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\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
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\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}
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\lhead{\color{blue} \Large Lecture Notes}
\chead{\color{black} \LARGE Equations and Inequalities}
\rhead{\large page   \ \thepage}
\cfoot{}
\lfoot{\small   \copyright $\;$ copyright  Hidegkuti,  Powell,  2008}
\rfoot{\small Last revised:  June 26, 2012}
\textwidth 7.6in 
\textheight 9.7in 
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\begin{document}


\begin{center}
{\Large Sample Problems}\bigskip
\end{center}

\begin{enumerate}
\item Consider the equation $2x^{2}+x+34=21x-8$. \ In case of each number
given, determine whether it is a solution of the equation or not.%
%TCIMACRO{\TeXButton{5 col begin}{\begin{multicols}{5}}}%
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a) \ $x=1$

b) \ $x=3$

c) \ $x=4$

d) \ $x=7$ \ 
%TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}%
%BeginExpansion
\end{multicols}%
%EndExpansion

\item Consider the equation $x^{2}-10x+x^{3}-4=4\left( x+5\right) $. \ In
case of each number given, determine whether it is a solution of the
equation or not.%
%TCIMACRO{\TeXButton{5 col begin}{\begin{multicols}{5}}}%
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\begin{multicols}{5}%
%EndExpansion

a) $x=0$

b) $\ x=-2$

c) $\ x=-3$

d) \ $x=2$ 
%TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}%
%BeginExpansion
\end{multicols}%
%EndExpansion

\item Consider the equation \ $3a-2b-1=\left( a-b\right) ^{2}+4$. \ \ In
case of each pair of numbers given, determine whether it is a solution of
the equation or not.%
%TCIMACRO{\TeXButton{3 col begin}{\begin{multicols}{3}}}%
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\begin{multicols}{3}%
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a) $\ a=8$ and $b=5$

b) \ $a=10$ and $b=7$ 
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\end{multicols}%
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\item Consider the inequality $3\left( 2y-1\right) +1\leq 5y-7$. \ In case
of each number given, determine whether it is a solution of the inequality
or not.%
%TCIMACRO{\TeXButton{5 col begin}{\begin{multicols}{5}}}%
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\begin{multicols}{5}%
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a) \ $y=-10$

b) \ $y=3$

c) \ $y=-5$

d) \ $y=0$ 
%TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}%
%BeginExpansion
\end{multicols}%
%EndExpansion

\item Consider the inequality $\dfrac{2x+1}{3}+5<\dfrac{3x-1}{2}$. \ In case
of each number given, determine whether it is a solution of the inequality
or not.%
%TCIMACRO{\TeXButton{5 col begin}{\begin{multicols}{5}}}%
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\begin{multicols}{5}%
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a) \ $x=1$

b) \ $x=13$

c) \ $x=7$

d) \ $x=-5$ 
%TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}%
%BeginExpansion
\end{multicols}%
%EndExpansion
\pagebreak 
\end{enumerate}

\begin{center}
{\Large Practice Problems}\bigskip
\end{center}

\begin{enumerate}
\item Consider the equation $\dfrac{2x^{2}-11x-21}{2x+3}=3x-\left(
2x+7\right) $. \ In case of each number given, determine whether it is a
solution of the equation or not.%
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\begin{multicols}{4}%
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a) \ $x=8$

b) \ $x=13$

c) \ $x=10$ 
%TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}%
%BeginExpansion
\end{multicols}%
%EndExpansion

\item Consider the equation $-x^{2}-2x\left( 3-x^{2}\right) =-x+2$. \ In
case of each number given, determine whether it is a solution of the
equation or not.%
%TCIMACRO{\TeXButton{5 col begin}{\begin{multicols}{5}}}%
%BeginExpansion
\begin{multicols}{5}%
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a) \ $x=0$

b) $\ x=1$

c) \ $x=-1$

d) \ $x=2$

e) \ $x=-2$  
%TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}%
%BeginExpansion
\end{multicols}%
%EndExpansion

\item Consider the equation $y=\dfrac{5x-3}{2}$. \ \ In case of each pair of
numbers given, determine whether it is a solution of the equation or not.%
%TCIMACRO{\TeXButton{4 col begin}{\begin{multicols}{4}}}%
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\begin{multicols}{4}%
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a) \ $x=1$ and $y=1$

b) \ $x=9$ and $y=4$

c) \ $x=3$ and $y=6$

d) \ $x=17$ and $y=41$ \ 
%TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}%
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\end{multicols}%
%EndExpansion

