Помогите пожалуйста алгебра. С.м работа

Question 1

Question 2

Решите задачу:

$1. \left \{ {{x^2-5y^2}=11 \atop {x^2+5y^2=61}}+ \right. \\ ------- \\ \left \{ {{} 2x^2=72;|:2 }} \right. \left \{ {{} x^2=36; }} \right. ; \\ x_1=6; \ x_2=-6 =\ \textgreater \ \\ (-6)^2-5y^2=11; \ 36-5y^2=11; \ -5y^2=-25; \ y_1=5, y_2=-5; \\ \left \{ {{x_{1,2}=+-6} \atop {y_{1:2}=+-5}} \right. \\ 2. \ \left \{ {{x+y=6} \atop {xy=8}} \right. ; \ \left \{ {{x=-y+6} \atop {(-y+6)*y=8}}; \right. ; \\ -y^2-6y-8=0; |*(-1); \ y^2+6y+8=0 ; \ (y+2)(y+4)=0; \\ y_1=-2 ; \ y_2=-4; \\ ]x_1=-(-2)+6; \ x_1=8; \ x_2=-(-4)+6; x_2=10; \\ \left \{ {{x_{1}=8, \ x_2=10 \atop {y_1=-2, \ y_2=-4}} \right. ; \\$
$\left \{ {{x+y=-2} \atop {x-y=-8}}+ \right. \\ ----- \\ \left \{ {{2x=-10; } \right.; \\ \left x=-5; \right. \\ -5-y=-8; \ -y=-3; \ y=3; \\ \left \{ {{x=-5} \atop {y=3}} \right. \left \{ {{x^2-y^2=6} \atop {x^2+y^2=10}}+ \right. \\ ----- \\ 2x^2=16; \\ x^2=8; \\ x_{1.2}=-+ \sqrt{8}; \\ ( \sqrt{8})^2-y^2=6; \\ -y^2=-2; \\ y_{1,2}=+- \sqrt{2}; \\ \left \{ {{x_{1,2}=+- \sqrt{5} } \atop {y_{1,2}=+- \sqrt{2}} \right. \\ \left \{ {{3x+y=2} \atop {x^2-xy+6y=-4}} \right.; \ \left \{ {{y=-3x+2} \atop {(-3x+2)^2-x(-3x+2)}+6(-3x+2)=-4} \right. ; \\ 12x^2-32+16=-4; \\ 12x^2-32+20=0;|:2 \\ 6x^2-16+10=0; \\ D=b^2-4ac; \ 16^2-4*6*10; \ 256-240=16, D\ \textgreater \ 0; \\ x_{1,2}= \frac{-b+- \sqrt{D} }{2a}; \\ x_1= \frac{16+4}{12}; \ x_1= 1\frac{2}{3}; \\ x_2= \frac{12}{12}=1; \\ 3*( \frac{5}{3})+y=2; \\ y_1=-3; \\ 3*1+y=2; \\ y_2=-1; \\ \left \{ {{x_{1}=1 \frac{2}{3} };x_2=1 \atop {y_1=-3; y_2=-1}} \right.$

Question 3

A - не пишите, это ошибка при вводе кода.