simultaneous equations ...
$\displaystyle x+y = 5$ ... this is a linear equation
$\displaystyle x^2+y^2 = 13$ ... this is the equation of a circle
using the linear equation, note that $\displaystyle y = 5-x$
substitute $\displaystyle (5-x)$ for $\displaystyle y$ in the circle equation ...
$\displaystyle x^2 + (5-x)^2 = 13
$
solve the quadratic equation for x , then you can determine y.
$\displaystyle x+y=5$
$\displaystyle x^2+y^2=13$
$\displaystyle (x+y)^2=(x+y)(x+y)=x(x+y)+y(x+y)=x^2+xy+xy+y^2$
$\displaystyle (x+y)^2=x^2+y^2+2xy=5^2=25$
$\displaystyle (x+y)^2-\left(x^2+y^2\right)=2xy=25-13=12\Rightarrow\ xy=6$
So you could also solve
$\displaystyle x+y=5$
$\displaystyle xy=6$