Question 1075047
{{{x+y=8}}} first part.


If x and y is each used as perimeter to make a square of each, then sum of the areas is  {{{(x/4)^2+(y/4)^2=2}}}
or {{{x^2/16+y^2/16=2}}}

{{{x^2+y^2=32}}}




Solve the system  {{{system(x+y=8,x^2+y^2=32)}}}.


{{{x^2-(8-x)^2=32}}}
{{{2x^2-16x+64-32=0}}}
{{{x^2-8x+16=0}}}
{{{(x-4)^2=0}}}



{{{highlight(y=x=4)}}}