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:

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