\item Consider the equation \ $\left( p-q\right) ^{2}+\dfrac{3p-1}{6-q}%
=4\left( p+1\right) $. \ \ In case of each pair of numbers given, determine
whether it is a solution of the equation or not.%
%TCIMACRO{\TeXButton{3 col begin}{\begin{multicols}{3}}}%
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\begin{multicols}{3}%
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a) \ $p=8$ and $q=5$

b) \ $p=7$ and $q=1$ 
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\end{multicols}%
%EndExpansion

\item Consider the inequality $-x+2<-x^{2}+2\left( x+6\right) $. \ In case
of each number given, determine whether it is a solution of the inequality
or not.%
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\begin{multicols}{5}%
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a) \ $x=-5$

b) \ $x=-2$

c) \ $x=0$

d) \ $x=3$

e) \ $x=7$  
%TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}%
%BeginExpansion
\end{multicols}%
%EndExpansion

\item Consider the inequality $\dfrac{x}{3}+1\geq \dfrac{x+1}{2}-1$. \ In
case of each number given, determine whether it is a solution of the
inequality or not.%
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%BeginExpansion
\begin{multicols}{5}%
%EndExpansion

a) \ $x=-9$

b) \ $x=-3$

c) \ $x=27$

d) \ $x=15$

e) \ $x=-15$ 
%TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}%
%BeginExpansion
\end{multicols}%
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\pagebreak 
\end{enumerate}

\begin{center}
{\Large Answers - Sample Problems}\bigskip 
\end{center}

\begin{enumerate}
\item a) \ $37\not=13$ \ \ no \ \ \ \ b) \ $55=55$ \ \ yes \ \ \ \ c) \ $%
70\not=76$ \ \ no \ \ \ \ d) \ $139=139$ \ \ yes

\item a) $-4\not=20$ \ \ no \ \ \ \ b) $\ 12=12$ \ \ yes \ \ \ \ c) $\ 8=8$
\ \ yes \ \ \ \ d) \ $-12\not=28$ \ \ no

\item a) $\ 13=13$ \ yes \ \ \ \ \ b) \ $15\not=13$ \ no

\item a) \ $-62\leq -57$ \ \ yes \ \ \ \ b) \ $16\not\leq 8$ \ \ \ no \ \ \
\ c) \ $-32\leq -32$ \ \ yes \ \ \ \ d) \ $-2\not\leq -7$ \ no

\item a) \ $6\not<1$ \ \ no \ \ \ \ b) \ $14<19$ \ \ yes \ \ \ \ c) \ $10<10$
\ \ no \ \ \ \ d) \ $2<-8$ \ no\bigskip 
\end{enumerate}

\begin{center}
{\Large Answers - Practice Problems}\bigskip 
\end{center}

\begin{enumerate}
\item a) \ $1=1$ \ \ \ yes \ \ \ \ \ b) \ $6=6$ \ \ yes \ \ \ \ c) \ $3=3$ \
\ yes 

\item a) \ $0\not=2$ \ \ \ \ no \ \ \ \ b) $\ -5\not=1$ \ \ no \ \ \ \ \ c)
\ $3=3$ \ \ yes \ \ \ \ \ d) \ $0=0$ \ \ \ yes \ \ \ \ \ e) \ $-8\not=4$ \ \
\ no

\item a) \ $1=1$ \ \ \ yes \ \ \ \ \ b) \ $4\not=21$ \ \ \ \ no \ \ \ \ c) \ 
$6=6$ \ \ \ \ yes \ \ \ \ \ d) \ $41=41$ \ \ \ no

\item a) \ $32\not=36$ \ \ \ no \ \ \ \ \ \ \ b) \ $40\not=32$ \ \ \ \ no

\item a) \ $7\not<-23$ \ \ no \ \ \ \ b) \ $4\not<4$ \ \ no \ \ \ \ c) \ $%
2<12$ \ \ yes \ \ \ \ d) \ $-1<9$ \ \ yes \ \ \ \ e) \ $-5\not<-23$ \ \ no

\item a) \ $-2\geq -5$ \ \ yes \ \ \ \ b) \ $0\geq -2$ \ \ yes \ \ \ \ c) \ $%
10\not\geq 13$ \ \ \ no \ \ \ \ d) \ $6\not\geq 7$ \ \ \ \ \ no \ \ \ \ e) \ 
$-4\geq -8$ \ \ \ yes\bigskip 
\end{enumerate}

\begin{center}
{\Large Sample Problems - Solutions}\bigskip
\end{center}

\begin{enumerate}
\item Consider the equation $2x^{2}+x+34=21x-8$. \ In case of each number
given, determine whether it is a solution of the equation or not.

