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Find all ordered pairs (x,y) of real numbers such that x + y = 10 and x^2 + y^2 = 62 + 2xy.

 

For example, to enter the solutions $(2,4)$ and $(-3,9)$, you would enter "(2,4),(-3,9)"  (without the quotation marks).

 May 10, 2024
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\(\begin{cases} x + y = 10\\ x^2 + y^2 = 62 + 2xy \end{cases}\\ (x - y)^2 = 62\\ x - y = \pm \sqrt{62}\\ \)

When x - y = \(\sqrt{62}\)\((x, y) = \left(\dfrac{10 + \sqrt{62}}2, \dfrac{10 - \sqrt{62}}2\right)\).

Otherwise, \((x, y) = \left(\dfrac{10 -\sqrt{62}}2, \dfrac{10 + \sqrt{62}}2\right)\).

 

These 2 pairs are all the solutions.

 May 10, 2024

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