, they will still stay equal. \
This is the underlying principle of solving equations. \ We always apply the
same operations to both sides in an effort to bring the equation in a simple
form such as $x=-2$. \ The following equations are \textbf{one-step equations%
} because there is only one operation that separates us from the desired
form.\vspace{0.04in}

\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.)

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\textbf{Solution:} \ a) \ In order to isolate the unknown, we add $8$ to
both sides.%
\begin{eqnarray*}
x-8 &=&10\text{ \ \ \ \ \ \ \ \ \ add }8 \\
x &=&18
\end{eqnarray*}%
So the only solution of this equation is $18.$ \ We can also say that the
solution set is $\left\{ 18\right\} $. \ We should check;\ if $x=18,$ the
left-hand side is%
\begin{equation*}
\text{LHS}=x-8=18-8=10=\text{RHS }\checkmark
\end{equation*}%
So our solution, $\,$\fbox{$x=18$} is correct.\vspace{0.03in}\vspace{0.03in}

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b) \ In order to isolate the unknown, we divide both sides by $3$.%
\begin{eqnarray*}
3y &=&-12\text{ \ \ \ \ \ \ \ dividde by }3 \\
y &=&-4
\end{eqnarray*}%
So the only solution of this equation is $-4.$ \ We check; if $y=-4$, then%
\begin{equation*}
\text{LHS}=3y=3\left( -4\right) =-12=\text{RHS }\checkmark
\end{equation*}%
So our solution, $\,$\fbox{$y=-4$} is correct.

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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*}%
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 }\checkmark
\end{equation*}%
So our solution, $\,$\fbox{$x=-24$} is correct.\pagebreak

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d) \ In order to isolate the unknown, we we subtract $10$ from both sides.%
\begin{eqnarray*}
m+10 &=&-5\text{ \ \ \ \ \ \ \ \ \ \ \ subtract }10 \\
m &=&-15
\end{eqnarray*}%
So the only solution of this equation is $-15.$ \ We check; if $m=-15$, then%
\begin{equation*}
\text{LHS}=m+10=-15+10=-5=\text{RHS}
\end{equation*}%
So our solution, $\,$\fbox{$m=-15$} is correct.\vspace{0.03in}%
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\textbf{Discussion}: \ Solve each of the following equations. \ How are
these unusual?\vspace{0.07in}

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a) \ \ $5x=5$ \ \ \ \ \ \ \ b) \ $5x=0$\ \ \ \ \ \ \ \ c) \ $x-4=-4$ \ \ \ \
\ \ \ d) \ $\dfrac{x}{3}=0$ 
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\textbf{Example 3}. \ One side of a rectangle is $12$ feet long. \ Find the
length of the other side if the area of the rectangle is $60$ square-feet. \ 

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\textbf{Solution:} \ Let us denote the missing side by $x.$ \ We will write
and solve an equation expressing the area of the rectangle. \ 
\begin{eqnarray*}
12x &=&60\text{ \ \ \ \ \ \ \ \ divide by }12 \\
x &=&5
\end{eqnarray*}%
Thus the other side is \fbox{$5$ feet} long. \ Note that if we carry the
units in the computation, they will work out perfectly.\vspace{0.04in}

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\left( 12%
\unit{ft}\right) \,x=60\unit{ft}^{2}$ \ \ \ \ \ \ \ \ divide by $12\unit{ft}$
\ \ \ \ \ \ \ \ \ \ \ \ \ \ margin work:\ \ \ \ \ \ \ \ \ \ $\dfrac{60\unit{%
ft}^{2}}{12\unit{ft}}=5\dfrac{\unit{ft}\cdot \unit{ft}}{\unit{ft}}=5\unit{ft}
$\vspace{0.04in}\vspace{0.04in}

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ $x=5\unit{ft}$ \vspace{0.04in}\vspace{0.04in}

\bigskip

\begin{center}
{\Large Part 2 -- Two-Step Equations}
\end{center}

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Suppose we decide to hide a small object, say a coin. \ We put the coin on
the table, then place an envelope over it, and then, just to be sure, we
place a hat on top of the envelope. \ Let us find the coin! \ To do that,
what do we need to remove, and in what order? \ We would first remove the
hat and then the envelope, right? \vspace{0.07in}\ \newline
This is the basis of solving two-step equations. \ To isolate the unknown,
we will perform the inverse operations, in the reverse order. \ What
happened last can be undone first.\vspace{0.07in}

