\documentclass{article}
\usepackage[margin=2cm]{geometry}
\usepackage{tikz}
\usepackage{pgfplots}
\usepackage{amsmath}
\usepackage{hyperref}
\usetikzlibrary{calc}
%\pgfplotsset{compat=1.11}
\begin{document}
\underline{Quiz 21}
\begin{enumerate}
\item Solve the following system of equations by graphing each line (by finding two solutions to each and connecting the dots) and finding the intersection point in the given graph: $\left\{ \begin{array}{lll}
x+y &= 3 & \quad (i) \\
x-y &= 1 & \quad (ii)
\end{array} \right.$ \\
\textit{Solution:} First we will plot the lines by finding two solutions and connecting the dots (so our lines are entirely accurate). To draw $(i)$, first plug in $x=0$ into $(i)$ to get $y=3$ --- hence $(0,3)$ is a solution of $(i)$. Also plug in $y=0$ to get $x=3$ giving $(3,0)$ a solution of $(i)$. Plot these points and connect the dots to draw $(i)$. \\~\\
To draw $(ii)$, first plug $x=0$ into $(ii)$ to get $-y=1$ or $y=-1$ -- this means $(0,-1)$ is a solution of $(ii)$. Plug $y=0$ into $(ii)$ to get $x=1$ -- this means $(1,0)$ is a solution of $(ii)$. Plot these points and connect the dots to draw $(ii)$.
\begin{center}
\begin{tikzpicture}
\draw[step=1cm,gray,very thin] (-4,-4) grid (4,4);
\draw[very thick] (0,-4)--(0,4);
\draw[very thick] (-4,0)--(4,0);
\draw[fill=black] (0,3) circle (0.5ex);
\draw[fill=black] (3,0) circle (0.5ex);
\draw (-1,4) -- (4,-1);
\draw[fill=black] (0,-1) circle (0.5ex);
\draw[fill=black] (1,0) circle (0.5ex);
\draw (4,3)--(-3,-4);
\draw[fill=blue] (2,1) circle (0.5ex);
\end{tikzpicture} \\~\\
\noindent From the intersection of the lines, we see the solution of the system {\color{blue} \href{http://www.wolframalpha.com/input/?i=%7Bx%2By%3D3,x-y%3D1%7D}{is}} the point $(2,1)$.
\end{center}
\item Solve the following system of equation in any way you choose:
$$\left\{ \begin{array}{lll}
x-y &= 2 & \quad (i) \\
x-y &= 5 & \quad (ii).
\end{array} \right.$$
\textit{Solution:} Solve $(i)$ for $x$ to get
$$x=2+y.$$
Plug this into $(ii)$ to get
$$(2+y)-y=5,$$
or
$$2=5,$$
which is \textbf{always false}. Therefore there is {\color{blue}\href{http://www.wolframalpha.com/input/?i=%7Bx-y%3D2,x-y%3D5%7D}{no solution}} to this system.
\item Solve the following system of equations in any way you choose:
$$\left\{ \begin{array}{lll}
2x+3y &= 4 & \quad (i)\\
x-5y &= 7 & \quad (ii).
\end{array} \right.$$
\textit{Solution:} Solve $(ii)$ for $x$:
$$x = 7+5y.$$
Plug this into $(i)$ to get
$$2(7+5y)+3y=4.$$
Distribute the $2$ to get
$$14+10y+3y=4.$$
Simplify by combining like terms to get
$$14+13y=4.$$
Subtract $14$ to get
$$13y = -10.$$
Finally divide by $13$ to see $y = -\dfrac{10}{13}$. To get $x$ plug this value into either $(i)$ or $(ii)$ -- we will plug into $(ii)$ -- to get
$$x-5 \left( - \dfrac{10}{13} \right) = 7,$$
or equivalently
$$x + \dfrac{50}{13} = 7.$$
This yields
$$x = 7 - \dfrac{50}{13} = \dfrac{91}{13} - \dfrac{50}{13} = \dfrac{41}{13}.$$
Therefore the solution of the system {\color{blue}\href{http://www.wolframalpha.com/input/?i=%7B2x%2B3y%3D4,+x-5y%3D7%7D}{is}} $\left( \dfrac{41}{13}, -\dfrac{10}{13} \right)$.
\item Solve the following system of equations in any way you choose:
$$\left\{ \begin{array}{lll}
3x-2y &= 1 & \quad (i) \\
-6x+4y &= -2 & \quad (ii).
\end{array} \right.$$
\textit{Solution:} First we will solve $(i)$ for $x$. Add $2y$ to get
$$3x = 1+2y$$
and divide by $3$ to get
$$x = \dfrac{1}{3} + \dfrac{2}{3}y.$$
Plug this into $(ii)$ to get
$$-6 \left( \dfrac{1}{3} + \dfrac{2}{3} y \right) + 4y = -2.$$
Distribute the $-6$ into the sum to get
$$-\dfrac{6}{3} - \dfrac{12}{3}y + 4y = -2,$$
or equivalently
$$-2-4y+4y=-2,$$
or equivalently
$$-2=-2.$$
This equation is \textbf{always true} and so this system has infinitely many solutions (so to write them down would require us to basically write $(i)$ or $(ii)$ again).
\end{enumerate}
\end{document}