Question

Решите систему уравнений {x-y=1 {x2-y=3

Answer 1

X = y + 1
(y + 1)^2 - y - 3 = 0 *

y^2 + 2y + 1 - y - 3 = 0
y^2 + y - 2 = 0
D = 1 + 4*2 = 9
y1 = ( - 1 + 3)/2 = 1
y2 = ( - 1 - 3)/2 = - 2

x1 = y1 + 1 = 1 + 1 = 2
y1 = 1

x2  = y2 + 1 = - 2 + 1 = - 1 
y2 = - 2 

ответ
( - 1; - 2) ;
( 2; 1)

Answer 2

Найдем разность обеих частей, получится:
х-у-(х²-у)=1-3
х-у-х²+у=-2
х-х²=-2
-х²+х+2=0
х²-х-2=0
х₁*х₂=-2
х₁+х₂=1
х₁=2, х₂=-1
$1) \left \{ {{x=2} \atop {x-y=1}} \right. \\ \left \{ {{x=2} \atop {2-y=1}} \right.\\ \left \{ {{x=2} \atop {y=1}} \right.\\ 2) \left \{ {{x=-1} \atop {x-y=1}} \right. \\ \left \{ {{x=-1} \atop {-1-y=1}} \right. \\ \left \{ {{x=-1} \atop {y=-2}} \right. \\ OTBET: (2;1),(-1;-2).$