methods to solve : $6 = x+y, 65 = x^2 + y^2$

Question

what's the neatest way you can solve this :

$$ \begin{cases} x+y=6 \\ x^2 + y^2 =65 \end{cases} $$ without direct substitution? I smell some neat trick in the air here.

Answer 1

$$65=x^2+y^2$$ $$\Rightarrow 65=(x+y)^2-2xy$$ $$\Rightarrow 65=36-2xy$$ $$\Rightarrow 2xy=-29$$ $$(x-y)^2=x^2+y^2-2xy=65+29=94$$ $$(x-y) =\pm \sqrt {94}$$

Now solve simultaneous linear equations to get answers

Alternatively

$(x+y)=65$ and $xy=\frac {-29}{2}$

Hence $x , y$ are the roots of the quadratic $$2m^2-12m-29=0$$

Now solve using quadratic formula

Answer 2

$$2xy=(x+y)^2-(x^2+y^2)=36-65=-29$$

So, $x,y$ are the roots of $$t^2-6t-\dfrac{29}2=0$$