solve the simultaneous equation

I assume we have :

9^x=27^y     (1)

64^(xy) = 512^(x+1)   (2)

(1)  can be written as

(3^2)^x  = (3^3)^y    → (3)^(2x)  = (3)^(3y)  ... equate exponents  →   2x = 3y  →  y  = (2/3)x

Sub this into (2)

64^(x* (2/3)x )  =  512^(x + 1)

(2^6)^[ (2/3)x^2]  = (2^9)^(x + 1)

(2)^[4x^2]  = (2)^[9x + 9]     equate exponents again

4x^2  = 9x + 9

4x^2 - 9x - 9  = 0      factor

(4x + 3)(x - 3)  = 0   set each factor to 0   

And

x = -3 / 4    → y  = (2/3)(-3/4)  =  -1/2 

Or

x  = 3    →  y  = (2/3)(3)   =  2

So....the solutions are   (x, y )  =    ( - 3/4, -1/2)     or  (3, 2)