Question

Решите систему уравнений х²+у²=25 y-x=7

Answer 1

$\left \{ {{x^2+y^2=25} \atop {y-x=7}} \right. \left \{ {{x^2+y^2=25} \atop {y=7+x}} \right. \left \{ {{x^2+y^2=25} \atop {y^2=49+14x+x^2}} \right. \left \{ {{x^2+49+14x+x^2=25} \atop {y^2=49+14x+x^2}} \right. \\\left \{ {{2x^2+14x+24=0} \atop {y^2=49+14x+x^2}} \right. \\2x^2+14x+24=0\\x^2+7x+12=0\\D=b^2-4ac=49-4*12=1\\x_{1,2}= \frac{-bб \sqrt{D} }{2a} \\ |{ {{x= \frac{-7+1}{2} =-3} \atop {x= \frac{-7-1}{2}=-4}} \right. \\ \left \{ {{x=-3} \atop {y^2=49+52+25}} \right. \left \{ {{x=-3} \atop {y^2=126}} \righ$
$\left \{ {{x=-4} \atop {y^2=49+56+9}} \right. \left \{ {{x=-4} \atop {y^2=114}} \right. \left \{ {{x=-4} \atop {y=\sqrt{114}}} \right.$
ответ: $(-3; \sqrt{126} );(-4;\sqrt{114})$