решите систему {х-у=1 х²-у=3

x = y + 1(y + 1)^2 - y - 3 = 0 **y^2  + 2y + 1 - y - 3 = 0 y^2 + y - 2 = 0 D = 1 +  8 = 9y1 = ( - 1 + 3)/2 = 2/2 = 1y2 = ( - 1  - 3)/2 = -  4/2 = - 2x1 = y1 + 1 = 1  + 1 = 2y1 = 1 x2 = y2 + 1 = - 2 + 1 = - 1y2 = - 2 Ответ( 2; 1) ; ( - 1; - 2)

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