Question 395883
It's not really a systems of equations problem...I have changed it to "expressions involving variables."


Anyway, if we multiply by x-1 we get {{{x+1 = y(x-1) = yx - y}}}. Isolating the x terms we get


{{{y+1 = yx - x}}}


{{{y+1 = x(y-1)}}}


{{{x = (y+1)/(y-1)}}}


Interesting, each equation is equal to its inverse function, so the two equations are symmetric about the line y = x.


{{{graph(300, 300, -10, 10, -10, 10, (x+1)/(x-1), x)}}}