solve the system of equations

Hello ballhopper!

{ y^2 + x^2 = 10

y+x = 4

$y^{2}+x^{2}=10$ 

$y+x=4$

$y=4-x$

$\left(4-x\right) ^{2}+x^{2}=10$

$16-8x+x²+x²=10$

$8-4x+x² =5$

$x²-4x+3=0$

$ax²+bx+c=0$

$a=1; \ b=-4; \ c=3$

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

$x = {4 \pm \sqrt{16-12} \over 2}$ 

$x_{1} = {4 + \sqrt{16-12} \over 2}=3$

$x_{2} = {4 - \sqrt{16-12} \over 2}=1$

$solution\ set = \{3 ;1\}$

Greeting asinus :- )  !