Решить систему уравнений:№1у - 2х = 9х^2 + 2у = 14№24х - у + 1 = 0х^2 + ху + у^2 = 13№3х^2 - у^2 = -24х - у = 2

[latex]1)\left \{ {{y-2x=9} \atop {x^2+2y=14}} \right. \left \{ {{y=9+2x} \atop {x^2+2(9+2x)=14}} \right.\\x^2+4x+4=0\\D=0\\x=-2;y=5 \\2) \left \{ {{4x-y+1=0} \atop {x^2+xy+y^2=13}} \right. \left \{ {{y=1+4x} \atop {x^2+x(1+4x)+(1+4x)^2=13}} \right.\\ 7x^2+3x-4=0\\D=11^2\\x_1=-1;x_2=- \frac{4}{7} \\y_1=-3;y_2=-1 \frac{9}{7} \\3) \left \{ {{x-y=2} \atop {x^2-y^2=-24}} \right. \left \{ {{x=2+y} \atop {(2+y)^2-y^2=-24}} \right. \left \{ {{x=-5} \atop {y=-7}} \right. [/latex]