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\textbf{Example 4}. \ Solve each of the given equations. \ Make sure to
check your solutions.\vspace{0.04in}

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a) \ $10=3x-11$ \ \ \ \ \ \ \ \ \ \ b) \ $3x+8=-7$ \ \ \ \ \ \ \ \ \ \ \ c)
\ $\dfrac{t-7}{2}=-8$ \ \ \ \ \ \ \ \ \ \ \ \ d) \ $\dfrac{x}{-3}+4=15$%
\vspace{0.05in}

\pagebreak

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\textbf{Solution:} \ a) \ The equation \ $10=3x-11$ looks unusual in the
sense that two-step equations often contain the unknown on the\ left-hand
side. \ We are always allowed to swap two sides of an equation. \ If $A=B$,
then clearly, also $B=A$. \ We will do this first. \ This is an optional
step that is always available.%
\begin{eqnarray*}
10 &=&3x-11\text{ \ \ \ \ we swap the two sides} \\
3x-11 &=&10
\end{eqnarray*}%
We now look at the side that contains $x$ and ask: \textit{What happened to
the unknown}? \ The answer is:\ \textit{Multiplication by }$3$\textit{\ and
then subtraction of }$11$. \ We need to apply the inverse operations, in
reverse order. In this case, this means that we will add $11$ to both sides
and then divide both sides by $3$.%
\begin{eqnarray*}
3x-11 &=&10\text{ \ \ \ \ \ \ \ add }11 \\
3x &=&21\text{ \ \ \ \ \ \ \ divide by }3 \\
x &=&7
\end{eqnarray*}%
So the only solution of this equation is $7$. \ We check: if $x=7$, then%
\begin{equation*}
\text{LHS}=3x-11=3\cdot 7-11=21-11=10=\text{RHS }\checkmark
\end{equation*}%
So our solution, $\,$\fbox{$x=7$} is correct.\vspace{0.06in}

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b) \ As we look at the equation $3x+8=-7$, we ask: \textit{What happened to
the unknown}? \ The answer is:\ \textit{Multiplication by }$3$\textit{\ and
then addition of }$8$. \ We need to apply the inverse operations, in a
reverse order. \ In this case, this means that we will subtract $8$ from
both sides and then divide both sides by $3$.%
\begin{eqnarray*}
3x+8 &=&-7\text{ \ \ \ \ \ \ \ \ \ subtract }8 \\
3x &=&-15\text{ \ \ \ \ \ \ \ divide by }3 \\
x &=&-5
\end{eqnarray*}%
So the only solution of this equation is $-5$. \ We check: if $x=-5$, then%
\begin{equation*}
\text{LHS}=3x+8=3\left( -5\right) +8=-15+8=-7=\text{RHS }\checkmark
\end{equation*}%
So our solution, $\,$\fbox{$x=-5$} is correct.\vspace{0.06in}

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c) \ What happened to the unknown? \ On the left-hand side, there was a
subtraction of $7$ and then a division by $2$. \ To reverse that, we will
multiply both sides by $2$ and \ then add $7$ to both sides.%
\begin{eqnarray*}
\dfrac{t-7}{2} &=&-8\text{ \ \ \ \ \ \ \ \ \ \ \ multiply by }2 \\
t-7 &=&-16\text{ \ \ \ \ \ \ \ \ \ add }7 \\
t &=&-9
\end{eqnarray*}%
So the only solution of this equation is $-9$. \ We check: if $t=-9$, then%
\begin{equation*}
\text{LHS}=\dfrac{t-7}{2}=\dfrac{-9-7}{2}=\dfrac{-16}{2}=-8=\text{RHS }%
\checkmark
\end{equation*}%
So our solution, $\,$\fbox{$t=-9$} is correct.\vspace{0.07in}

