решите систему уравнений {(3x+y)^2+2(x-y)^2=96{3x+y=2(x-y)

9х^2 + 6ху + у^2 +2х^2 -4ху +2у^2 =963х+у =2х -2у11х^2 +2ху +3у^2 = 96х+3у = 011х^2 +2ху +3у^2 = 96х=-3у99у^2 - 6у^2 +3у^2 = 96х = -3у96у^2 = 96х = -3уу^2 =1х = -3уу1=1у2= -1х1 = -3х2= 3