Помогите, пожалуйста решить

Question 1

Question 2

:)

Question 3

Решите задачу:

$1)\; \left \{ {{xy=-12} \atop {x^2+y^2=25}} \right. \; \left \{ {{2xy=-24} \atop {x^2+2xy+y^2=-24+25}} \right. \; \left \{ {{xy=-12} \atop {(x+y)^2=1}} \right. \; \left \{ {{xy=-12} \atop {x+y=\pm 1}} \right. \\\\a)\; \; \left \{ {{xy=-12} \atop {x+y=1}} \right. \; \left \{ {{x=-\frac{12}{y}} \atop {-\frac{12}{y}+y=1}} \right. \; \left \{ {{x=\frac{-12}{y}} \atop {y^2-y-12=0,\; y\ne 0}} \right. \; y_{1,2}=-3,4\\\\x_{1,2}=4,-3$

$b)\; \left \{ {{xy=-12} \atop {-\frac{12}{y}+y=-1}} \right. \; \left \{ {{x=\frac{-12}{y}} \atop {\frac{-12}{y}+y=-1}} \right. \; \left \{ {{x=\frac{-12}{y}} \atop {y^2+y-12=0}} \right. \; ;\; y_{1,2}=3,-4\\\\x_{1,2}=-4,3\\\\Otvet:\; (-3,4),(4,-3),(3,-4),(-4,3).$

$3)\; \left \{ {\frac{x+3y}{ 2}=\frac{x-2y}{3}+31\, |\cdot 6\atop {x+y=\frac{3}{4}(x-y)+27\, |\cdot 4}} \right. \; \left \{ {{3(x+3y)=2(x-2y)+186} \atop {4(x+y)=3(x-y)+108}} \right. \; \left \{ {{x+13y=186} \atop {x+7y=108}} \right. \\\\ \left \{ {{13y-7y=186-108} \atop {x+7y=108}} \right. \; \left \{ {{6y=78} \atop {x=108-7y}} \right. \; \left \{ {{y=13} \atop {x=17}} \right. \\\\Otvet:\; (17,13).$

$2)\; \left \{ {{3(3x-y)+\frac{2}{5}(x-2y)=17\, |\cdot 5} \atop {5x+y=3}} \right. \; \left \{ {{15(3x-y)+2(x-2y)=85} \atop {y=3-5x}} \right. \; \left \{ {{47x-19y=85} \atop {y=3-5x}} \right. \\\\ \left \{ {{47x-19(3-5x)=85} \atop {y=3-5x}} \right. \; \left \{ {{142x=142} \atop {y=3-5x}} \right. \; \left \{ {{x=1} \atop {y=-2}} \right. \\\\Otvet:\; (1,-2)$

Question 4

Спасибо!