{ y = x + 1 + 4 x − 1 − y 4 + 2 0 = x 2 + 2 x ( y − 1 ) + y 2 − 6 y + 1
Let ( x 1 , y 1 ) , ( x 2 , y 2 ) , … , ( x n , y n ) be the solution set of the above system of equations for real x , y
Evaluate n + i = 1 ∑ n ( x i + y i )
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According to second equation, having: x = 2 y − ( y − 1 ) Replacing to first equation: − y + 2 y + 2 + ( − y + 2 y ) 4 1 = y 4 + 2 + y ⇒ 0 ≤ y ≤ 1 It is easy to get two roots of (x, y): (1, 0), (2, 1) n + Σ i = 1 n ( x i + y i ) = 6