Let x + y = 5 and x2 + y2 = 13. Find x and y.
2 and 3
From \((x+y)^2 = x^2+y^2+2xy\): \(25 = 13 + 2xy \Rightarrow xy = 6\).
So x and y are roots of \(t^2 - (x+y)t + xy = 0\), i.e. \(t^2 - 5t + 6 = 0\).
Factoring: \((t-2)(t-3) = 0 \Rightarrow t = 2, 3\).
The values of x and y are 2 and 3.
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