Question

Системы:х+у/2=1 
х-у=3.

ху=-14
х-у=9.

Answer 1

$begin{cases}x+frac{y}{2}=1|*2\x-y=3end{cases}<=>begin{cases}2x+y=2\+\x-y=3end{cases}\\3x=5\x=frac{5}{3}\frac{5}{3}-y=3\y=-1frac{1}{3}=-frac{4}{3}\OTBET:(frac{5}{3};-1frac{1}{3})$

$begin{cases}xy=-14\x-y=9=>x=9+yend{cases}\y(9+y)=-14\y^2+9y+14=0\y_{1,2}=frac{-9^+_-5}{2}\y_1=-7 ;y_2=-2\x+7=9 x+2=9\x_1=2 x_2=7\OTBET:(2;-7);(7;-2)$