= 3     (1)

 = 9     (2)



Right side of (2) is 9 = 3*3.  Replace one factor of 3 by the left side of (1).  You will get


 = ,   or, equivalently


 = 0.    (3)  


Now factor left side of (3) to get

(x-2y)*(3x+y) = 0.


Thus EITHER  x-2y = 0  OR  3x+y = 0.


Lets consider each case separately.



1)  x-2y = 0  ====>  x = 2y   ====>  substitute it into eq(1)  ====>   = 3  ====>   = 3  

              ====>   =   ====>  y = +/-  = +/- .


              If y =    <--->  x = ,    and both equations (1) and (2) are satisfied.

              If y =   <--->  x = ,   and both equations (1) and (2) are satisfied.



2)  3x+y = 0  ====>  y = -3x  ====>  substitute it into eq(1)  ====>   = 3    ====>   = 3  

              ====>   = 3/4  ====>  x = +/- .

              
              If x =    <--->  y = ,    and both equations (1) and (2) are satisfied.

              If x =   <--->  y = ,     and both equations (1) and (2) are satisfied.


Answer.  The system (1), (2) has four solutions  (x,y) = (,),  (,),  (,),  (,).