Question 465558
3^x * 9^2y = 1

=> 3^(x+4y) = 1, [ because  9 = 3^2]

so, x+ 4y =0,[ because 3^0 =1  ]


5 ^(3x+2y) = 5^(-3)

so, 3x+2y= -3

now, multiply the second equation by 2 and subtract from first equation, 

x -6x = 0 - (-6)

=> -5x = 6 

so, x = -6/5 

and y = 6/20 


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