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

As Sean Lee and InterstellarProbe note, dividing the second equation by the first leaves
$${x-2\over y-2}={6\over21}={2\over7}$$
It follows that $x-2=2k$ and $y-2=7k$ for some $k\not=0$.  Plugging $x=2k+2$ and $y=7k+2$ into $6x^2-6x=y^2-y$, we have
$$6(2k+2)^2-6(2k+2)=(7k+2)^2-(7k+2)$$
which expands and simplifies initially to
$$24k^2+36k+12=49k^2+21k+2$$
and then to
$$25k^2-15k-10=5(5k+2)(k-1)=0$$
This has two solutions, $k=1$ and $k=-2/5$.  The first gives a meaningful solution in integers, $x=4$, $y=9$.  The second gives a meaningless non-solution, $x=1/5$, $y=-4/5$.  Thus $9$ marbles (with $4$ of them blue) is not only the fewest number of marbles you can have to start with, it's the only number.
