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Find all pairs (x,y) of real numbers such that x + y = 10 and x^2 + y^2 = 56 + 2xy.

 Jun 12, 2022
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We have: 

 

\(x+y = 10\)                   (1)

\(x^2 + y^2 =56 + 2xy\)   (2)

 

In the second equation, subtract \(2xy\) from both sides to give: \(x^2 + y^2 - 2xy = 56\) 

 

Now, recall the identity: \(x^2 + y^2 = (x+y)^2 - 2xy\). Substituting this in, along with the first equation, gives us: \(100 - 4xy = 56\). Thus. \(xy = 11\)

 

This gives us a new system, which we can solve for directly:

 

\(x+y=10\)                   (1)

\(xy = 11\)                        (2)

 

The first equation gives us \(x = 10 - y\), which we can substitute in the second equation. 

 

This yields \(-y^2 + 10y = 11\). Subtracting 11 from both sides to make a quadratic gives us: \(-y^2 +10y -11 = 0 \)

 

Now, plugging in the quadratic formula, we get: \(y = {-10 \pm \sqrt{10^2-4 \times -1 \times -11} \over -2} = {-10 \pm \sqrt{56} \over -2} = 5 \pm \sqrt {14}\)

 

This means our x-coordinates are \(10 - (5 \pm \sqrt {14}) = 5 \pm \sqrt {14}\)

 

So, our final coordinates are \(\color{brown}\boxed{(5 + \sqrt {14} , 5- \sqrt{14}), (5 + \sqrt {14} , 5- \sqrt{14})}\)

 Jun 12, 2022

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