a) \ $x=1$\newline
Solution: \ We need to substitute $1$ for $x$ into both sides of the
equation and evaluate those algebraic expressions to see whether the
left-hand side equals to the right-hand side. \ The left-hand side:%
\begin{equation*}
\text{LHS}=2x^{2}+x+34=2\left( 1\right) ^{2}+\left( 1\right) +34=2\cdot
1+1+34=2+1+34=3+34=37
\end{equation*}%
The right-hand side:%
\begin{equation*}
\text{RHS}=21x-8=21\left( 1\right)
-8=21-8=13~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation*}%
When $x=1$, the two expressions are not equal. \ Thus\ $1$ is not a solution
of the equation.

b) \ $x=3$\newline
Solution: \ We need to substitute $3$ for $x$ into both the left-hand side
and right-hand side of the equation and evaluate those algebraic expressions
to see whether the left-hand side equals to the right-hand side.\newline
The left-hand side:%
\begin{equation*}
\text{LHS}=2x^{2}+x+34=2\left( 3\right) ^{2}+\left( 3\right) +34=2\cdot
9+3+34=18+3+34=21+34=55
\end{equation*}%
The right-hand side:%
\begin{equation*}
\text{RHS}=21x-8=21\left( 3\right)
-8=63-8=55~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation*}%
When $x=3$, the two expressions are equal. Thus\ $3$ is a solution of the
equation.

c) \ $x=4$\newline
Solution: \ We need to substitute $4$ for $x$ into both sides of the
equation and evaluate those algebraic expressions to see whether the
left-hand side equals to the right-hand side.\newline
The left-hand side:%
\begin{equation*}
\text{LHS}=2x^{2}+x+34=2\left( 4\right) ^{2}+\left( 4\right) +34=2\cdot
16+4+34=32+4+34=36+34=70
\end{equation*}%
The right-hand side:%
\begin{equation*}
\text{RHS}=21x-8=21\left( 4\right)
-8=84-8=76~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation*}%
Since the two expressions are not equal when $x=4$, \ $4$ is not a solution
of the equation.

d) \ $x=7$\newline
Solution: \ We need to substitute $7$ for $x$ into both sides of the
equation and evaluate those algebraic expressions to see whether the
left-hand side equals to the right-hand side.\newline
The left-hand side:%
\begin{equation*}
\text{LHS}=2x^{2}+x+34=2\left( 7\right) ^{2}+\left( 7\right) +34=2\cdot
49+7+34=98+7+34=105+34=139
\end{equation*}%
The right-hand side:%
\begin{equation*}
\text{RHS}=21x-8=21\left( 7\right)
-8=147-8=139~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation*}%
Since the two expressions are equal when $x=7$, $7$ is a solution of the
equation.

\item Consider the equation $x^{2}-10x+x^{3}-4=4\left( x+5\right) $. \ In
each case, determine whether the number given is a solution of the equation
or not.

a) \ $x=0$\newline
Solution: \ We simply evaluate both sides of the equation when $x=0$.%
\begin{equation*}
\text{LHS}=\left( 0\right) ^{2}-10\left( 0\right) +\left( 0\right)
^{3}-4=0-10\cdot 0+0-4=0-0+0-4=-4
\end{equation*}%
\begin{equation*}
\text{RHS}=4\left( \left( 0\right) +5\right) =4\left( 0+5\right) =4\cdot
5=20~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation*}%
Since $-4\not=20$, \ $x=0$ \ is not a solution of this equation.

b) $\ x=-2$\newline
Solution: \ We simply evaluate both sides of the equation when $x=-2$.%
\begin{eqnarray*}
\text{LHS} &=&\left( -2\right) ^{2}-10\left( -2\right) +\left( -2\right)
^{3}-4=4-10\left( -2\right) +\left( -8\right) -4=4-\left( -20\right) +\left(
-8\right) -4 \\
&=&24+\left( -8\right) -4=16-4=12
\end{eqnarray*}%
\begin{equation*}
\text{RHS}=4\left( \left( -2\right) +5\right) =4\cdot
3=12~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation*}%
Since $12=12$, \ $x=-2$ \ is a solution of this equation.

c) \ $x=-3$\newline
Solution: \ We simply evaluate both sides of the equation when $x=-3$.%
\begin{eqnarray*}
\text{LHS} &=&\left( -3\right) ^{2}-10\left( -3\right) +\left( -3\right)
^{3}-4=9-10\left( -3\right) +\left( -27\right) -4=9-\left( -30\right)
+\left( -27\right) -4 \\
&=&39+\left( -27\right) -4=12-4=8
\end{eqnarray*}%
\begin{equation*}
\text{RHS}=4\left( \left( -3\right) +5\right) =4\left( -3+5\right) =4\cdot
2=8~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation*}%
Since $8=8$, \ $x=-3$ \ is a solution of this equation.