\pagebreak

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d) \ What happened to the unknown? \ On the left-hand side, there was a
division by $-3$ and then an \newline
addition of $4$.\ \ To reverse that, we will subtract $4$ from both sides by
and then multiply both sides \newline
by $-3$.%
\begin{eqnarray*}
\dfrac{x}{-3}+4 &=&15\text{ \ \ \ \ \ \ \ \ subtract }4 \\
\dfrac{x}{-3} &=&11\text{ \ \ \ \ \ \ \ \ multiply by }-3 \\
x &=&-33
\end{eqnarray*}%
So the only solution of this equation is $-33$. \ We check: if $x=-33$, then%
\begin{equation*}
\text{LHS}=\dfrac{x}{-3}+4=\dfrac{-33}{-3}+4=11+4=15=\text{RHS }\checkmark
\end{equation*}%
So our solution, $\,$\fbox{$x=-33$} is correct.\vspace{0.03in}

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\textbf{Example 6}. \ The sum of three times a number and seven is $-5$. \
Find this number. \ \vspace{0.05in}

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\textbf{Solution:} \ Let us denote our mystery number by $x$. \ The equation
will be just the first sentence, translated to algebra. \ The sum of three
times the number and seven is $3x+7$. \ So our equation is $3x+7=-5$. \ We
will know the number if we solve this equation.%
\begin{eqnarray*}
3x+7 &=&-5\text{ \ \ \ \ \ \ \ \ \ \ subtract }7 \\
3x &=&-12\text{ \ \ \ \ \ \ \ \ \ divide by }3 \\
x &=&-4
\end{eqnarray*}%
Good news! \ We do not need to check if $-4$ is indeed the solution of the
equation. \ What if we correctly solved the \textit{wrong} equation? \
Recall that \textit{we} came up with the equation, it was not given. \
Instead of checking the number against the equation, we should check if our
solution satisfies the conditions stated in the problem. \ Is it true the
sum of three times $-4$ and seven is $-5$? \ Indeed, \newline
$3\left( -4\right) +7=-12+7=-5$. \ Thus our solution, \fbox{$-4$}, is
correct.

\pagebreak

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\vspace{0.1in}\ \ {\LARGE Sample Problems}

Solve each of the following equations. Make sure to check your solutions.

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\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 $-4x+6=-18$
\end{enumerate}

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Solve each of the following application problems.%
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\begin{enumerate}
\item Paul invested his money on the stock market. \ First he bet on a risky
stock and lost half of his money. \ Then he became a bit more careful and
invested money in more conservative stocks that involved \ less risk but
also less profit. \ His investments made him $80$ dollars. \ If he has $250$
dollars in the stock market today, with how much money did he start
investing?

\item In a hotel, the first night costs $45$ dollars, and all additional
nights cost $35$ dollars. \ How long did Mr. Williams stay in the hotel if
his bill was $325$ dollars?
\end{enumerate}

\vspace{0.2in}

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\ \ {\LARGE Practice Problems }\bigskip

Solve each of the following equations. Make sure to check your
solutions.\bigskip

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\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}

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Solve each of the following application problems.%
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\begin{enumerate}
\item Three times the difference of $x$ and $7$ is $-15$. \ Find $x$.

\item Ann and Bonnie are discussing their financial situation. \ Ann said: \ 
\textit{If you take }$50$\textit{\ bucks from me and then doubled what is
left, I would have }$\$300.$ \ Bonnie answers: \ \textit{That's funny. \ If
you doubled my money first and then took }$\$50,$\textit{\ then I would have 
}$\$300$! \ How much do they each have?

\item Susan was asked about her age. \ She answered as follows: \ My big
brother's age is six less than three times my age. \ How old is Susan if her
big brother is $21$ years old?

