Question

решите систему уравнений { х^2=4y+1  
                                           x^2+3=4y+y^2

Answer 1

X^2 = 4Y + 1 
X^2 = Y^2 + 4Y - 3 
---------------------
4Y + 1 = Y^2 + 4Y - 3 
Y^2 + 4Y - 3 - 4Y - 1 = 0 
Y^2 = 4
Y1 = + 2 
Y2 = - 2 
----------------
X^2 = 4Y + 1  ( X^2 > 0 )
X1 ^2 = 8 + 1 = 9 
X1 = + 3 
X2 = - 3 
---------------
X2 ^2 = - 8 + 1 = - 7 ( < 0 ) 
----------------------