[Joe Ross]

Originally Posted by whytheheckme

How is this step performed?

x - x(e^y) = 1 + e^y
x - 1 = (x + 1)(e^y)

I performed it as two steps


x - x(e^y) = 1 + e^y
x - 1 = xe^y + e^y
factor the e^y out
x -1 = (x + 1)(e^y)