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
 A: 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).
A: Divide the second equation by the first:
$$\frac{x-2}{y-2} = \frac{6}{21} = \frac{2}{7}$$
Which when simplified gives:
$$7x = 2y + 10$$
What are the smallest integer values of x and y that solve this equation? Well, observe that a possible solution is $x = 2, y = 2$, but this is clearly wrong since $x = y$ in this case. Since $x$ cannot be odd, we go on to the next number, which is 4. Luckily, we can observe that a solution will be given by: $$x = 4, y = 9$$
which happens to be the correct answer. 
A: 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.
