Volume of a frustum given the bottom radius and the top cone height. 
A cone with base radius 12 cm is sliced parallel to its base, as shown,
  to remove a smaller cone of height 15 cm. If the height of the smaller
  cone is three-fourths that of the original cone, what is the volume of the remaining frustum?


I set the frustum's top radius as x and its height as y. Using the cone volume formula, I have $$\frac{144\pi\cdot(y+15)}{4}=\frac{15x^2\pi}{3}$$$$\implies 36 y+540=5x^2$$
I am stuck here. How should I continue?

Hint:
The smaller cone is homothetic of the larger cone in a homothety with centre the vertex of the cones and ratio $3/4$. So the height $h$ of the smaller cone is $h=3/4$ of the height $H$ of the larger cone, its base area is $b=9/16B$ and its volume is $v=27/64V$. So the remaining frustum has volume 
$$\mathcal V=V-v =\frac{37}{64}V.$$
Can you calculate the volume of the given cone?
