Question

Решите систему уравнений:
x^2-4xy+4y^2=25
x+2y=3

Answer 1

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Answer 2

X^2-4xy+4y^2=25
x+2y=3 

(х-2y)² = 5²
x= 3-2y

(x-2y)²-5² = 0
x = 3-2y

(x-2y-5)*(x-2y+5) = 0
x = 3-2y

(3-2y-2y-5)*(3-2y-2y+5) = 0
x = 3-2y

(-4y-2)*(-4y+8) = 0
x= 3-2y

-4y-2=0
-4y=2
y1=-1\2

-4y+8 = 0
-4y = -8
y2 = 2
x1= 3-2*(-1\2) = 4
x2 = 3-2*2= -1

Ответ: (4;-1\2) (-1;2)