\pagebreak
\end{enumerate}

{\LARGE 
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\vspace{0.1in}\ \ \ 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 \vspace{0.1in}}

{\large Sample Problems}

\begin{enumerate}
\item $11$ \ \ \ \ \ \ 2. \ $-5$ \ \ \ \ 3. \ $24$\ \ \ \ 4. \ $53$\ \ \ \
5. \ $2$\ \ \ \ 6. \ $-14$\ \ \ \ \ 7. \ $-30$ \ \ \ \ 8. $0$\ \ \ \ \ \ \
9. $5$\ \ \ \ \ \ 10. \ $0$ \ \ \ \ 11. \ $-4$ \ \ \ \ \ 

\item[12.] $\$340$ \ \ \ \ \ 13. \ $9$ nights{\LARGE \vspace{0.1in}}\bigskip
\end{enumerate}

{\large Practice Problems}

\begin{enumerate}
\item $-4$ \ \ \ \ \ 2. \ $-5$ \ \ \ \ \ 3. \ $4$ \ \ \ \ \ 4. \ $-11$ \ \ \
\ \ 5. \ $14$ \ \ \ \ 6. \ $-4$ \ \ \ \ \ 7. \ $-37$ \ \ \ \ 8. \ $0$ \ \ \
\ \ \ 9. \ $-14$ \ \ \ \ 10. \ $12$

\item[11.] $-10$ \ \ \ 12. \ $-2$\ \ \ \ \ \ \ \ 13. \ $-6$ \ \ \ \ 14. \ $%
-6 $ \ \ \ 15. \ $42$ \ \ \ \ 16. \ $2$ \ \ \ 17. \ Ann has\ $\$200$ and
Bonnie has $\$175$\vspace{0.1in}\vspace{0.15in}
\end{enumerate}

\pagebreak

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{\LARGE Sample Problems - Solutions}\bigskip
\end{center}

Solve each of the following equations. \ Make sure to check your solutions.%
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\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.

\pagebreak

\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.

\pagebreak

\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.

\pagebreak

\item Paul invested his money on the stock market. \ First he bet on a risky
stock and lost half of his money. \ Then he became a bit more careful and
invested money in more conservative stocks that involved \ less risk but
also less profit. \ His investments made him $80$ dollars. \ If he has $250$
dollars in the stock market today, with how much money did he start
investing?

Solution: \ Let us denote the amount of money with which Paul started to
invest by $x$. First he lost half of his money, so he had $\dfrac{x}{2}$. \
Then he gained $80$ dollars end ended up with $250$ dollars. \ So, we can
write the equation $\dfrac{x}{2}+80=250$. \ We will solve this two-step
equation for $x.$ \ What happened to the unknown was first division by $2$
and then addition of $80$. \ To reverse these operations, we will first
subtract $80$ and then multiply by $2$. 
\begin{eqnarray*}
\dfrac{x}{2}+80 &=&250\text{ \ \ \ \ \ \ \ \ \ \ subtract }80 \\
\dfrac{x}{2} &=&170\text{ \ \ \ \ \ \ \ \ \ \ multiply by }2 \\
x &=&340
\end{eqnarray*}%
So Paul started with $340$ dollars. \ We check: \ If we lose half of $340,$%
dollars we have $170$ dollars left. \ Then when we add $80,$dollars we
indeed end up with $250$ dollars. \ So our solution is correct, \fbox{Paul
started with $340$ dollars}.

\item In a hotel, the first night costs $45$ dollars, and all additional
nights cost $35$ dollars. \ How long did Mr. Williams stay in the hotel if
his bill was $325$ dollars?

Solution: \ Suppose that Mr. Williams stayed for the first night and the an
additional $x$ many nights. \ Then the bill would be $45+x\cdot 35$ or $%
35x+45$. \ So we write and then solve the equation \thinspace $35x+45=325$.%
\begin{eqnarray*}
35x+45 &=&325\text{ \ \ \ \ \ \ \ \ \ \ \ subtract }45 \\
35x &=&280\text{ \ \ \ \ \ \ \ \ \ \ \ divide by }35 \\
x &=&8
\end{eqnarray*}%
Thus Mr. Williams stayed in the hotel for $9$ nights. \ Why not $8$ if we
got $x=8$? \ Remember, the first night was counted separately; there was the
first night and then $x=8$ additional nights. \ This is why\vspace{0.03in}
it is a good idea to read the text of the problem one more time before we
state our final answer. \ So \fbox{Mr. Williams stayed $9$ nights in the
hotel}. \ We\vspace{0.03in} check: the bill for $9$ nights would be $%
45+8\left( 35\right) =\allowbreak 325$, and so our solution is correct. \ 
\vspace{0.25in}\vspace{2.9in}
\end{enumerate}

{\Large 
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\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}