пусть (Х0; У0) - решение системы х- 2у= -3
у квадрат -2х=3.
найдите Х0+2У0

X - 2y = - 3 y^2 - 2x = 3 ••••••X = 2y - 3 2x = 4y - 6 2x = y^2 - 3 ••••••••4y - 6 = y^2 - 3 y^2 - 4y + 3 = 0 D = 16 - 12 = 4 = 2^2 y1 = ( 4 + 2 ) : 2 = 3 y2 = ( 4 - 2 ) : 2 = 1 ••••••••••X = 2y - 3 X1 = 6 - 3 = 3 X2 = 2 - 3 = - 1 ____________1) 3 + 2*3 = 9 2) - 1 + 2*1 = 1 Ответ 9; 1 

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