Multiplying the denominator $$\frac{2}{3} \gt -4y - \frac{25}{3}$$
My question is to solve this problem we need to multiply both sides by $3$. The result will be 
$$2 \gt -12y - 25.$$
Why doesn't the numerator of both $\frac{2}{3}$ and $\frac{25}{3}$ get multiplied?
 A: As an expansion of what I said in the comments:
When we multiply the equation $\frac{2}{3} > -4y - \frac{25}{3}$  by $3$ we get the following equation
$\frac{2\times3}{3} > (-4y)\times3 - \frac{25\times3}{3}$. Simplifying the following we get
$\frac{6}{3} > -12y - \frac{75}{3}$.
Now $\frac{6}{3} =2$ and $\frac{75}{3} = 25$. Thus utilizing these known facts we can simplify the above equation giving us
$2 > -12y - 25$. So the numerators got multiplied by $3$, but then we divided by $3$, which amounts to returning to the original numerator. This is because multiplication and division are inverse operators. 
From your comment, if you changed the $3$ in the denominator to a $6$ you would not get the same result. Instead of ending up with $25$, you would get $\frac{25}{2}$.
A: $$\begin{align}
\frac{2}{3}
&>
-4y-\frac{25}{3} \\
3\left(\frac{2}{3}\right)
&>
3\left(-4y-\frac{25}{3}\right) \\
3·\frac{2}{3}
&>3·\left(-4y\right)-3·\frac{25}{3} \\
\require{cancel} \cancel{3}·\frac{2}{\require{cancel} \cancel{3}} 
&>
3·\left(-4y\right) -\require{cancel} \cancel{3} ·\frac{25}{\require{cancel} \cancel{3}} \\
2
&>
-12y -25 \\
\end{align}$$
A: If you have something like $\frac{a}{b}>c-\frac{d}{b}$ and you are trying to get rid of the fraction on the left side, then we have
$$\left(\frac{a}{b}\right)\left(b\right)>c(b)-\left(\frac{d}{b}\right)\left(b\right)$$
$$\frac{ab}{b}>c(b)-\frac{db}{b}$$
$$a\cdot\frac{b}{b}>c(b)-d\cdot\frac{b}{b}$$
$$a\cdot1>c(b)-d\cdot1$$
$$a>c(b)-d$$
So as you can see, the numerators are in fact multiplied by the denominator you are trying to remove, $b$, but it gets "cancelled out" because $b$ divided by itself is $1$ given that $b\ne 0$
