Yeah there is a way
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
Thus
[tex]f=\frac{ab}{b-a}[/tex]
I hope I didn't make any mistakes