Formula for finding $a$ by $b$ and $c$ I have the formula:
$$ c = \frac{a}{100(a + b)} $$
How to find $a$ by $b$ and $c$?
$$ a = \text{?} $$
 A: First, multiply both sides by $100(a+b)$ to get
$$ c \cdot 100(a+b) = a.$$
Then, distribute the $c$ and $100$ to $(a+b)$ on the left side to get
$$100ca + 100cb = a.$$
Now, bring every term with an $a$ in it to the left side. We do this by subtracting $a$ from both sides and then subtracting $100cd$ from both sides. This will give you
$$100ca - a = -100cb.$$
From here, factor out the $a$ on the left side to get
$$a(100c - 1) = -100cb.$$
Can you take it from here to solve for $a$?
A: The first thing is to get rid of denominators containing $a$:
$$
c = \frac{a}{a + b}\cdot 100\\
c\cdot(a + b) = a\cdot 100
$$
The next thing is to multiply out parentheses containing $a$, and group all terms containing $a$ on one side:
$$
c\cdot(a + b) = a\cdot 100\\
ca + cb = a\cdot 100\\
ca - a\cdot 100 = -cb
$$
Then you factor out the $a$, leaving you with only on $a$ in the entire equation (this is a good thing):
$$
ca - a\cdot 100 = -cb\\
a(c - 100) = -cb
$$
The last thing to do is to divide by whatever stands next to $a$:
$$
a(c - 100) = -cb\\
a = \frac{-cb}{c-100}
$$
some might prefer to get rid of as many minuses as they can, so they rewrite
$$
a = \frac{-cb}{c-100}\\
a = \frac{cb}{100 - c}
$$
but this is not strictly necessary.
A: Multiply by $100$ and take reciprocals to get $\frac{1}{100c}=1+\frac{b}{a}$.  Now subtract $1$ and take reciprocals again: 
$$\frac{1}{\frac{1}{100c}-1}=\frac{a}{b}.$$
Now multiply by $b$ to get
$$a=\frac{b}{\frac{1}{100c}-1}=\frac{100bc}{1-100c}.$$