\item Consider the equation \ $3a-2b-1=\left( a-b\right) ^{2}+4$. \ \ In
case of each pair of numbers given, determine whether it is a solution of
the equation or not.

a) \ $a=8$ and $b=5$\newline
Solution: \ We need to substitute $a=8$ \ \ and $b=5$ into both sides of the
equation and evaluate those algebraic expressions to see whether the
left-hand side equals to the right-hand side. \ The left-hand side:%
\begin{eqnarray*}
\text{LHS} &=&3a-2b-1=3\left( 8\right) -2\left( 5\right) -1=3\cdot 8-2\cdot
5-1=24-2\cdot 5-1=24-10-1 \\
&=&14-1=13
\end{eqnarray*}%
The right-hand side:%
\begin{equation*}
\text{RHS}=\left( a-b\right) ^{2}+4=\left( \left( 8\right) -\left( 5\right)
\right) ^{2}+4=\left( 8-5\right) ^{2}+4=3^{2}+4=9+4=13~~~~~~~~~~~~~~~~
\end{equation*}%
Since the two expressions are equal when $a=8$ \ \ and $b=5$, \ this pair is
a solution of the equation.

b) \ $a=10$ and $b=7$\newline
Solution: \ We need to substitute $a=10$ \ \ and $b=7$ into both sides of
the equation and evaluate those algebraic expressions to see whether the
left-hand side equals to the right-hand side.\newline
The left-hand side:%
\begin{eqnarray*}
\text{LHS} &=&3a-2b-1=3\left( 10\right) -2\left( 7\right) -1=3\cdot
10-2\cdot 7-1=30-2\cdot 7-1=30-14-1 \\
&=&16-1=15
\end{eqnarray*}%
The right-hand side:%
\begin{equation*}
\text{RHS}=\left( a-b\right) ^{2}+4=\left( \left( 10\right) -\left( 7\right)
\right) ^{2}+4=\left( 10-7\right) ^{2}+4=3^{2}+4=9+4=13~~~~~~~~~~~~~~~~~
\end{equation*}%
Since the two expressions are not equal when $a=8$ \ \ and $b=5$, \ this
pair is NOT a solution of the equation.

\item Consider the inequality $3\left( 2y-1\right) +1\leq 5y-7$. \ In case
of each number given, determine whether it is a solution of the inequality
or not.

a) \ $y=-10$\newline
Solution: \ We need to substitute $y=-10$ into both sides of the inequality
and evaluate those algebraic expressions to see whether the left-hand side
is indeed less than or equal to the right-hand side. \ The left-hand side:%
\begin{equation*}
\text{LHS}=3\left( 2\left( -10\right) -1\right) +1=3\left( -20-1\right)
+1=3\left( -21\right) +1=-63+1=-62
\end{equation*}%
The right-hand side:%
\begin{equation*}
\text{RHS}=5\left( -10\right)
-7=-50-7=-57~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation*}%
So the statement $3\left( 2y-1\right) +1\leq 5y-7$ becomes $-62\leq -57$. \
Since this is a true statement, $y=-10$ is a solution of the inequality.

b) \ $y=3$\newline
Solution: \ We need to substitute $y=3$ into both sides of the inequality
and evaluate those algebraic expressions to see whether the left-hand side
is indeed less than or equal to the right-hand side. \ The left-hand side:%
\begin{equation*}
\text{LHS}=3\left( 2\left( 3\right) -1\right) +1=3\left( 6-1\right)
+1=3\left( 5\right) +1=15+1=16~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation*}%
The right-hand side:%
\begin{equation*}
\text{RHS}=5\left( 3\right)
-7=15-7=8~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation*}%
So the statement $3\left( 2y-1\right) +1\leq 5y-7$ becomes $16\leq 8$. \
Since this is a false statement, $y=3$ is not a solution of the
inequality.\pagebreak 

c) \ $y=-5$\newline
Solution: \ We need to substitute $y=-5$ into both sides of the inequality
and evaluate those algebraic expressions to see whether the left-hand side
is indeed less than or equal to the right-hand side. \ The left-hand side:%
\begin{equation*}
\text{LHS}=3\left( 2\left( -5\right) -1\right) +1=3\left( -10-1\right)
+1=3\left( -11\right) +1=-33+1=-32
\end{equation*}%
The right-hand side:%
\begin{equation*}
\text{RHS}=5\left( -5\right)
-7=-25-7=-32~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation*}%
So the statement $3\left( 2y-1\right) +1\leq 5y-7$ becomes $-32\leq -32$. \
Since this is a true statement, $y=-5$ is a solution of the inequality.

