Question

решите систему уравнений х^2+y^2=50 и xy=7

Answer 1

$\displaystyle \left \{ {{x^2+y^2=50} \atop {xy=7 |*2}} \right. \\\\\left \{ {{x^2+y^2=50} \atop {2xy=14}} \right.$

сложим первое и второе уравнение

$\displaystyle x^2+2xy+y^2=50+14\\\\(x+y)^2=64\\\\\left \{ {{x+y=8} \atop {x+y=-8}} \right.$

рассмотрим первый случай

$\displaystyle \left \{ {{x+y=8} \atop {xy=7}} \right.\\\\\left \{ {{x=8-y} \atop {(8-y)*y=7}}\right. \\\\\left \{ {{x=8-y} \atop {y^2-8y+7=0}} \right.\\\\y_1=7; y_2=1\\\\x_1=1; x_2=7$

рассмотрим второй случай

$\displaystyle \left \{ {{x+y=-8} \atop {xy=7}} \right.\\\\\left \{ {{x=-8-y} \atop {(-8-y)y=7}} \right.\\\\\left \{ {{x=-8-y} \atop {y^2+8y+7=0}} \right.\\\\y_3=-7; y_4=-1 \\\\x_3=-1; x_4=-7$