System

x + xy = 250y

x - xy = 240y

Enter all possible values of x, separated by commas.

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\(x + xy = 250y\\ x(1+y)=250y\\ x - xy = 240y\\ x(1-y)=240y \)

\(\frac{250y}{1+y}=\frac{240y}{1-y}\\ 250y-250y^2=240y+240y^2\\ 10y=490y^2\\ 0=y(490y-10)\)

\(\large y\in \{0,\dfrac{1}{49}\}\)

\(x=\dfrac{250y}{1+y}\)

\(x=\dfrac{250\cdot \dfrac{1}{49} }{1+\dfrac{1}{49}}=\dfrac{250}{49}\cdot \dfrac{49}{50}=5\)

\(\large x\in \{0,5\}\)

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