Known: x 2 + XY + y = 14, y 2 + XY + x = 28, find the value of X + y

Known: x 2 + XY + y = 14, y 2 + XY + x = 28, find the value of X + y

∵ x2 + XY + y = 14 (1), Y2 + XY + x = 28 (2), ∵ ① + 2, we get: x2 + 2XY + Y2 + X + y = 42, ∵ (x + y) 2 + (x + y) - 42 = 0, ∵ (x + y + 7) (x + y-6) = 0, ∵ x + y + 7 = 0 or x + y-6 = 0, the solution is: x + y = - 7 or x + y = 6

Given that x2 + xy = 2, Y2 + xy = 5, what is the value of half x2 + XY + Half Y2
X2 is the square of X

Idea: both sides of the equation multiply by a natural number, the equation does not change, so you might as well multiply the two sides of the equation by 0.5
0.5 (half) * (x2 + XY) = 0.5 * 2. (1)
0.5 (half) * (Y2 + XY) = 0.5 * 2. (2)
(1) + (2): 0.5 (half) x2 + XY + 0.5 (half) Y2
=3.5 (7 / 2)
*It's a multiple sign

Find all integer solutions of the equation x + YX2 − XY + y2 = 37

Taking x as the main element, the equation is arranged as 3x2 - (3Y + 7) x + (3y2-7y) = 0, ∵ x is an integer, ∵ x = [- (3Y + 7)] 2-4 × 3 (3y2-7y) ≥ 0, ∵ 21 − 1439 ≤ y ≤ 21 + 1439, ∵ integer y = 0, 1, 2, 3, 4, 5. By substituting the value of Y into the original equation, we know that only when y = 4 or 5, the equation has integer solution, that is, when y = 4, x = 5 or x = 43 (rounding off), when y = 5, x = 4 or 103 (rounding off) ）So all integer solutions of the equation are x = 5Y = 4, x = 4Y = 5