Question 825271
This is somewhat ambiguous, but here is a try.


x and y are the natural numbers to find.
{{{x+y=4}}};
{{{x^2+y^2-9(x+y)=8}}}.


The sum of just the two numbers can be substituted to give {{{x^2+y^2-9*4=8}}},
{{{x^2+y^2-36-8=0}}}
{{{x^2+y^2=44}}}
Going back to x+y=4, y=4-x, so substitute into the circle equation:
{{{x^2+(4-x)^2=44}}}
{{{x^2+16-8x+x^2=44}}}
{{{2x^2-8x+16-44=0}}}
{{{2x^2-8x-28=0}}}
{{{highlight_green(x^2-4x-14=0)}}}
'
{{{x=(4+- sqrt(72))/2}}};  unless I made a mistake, this value will not be a Natural Number.  Neither will be y Natural.