Question

а) {xy = -2, x+y=1; б) {5x в квадрате + 2y = -3, x-y=5;

Answer 1

б) $left { {5x^{2}+2y=-3} atop {x-y=5}} right$

 

y=$y=5+x$

 

$5x^{2}+2(5+x)=-3$

 

$5x^{2}+2x+13=0$

 

дальше решите это уравнение и найдите у

Answer 2

a)

$left { {{xy = -2} atop {x+y=1}} right.\ left { {{xy = -2} atop {x=1-y}} right.\ left { {{(1-y)*y = -2} atop {x=1-y}} right.\ left { {{-y^2+y+2=0} atop {x=1-y}} right.\ 1) -y^2+y+2=0;\ D=1+8=9;\ y1=(-1+3)/-2=-1;\ y2=(-1-3)/-2=2;\ 2) x1=1-(-1);\ x1=2;\ x2=1-2\ x2=-1;$

Ответ:

x1=-1;y1=2;

x2=2;y2=-1;

б)

$left { {{5x^2+2y=-3} atop {x-y=5}} right.\ left { {{5x^2+2y=-3} atop {x-5=y}} right.\ left { {{5x^2+2*(x-5)=-3} atop {y=x-5}} right.\ 1) 5x^2+2x-10+3=0;\ 5x^2+2x-7=0\ D=4+140=144;\ x1=(-2+12)/10=1;\ x2=(-2-12)/10=-1,4;\ y1=1-5;\ y1=-4;\ y2=-1,4-5;\ y2=-6,4;$

Ответ:

x1=-7/5(-14/10);y1=-32/5;(-64/10);

y1=1;y2=-4;