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

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. 
