<s>[INST] What are the steps to solve this simple algebraic equation? This is the equation that I use to calculate a percentage margin between cost and sales prices, where $x$ = sales price and $y$ = cost price:
\begin{equation}
z=\frac{x-y}{x}*100
\end{equation}
This can be solved for $x$ to give the following equation, which calculates sales price based on cost price and margin percentage:
\begin{equation}
x=\frac{y}{1-(\frac{z}{100})}
\end{equation}
My question is, what are the steps involved in solving the first equation for $x$? It's been 11 years since I last did algebra at school and I can't seem to figure it out.
I'm guessing the first step is to divide both sides by $100$ like so:
\begin{equation}
\frac{z}{100}=\frac{x-y}{x}
\end{equation}
Then what? Do I multiply both sides by $x$? If so how to I reduce the equation down to a single $x$?
 [/INST] First, clear the denominator by multiplying both sides by $x$:
\begin{align*}
z &= \frac{100(x-y)}{x}\\
zx &= 100(x-y)
\end{align*}
Then move all the terms that have an $x$ in it to one side of the equation, all other terms to the other side, and factor out the $x$:
\begin{align*}
zx &= 100x - 100y\\
zx - 100x &= -100y\\
x(z-100) &= -100y
\end{align*}
Now divide through by $z-100$ to solve for $x$; you have to worry about dividing by $0$, but in order for $z-100$ to be $0$, you need $z=100$; the only way for $z$ to be equal to $100$ is if $\frac{x-y}{x}=1$, that is, if $x-y=x$, that is, if $y=0$. Since, presumably, you don't get the things for free, you can assume that $y\neq 0$ so this division is valid. You get:
$$x = \frac{-100y}{z-100}.$$
Now, to get it into nicer form, use the minus sign in the numerator to change the denominator from $z-100$ to $100-z$. Then divide both the numerator and the denominator by $100$ to get it into the form you have:
\begin{align*} 
x & = \frac{-100y}{z-100}\\
x &= \frac{100y}{100-z}\\
x &= \frac{\frac{1}{100}\left(100 y\right)}{\frac{1}{100}(100-z)}\\
x &= \frac{y}{1 - \frac{z}{100}}.
\end{align*}
Added: Alternatively, following Myself's very good point, you can go "unsimplify" $\frac{x-y}{x}$ to $1 - \frac{y}{x}$, to go from 
$$\frac{z}{100} = \frac{x-y}{x} = 1 - \frac{y}{x}$$
to
$$\frac{y}{x} = 1 - \frac{z}{100}.$$
Taking reciprocals and multiplying through by $y$ gives
\begin{align*}
\frac{x}{y} = \frac{1}{1 - \frac{z}{100}}\\
x = \frac{y}{1-\frac{z}{100}}
\end{align*}
which is probably how the particular expression you had (as opposed to $\frac{100y}{100-z}$) arose in the first place. 
</s>