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\begin{Large}
\section{\bf Apendix A: Algebra Review}
\end{Large}

Because this is a freshman laboratory, we do not use high-level mathematics.
But we do sometimes encounter a little basic algebra and we need to briefly
review the main concepts. Algebra deals with equations and ``unknowns''.
Unknowns, or ``variables'', are usually represented as a letter in an equation:
$y$ = 3$x$ + 7. In this equation both ``$x$'' and ``$y$'' are variables. You
do not know what the value of $y$ is until you assign a value to $x$. For
example, if $x$ = 2, then $y$ = 13 ($y$ = 3$\times$2 + 7 = 13).
Here are some additional examples:
\vspace{0.10 in}
\begin{flushleft}
y = 5x + 3, if x=1, what is y? Answer: y = 5$\times$1 + 3 = 5 + 3 = 8\\
\vspace{0.3in}
q = 3t + 9, if t=5, what is q? Answer: q = 3$\times$5 + 9 = 15 + 9 = 24\\
\vspace{0.3in}
y = 5x$^{\rm 2}$ + 3, if x=2, what is y? Answer: y = 5$\times$(2$^{\rm 2}$) + 3
= 5$\times$4 + 3 = 20 + 3 = 23\\
\vspace{0.3in}
What is y if x = 6 in this equation: y = 3x + 13 =\\
\end{flushleft}

\subsection{Solving for X}

These problems were probably easy for you, but what happens when you have this
equation: y = 7x + 14, and you are asked to figure out what x is if y = 21?
Let's do this step by step, first we re-write the equation:
\begin{flushleft}
y = 7x + 14\\
\vspace{0.2in}
We now substitute the value of y (y = 21) into the equation:\\
\vspace{0.2in}
21 = 7x + 14\\
\vspace{0.3in}
Now, if we could get rid of that 14 we could solve this equation! Subtract
14 from both sides of the equation:\\
\vspace{0.2in}
21 $-$ 14 = 7x + 14 $-$ 14 \hspace{0.25in}(this gets rid of that pesky 14!)\\
\vspace{0.2in}
7 = 7x \hspace{0.25in}(divide both sides by 7)\\
\vspace{0.2in}
x = 1\\
\end{flushleft}
Ok, your turn: If you have the equation y = 4x + 16, and y = 8, what is x?
\vspace{0.4in}

We frequently encounter more complicated equations, such as y= 3x$^{\rm 2}$ +
2x $-$ 345, or p$^{\rm 2}$ = a$^{\rm 3}$. There are ways to solve such
equations, but that is beyond the scope of our introduction. However, you
do need to be able to solve equations like this: y$^{\rm 2}$ = 3x + 3 (if
you are told what ``x'' is!). Let's do this for x = 11:
\begin{flushleft}
Copy down the equation again:\\
\vspace{0.2in}
y$^{\rm 2}$ = 3x + 3 \\
\vspace{0.2in}
Substitute x = 11:\\
\vspace{0.2in}
y$^{\rm 2}$ = 3$\times$11 + 3 = 33 + 3 = 36\\
\vspace{0.2in}
Take the square root of both sides:\\
\vspace{0.2in}
(y$^{\rm 2}$)$^{\rm 1/2}$ = (36)$^{\rm 1/2}$\\
\vspace{0.2in}

\vspace{0.2in}
y = 6
\end{flushleft}

Did that make sense? To get rid of the square of a variable you have to
take the square root: (y$^{\rm 2}$)$^{\rm 1/2}$ = y. So to solve for y$^{\rm 2}$,
we took the square root of both sides of the equation.

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