d) \ $y=0$\newline
Solution: \ We need to substitute $y=0$ into both sides of the inequality
and evaluate those algebraic expressions to see whether the left-hand side
is indeed less than or equal to the right-hand side. \ The left-hand side:%
\begin{equation*}
\text{LHS}=3\left( 2\left( 0\right) -1\right) +1=3\left( 0-1\right)
+1=3\left( -1\right) +1=-3+1=-2~~~~~~~~~
\end{equation*}%
The right-hand side:%
\begin{equation*}
\text{RHS}=5\left( 0\right)
-7=0-7=-7~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation*}%
So the statement $3\left( 2y-1\right) +1\leq 5y-7$ becomes $-2\leq -7$. \
Since this is a false statement, $y=0$ is not a solution of the inequality.

\item Consider the inequality $\dfrac{2x+1}{3}+5<\dfrac{3x-1}{2}$. \ In case
of each number given, determine whether it is a solution of the inequality
or not.

a) \ $x=1$\newline
Solution: \ We need to substitute $x=1$ into both sides of the inequality
and evaluate those algebraic expressions to see whether the left-hand side
is indeed less than the right-hand side. \ The left-hand side:%
\begin{equation*}
\text{LHS}=\dfrac{2\left( 1\right) +1}{3}+5=\dfrac{2+1}{3}+5=\dfrac{3}{3}%
+5=1+5=6~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation*}%
The right-hand side:%
\begin{equation*}
\text{RHS}=\dfrac{3\left( 1\right) -1}{2}=\dfrac{3-1}{2}=\dfrac{2}{2}%
=1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation*}%
So the statement $\dfrac{2x+1}{3}+5<\dfrac{3x-1}{2}$ becomes $6<1$. \ Since
this is a false statement, $x=1$ is not a solution of the inequality.

b) \ $x=13$\newline
Solution: \ We need to substitute $x=13$ into both sides of the inequality
and evaluate those algebraic expressions to see whether the left-hand side
is indeed less than the right-hand side. \ The left-hand side:%
\begin{equation*}
\text{LHS}=\dfrac{2\left( 13\right) +1}{3}+5=\dfrac{26+1}{3}+5=\dfrac{27}{3}%
+5=9+5=14~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation*}%
The right-hand side:%
\begin{equation*}
\text{RHS}=\dfrac{3\left( 13\right) -1}{2}=\dfrac{39-1}{2}=\dfrac{38}{2}%
=19~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation*}%
So the statement $\dfrac{2x+1}{3}+5<\dfrac{3x-1}{2}$ becomes $14<19$. \
Since this is a true statement, $x=13$ is a solution of the
inequality.\pagebreak 

c) \ $x=7$\newline
Solution: \ We need to substitute $x=7$ into both sides of the inequality
and evaluate those algebraic expressions to see whether the left-hand side
is indeed less than the right-hand side. \ The left-hand side:%
\begin{equation*}
\text{LHS}=\dfrac{2\left( 7\right) +1}{3}+5=\dfrac{14+1}{3}+5=\dfrac{15}{3}%
+5=5+5=10~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation*}%
The right-hand side:%
\begin{equation*}
\text{RHS}=\dfrac{3\left( 7\right) -1}{2}=\dfrac{21-1}{2}=\dfrac{20}{2}%
=10~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation*}%
So the statement $\dfrac{2x+1}{3}+5<\dfrac{3x-1}{2}$ becomes $10<10$. \
Since this is a false statement, $x=7$ is not a solution of the inequality.

d) \ $x=-5$\newline
Solution: \ We need to substitute $x=-5$ into both sides of the inequality
and evaluate those algebraic expressions to see whether the left-hand side
is indeed less than the right-hand side. \ The left-hand side:%
\begin{equation*}
\text{LHS}=\dfrac{2\left( -5\right) +1}{3}+5=\dfrac{-10+1}{3}+5=\dfrac{-9}{3}%
+5=-3+5=2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation*}%
The right-hand side:%
\begin{equation*}
\text{RHS}=\dfrac{3\left( -5\right) -1}{2}=\dfrac{-15-1}{2}=\dfrac{-16}{2}%
=-8~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation*}%
So the statement $\dfrac{2x+1}{3}+5<\dfrac{3x-1}{2}$ becomes $2<-8$. \ Since
this is a false statement, $x=-5$ is not a solution of the inequality.
\end{enumerate}

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