Suppose x and y are real numbers both greater than 1 that satisfy xy + x + y = 19 and x2y + y2x = 84. Find x2 + y2.
let a = xy and b = x+y. Then a+b = 19 by the first equation and ab = 84 by the second. Write a = 19-b and substitute:
b(19-b) = 84
b^2 - 19b + 84 = 0
(b-7)(b-12) = 0
b is 7 or 12. Similarly, a is 12 or 7 since the equations are symmetric in a and b.
So there are two cases to consider: xy = 7 and x+y=12 or xy=12 and x+y = 7, and these can be approached the same way as above.
Case 1
y(12-y) = 7
y^2 - 12y + 7 = 0
y = (12 +/- sqrt(144-28))/2 = 6+/-sqrt(29) (and x is the other)
But 6-sqrt(29) is less than 1 so the first constraint fails and there are no solutions
Case 2
y(7-y) = 12
y^2 - 7y + 12 = 0
(y-3)(y-4) = 0
y is 3 or 4 (and x is 4 or 3)
This seems to be the only solutions. If (x,y) = (3,4) in whatever order, then x^2 + y^2 = 3^2 + 4^2 = 25
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Posted by Paul
on 2025-01-13 11:25:08 |
Solution
