Let X = x^2 and Y = y^2
So we have 9X^2 - 4Y^2 = 3X+2Y
(3X+2Y) * ( 3X- 2Y) = 3X+2Y
3X-2Y = 1
3X =1+ 2Y
X =(1+ 2Y)/3 (or)
x^2 =(1+ 2y^2)/3
C IMO
x and y
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Note that for x = y = 1, the given relation holds.gmatblood wrote:If x and y are non-zero integers, and 9x^4 - 4y^4 = 3x^2 + 2y^2, which of the following could be the value of x^2 in terms of y?
Hence, the correct option is the one which is equal to x^2 = 1 for y = 1.
- A. -4(y^2)/3 = -4/3 ----> NO
B. -2y^2 = -2 ----> NO
C. (2[y^2]+1)/3 ----> (2 + 1)/3 = 1 ----> YES
D. 2y^2 = 2 ----> NO
E. [6y]^2/3 = (6^2)/3 = 12 ----> NO
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