+123 Find all pairs $(x,y)$ of real numbers $(x,y)$ such that $x + y = 10$ and $x^2 + y^2 = 56$.
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:
\((5+\sqrt{3},5-\sqrt{3})\) & \((5-\sqrt{3} ,5+\sqrt{3})\)
