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avatar+1351 

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).

 May 1, 2024
 #1
avatar+9664 
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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.

 May 1, 2024

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