Решите систему уравнений
{4х+4ху+у^2=25 {2х-у=3

{4х+4ху+у^2=25{2х-у=3у=(2х-3)4х+4х×(2х-3)+(2х-3)²=254х+8х²-12x+((2х)²-2×2x×3+3²)=258x²-8x+4x²-12x+9-25=012x²-20x-16=0|÷43x²-5x-4=0D=(-(-5))²-4×3×(-4)=25+48=72x1=(-(-5)-√72)/2×3=(5-8,485)/6=(-3,485)/6=-0,581x2=(-(-5)+√72)/2×3=(5+8,485)/6=13,485/6=2,247y1=(2x1-3)y1=(2×(-0,581)-3)y1=(-1,162-3)y1=-4,162y2=(2x2-3)y2=(2×2,247-3)y2=(4,494-3)y2=1,494(-0,581;-4,162); (2,247;1,494)