Question

невиконуючи побудови знайдіть координати точок перетину графіків рівнянь х-5у=3 і х²-2ху-у²=-1​

Answer 1

Ответ: (-2;-1),  (-4/7;-5/7).

Объяснение:

$\displaystyle\\\left \{ {{x-5y=3} \atop {x^2-2xy-y^2=-1}} \right. \ \ \ \ \left \{ {{x=3+5y} \atop {x^2-2xy-y^2+y^2=-1+y^2}} \right. \ \ \ \ \left \{ {{x=5y+3} \atop {(x-y)^2-y^2=y^2-1}} \right. \\\\\\\left \{ {{x=5y+3} \atop {(5y+3-y)^2-y^2=y^2-1}} \right. \ \ \ \ \left \{ {{x=5y+3} \atop {(4y+3)^2-2y^2+1=0}} \right.\ \ \ \ \left \{ {x=5y+3} \atop {16x^2+24y+9-2y^2+1=0}}\right.$

$\displaystyle\\\left \{ {{x=5y+3} \atop {14y^2+24y+10=0\ |:2}} \right. \ \ \ \ \left \{ {{x=5y+3} \atop {7y^2+12y+5=0}} \right. \ \ \ \ \left \{ {{x=5y+3} \atop {7y^2+7y+5x+5=0}} \right.\\\\\\\left \{ {{x=5y+3} \atop {7*y*(y+1)+5*(y+1)=0}} \right.\ \ \ \ \left \{ {{x=5y+3} \atop {(y+1)*(7y+5)=0}} \right. \ \ \ \ \left \{ {{x_1=-2\ \ \ \ x_2=-\frac{4}{7} } \atop {y_1=-1\ \ \ \ y_2=-\frac{5}{7} }} \right. .$