Fewest number of marbles in a bag such that drawing the probability of drawing 2 blue marbles is $\frac{1}{6}$. 
Two marbles are randomly selected without replacement from a bag
  containing blue and green marbles. Probability of drawing both blue is
  $\frac{1}{6}$. If three marbles are drawn, then the probability all
  three are blue is $\frac{1}{21}$. What is the fewest number of marbles
  that must have been in the bag before any were drawn?

I use the criteria to make two equations:
$$\frac{x}{y}\cdot\frac{x-1}{y-1}=\frac{1}{6}$$
for drawing 2 marbles$$\frac{x}{y}\cdot\frac{x-1}{y-1}\cdot\frac{x-2}{y-2}=\frac{1}{21}$$
Where $x$ is the number of blue marbles and $y$ is the total number of marbles.
Simplifying these equations gives:
$$6x^2-6x=y^2-y$$ for the first equation and $$21x^3-63x^2+42x=y^3-3y^2+2y$$
I can't seem to solve this set of equations. 
Is my approach correct? If so, how should I continue? If not, how would one solve this problem?
Thanks! Your help is appreciated!
Max0815

Let's look at the second equation:
$$\underbrace{\dfrac{x}{y}\cdot \dfrac{x-1}{y-1}}_{\dfrac{1}{6}}\cdot \dfrac{x-2}{y-2} = \dfrac{1}{21}$$
The first two terms are the same as the LHS of your first equation. So, we can substitute.
$$\dfrac{x-2}{y-2} = \dfrac{6}{21}$$
Reducing the RHS to lowest terms gives:
$$\dfrac{x-2}{y-2} = \dfrac{2}{7}$$
There you have $x-2=2, y-2=7$
So, $x=4, y=9$ should be the smallest possible values for which the problem should hold true (if there were smaller values, it would contradict $\dfrac{2}{7}$ being in lowest terms).
