РЕШИТЬ СИСТЕМУ УРАВНЕНИЯ! Срочнооо ; X в квадрате -5xy=10 X-5y=1

[latex] \left \{ {{x^2-5xy=10} \atop {x-5y=1}} \right. => \left \{ {{x=1-5y} \atop {(1-5y)^2-5y(1-5y)=10}} \right.=> \left \{ {{x=1-5y} \atop {(1-5y)(1-5y-5y)=10}} \right. =>\\ => \left \{ {{x=1-5y} \atop {-10y-5y+1-50y^2=0}} \right. => \left \{ {{x=1-5y} \atop {50y^2+15y-1=0}} \right. [/latex] 50y^2+15y-1=0 D=225+200=425 x1 = (245-15)/100=2.3 x2 = -260/100=-2.6 y1 = 1- 2.3 * 5= -10.5 y2= 1+2.6*5=14 S={(2.3;  -10.5);  (-2.6; 14 )}