Question

x + y + xy = 7
x^2 + y^2 = 10

Answer 1

$\displaystyle \left \{ {{x+y+xy=7} \atop {x^{2}+y^{2}=10 \ \ \ }} \right.\\\\\\\left \{ {{x(1+y)=7-y} \atop {x^{2}+y^{2}=10 \ \ \ \ \ }} \right.\\\\\\\left \{ {{x=\displaystyle\frac{7-y}{1+y}} \ \ \ \ \ \ \ \ \ \ y\neq-1 \atop {x^{2}+y^{2}=10 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }} \right.\\\\\\\frac{(7-y)^{2}}{(1+y)^{2}}+y^{2}=10\\\\(7-y)^{2}=(10-y^{2})\cdot(1+y)^{2}\\\\49-14y+y^{2}=10+20y+10y^{2}-y^{2}-2y^{3}-y^{4}\\\\y^{4}+2y^{3}-8y^{2}-34y+39=0\\\\y^{4}-4y^{3}+3y^{2}+6y^{3}-24y^{2}+18y+13y^{2}-42y+39=0$

$\displaystyle y^{2}(y^{2}-4y+3)+6y(y^{2}-4y+3)+13(y^{2}-4y+3)=0\\\\(y^{2}+6y+13)(y^{2}-4y+3)=0\\\\(y-3)(y-1)(y^{2}+6y+13)=0$

Так как  у² + 6у + 13  положительно при любых у, то:

$\displaystyle \\y_{1}=3 \ \ \ \ \ \ \ x_{1}=1\\\\y_{2}=1 \ \ \ \ \ \ \ x_{2}=3$

Ответ: {3; 1}, {1; 3}

Answer 2

$\displaystyle\left \{ {{x+y+xy=7} \atop {x^2+y^2=10}} \right. ~~\Leftrightarrow~~\left \{ {{x+y+xy=7~~~~~~~~~~~\big|\cdot 2} \atop {(x^2+y^2+2xy)-2xy=10}} \right.\\\\\\\left \{ {{2\big(x+y\big)+2xy=14} \atop {\big(x+y\big)^2-2xy=10}} \right.~~+\\\\\overline{~\big(x+y\big)^2+2\big(x+y\big)=24}$

Замена переменных   (x + y) = z

$z^2+2z-24=0\\(z+6)(z-4)=0\\z_1=-6;~~~z_2=4$

Обратная замена переменных

$\displaystyle1)~z=-6;~~x+y=-6\\\\~~~~\left \{ {{x+y=-6} \atop {x+y+xy=7}} \right.~~~~\left \{ {{y=-6-x} \atop {x+(-6-x)+x(-6-x)=7}} \right.\\\\\\~~~~\left \{ {{y=-6-x} \atop {-6-6x-x^2=7}} \right.~~~~\left \{ {{y=-6-x} \atop {x^2+6x+13=0}} \right.\\\\~~~~~x^2+6x+13=0\\~~~~~D=6^2-4\cdot13=-16<0~~\Rightarrow~~x\in\varnothing$

$\displaystyle2)~z_2=4;~~x+y=4\\\\~~~~\left \{ {{x+y=4} \atop {x+y+xy=7}} \right.~~~~\left \{ {{y=4-x} \atop {x+(4-x)+x(4-x)=7}} \right.\\\\\\~~~~\left \{ {{y=4-x} \atop {4+4x-x^2=7}} \right.~~~~\left \{ {{y=4-x} \atop {x^2-4x+3=0}} \right.\\\\\\~~~~\left \{ {{y=4-x} \atop {(x-3)(x-1)=0}} \right.~~~~\left \{ {{y_1=1;~y_2=3} \atop {x_1=3;~x_2=1}} \right.$

Ответ : (1;3);  (3;1)