y = (3x-7) / (x+1)
inverse
x = (3y-7) / (y+1)
x y + x = 3 y - 7
3 y - x y = x + 7
y (3 - x) = (x+7)
y = -(x+7)/(x-3)
f^-1(x) = -(x+7)/(x-3)
Let f(x)= 3x-7/x+1. Find the inverse f^-1(x). Thank you.
2 answers
x = (3y-7)/(y+1)
x(y+1) = 3y-7
xy + x = 3y - 7
y(x-3) = -(x+7)
y = (7+x)/(3-x)
x(y+1) = 3y-7
xy + x = 3y - 7
y(x-3) = -(x+7)
y = (7+x)/(3-x)