Помогите решить систему log 4(x) + log4(y)=1, y-2x=7

[latex]log_{4}x+lg_{4}y=1\\ y-2x=7\\\\ y-2x=7\\ y=7+2x\\\\ log_{4}x+lg_{4}y=1\\ lg_{4}(x*y)=1\\ lg_{4}(x(7+2x))=1\\ lg_{4}(x(7+2x))=lg_{4}4\\ x(7+2x)=4\\ 7x+2x^2-4=0\\ D=49-4*2*(-4)=81\\ x_{1,2}=\frac{-7\pm 9}{4} = 0,5; -4.\\ y_{1}=7+2(0,5)=8\\ y_{2}=7+2*(-4)=-1\\\\ x_{1}=0,5; \ x_{2}=-4; \ y_{1}=8; \ y_{2}=-1[/latex]