решите алгебру пожалуйста

{х+2у=1----------------> x=1-2y{ ху=-1 -----------------> (1-2y)*y = -1 -------------> y-2y^2 = -1 -----------> 2y^2-y-1 = 0=>
y1 = 2

y2 = -1

x1 = 1 -2y1 = 1-2*2 = -3

x2 = 1 - 2y2 = 1 - 2*(-1) = +3

{ х^2 + ху = 6{ х - у = 4 ---------------------> x = 4 + y

=>

(4 + y)^2 + (4+y)*y = 6

16 + 8y + y^2 + 4y + y^2 - 6 = 0

2y^2 + 12y + 10 = 0

y^2 + 6y + 5 = 0

y1 = -5

y2 = -1

=>

x1 = 4 + y1 = 4 + (-5) = -1

x2 = 4 + y2 = 4 + (-1) = 3

{ 4х^2 - ху = 26{ 3х^2 + ху = 2

=>

(4x^2 - xy) + (3x^2 + xy) = 26 + 2

7x^2 = 28

x^2 = 4

x1 = +2

x2 = -2

=>

3x^2 + xy = 2

3*(x1)^2 + (x1)*(y1) = 2

3*2 + 2*(y1) = 2

6 + 2*(y1) = 2----------------> y1 = (-4)/2 = -2

3*(x2) + (x2)*(y2) = 2

3*(-2) + (-2)*(y2) = 2

-6 - 2*(y2) = 2 ----------------> y2 = -8/2 = -4