+1072 Find all ordered pairs x, y of real numbers such that x + y = 10 and x^2 + y^2 = x^3 + y^3.
For example, to enter the solutions (2, 4) and (-3, 9), you would enter "(2,4),(-3,9)" (without the quotation marks).
+9664 Suppose \(xy = a\). Then,
\(x^2 + y^2 = (x^2 + 2xy + y^2) - 2xy = (x + y)^2 - 2xy = 100 - 2a\)
\(x^3 + y^3 = (x + y)^3 - 3xy(x + y) = 1000 - 30a\)
So the second equation is equivalent to
\(100 - 2a = 1000 - 30a\\ 28a = 900\\ a = \dfrac{225}7\\ xy = \dfrac{225}7\\ x = \dfrac{225}{7y}\)
Substituting this into x + y = 10, we have
\(y + \dfrac{225}{7y} = 10\\ 7y^2 - 70y + 225 = 0\)
Checking the discriminant of this equation, \(\Delta = (-70)^2 - 4(7)(225) = -1400 < 0\).
Therefore, there are no real solutions.
