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

 May 5, 2020
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Given \(x+y=10\) it follows, \(x=10-y\) thus, \(x^2=(10-y)^2=(100-20y+y^2)\) (1)

 

\(x^2+y^2=56\)  (2)

substitute (1) into (2)

\(100-20y+y^2+y^2=56\) simplify and subtract 56 (Setting the quadratic equation equals zero)

\(2y^2-20y+44=0\) Divide by 2

\(y^2-10y+22=0\)

Use the Quadratic formula 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)            (a=1, b=-10 , c=22)

\(x = {10 \pm \sqrt{(-10)^2-4*1*22} \over 2*1}=5+\sqrt{3}\)  and \(5-\sqrt{3}\)

Now using x+y=10 again, substitute the x value to find y

\(5+\sqrt{3}+y=10\)

\(y=10-(5+\sqrt{3})\)

\(y=5-\sqrt{3}\)

Substitute the other x value

\(5-\sqrt{3}+y=10\)

\(y=10-(5-\sqrt{3})\)

\(y=5+\sqrt{3}\)

 

Thus the pairs of real numbers that satisfy this:

\((5+\sqrt{3},5-\sqrt{3})\) & \((5-\sqrt{3} ,5+\sqrt{3})\)

.
 May 5, 2020

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