Х/3+у=1  у квадрат - ху=7

[latex] \left \{ {{\frac{x}{3} +y=1} \atop {y^2-xy=7}} \right. \\ \left \{ {{x+3y=3} \atop {y(y-x)=7}} \right. \\ \left \{ {{x=3-3y} \atop {y(y-3+3y)=7}} \right. \\ \\ y(4y-3)=7 \\ 4y^2-3y-7=0 \\ D=9+112=121 \\ \sqrt{D} =11 \\ \\ y1= \frac{3-11}{8} =-1 \\ x1=3-3*(-1)=3+3=6 \\ \\ y2= \frac{3+11}{8} = \frac{14}{8} =1.75 \\ x2=3-3*1.75=-2.25[/latex]