Algebra

For starters, \(y=0\), because then \(x=0\).

 

However, there is more than \(1\) solution. 

 

Here is the work in LaTex:

 

\(x+xy=250y\)

\(x=250y-xy\)

\(x=(250-x)y\)

\({x\over{250-x}} = y\)

\(x- {{x^2}\over{250-x}}=-240{x\over{250-x}}\)

\(x- {{x^2}\over{250-x}}={-240x\over{250-x}}\)

\(x={{-240x+x^2}\over250-x}\)

\(250x-x^2=-240x+x^2\) 

\(490x=2x^2\)

\(245x=x^2\)

\(x=245\) 

 

Plug it in to the equation: 

 

\(245+245y=250y\)

\(245 = 5y\) 

\(y=49\)

 

This means that the two answers are\(\color{brown}\boxed {y=0,y=49}\)