What's the inverse of \(h(y)=\dfrac{1+y}{2-y}\)
Let h(y) = x
x = ( 1 + y) / ( 2 - y) get y by itself
x ( 2 - y) = 1 + y
2x - 2y = 1 + y
3y = 2x - 1
y = ( 2x - 1) / 3 "swap" x and y
x = ( 2y - 1) / 3 = the inverse
