Assume the convex lens creates an image of S at the point S'. Now if you put a concave lens in front of the convex one then the image of S will move further and will be at point S''. Now consider a reverse path of light: S'' is now the object and you'll see it's image at S' which is created by the concave lens. Let the distance from the "center" of the concave lens to S'' be a and the distance from S' to the "center" of the concave lens be b. Then using the lens formula:
[tex] -\frac{1}{f}=\frac{1}{a}-\frac{1}{b} [/tex] notice there is a minus near b and f. It's because the image is imaginary
Oh... I thought I was supposed to do that only in the homework section, sorry. But it does make a lot more sense your way. I think I finally got the idea