The lens combination as shown in the figure, consists of two lenses, $L_1$ and $L_2$, of the focal lengths $+10$ cm and $-10$ cm, respectively. The position of the image formed is :

For the first lens $L_1$ (convex lens):
Object distance, $u_1 = -30$ cm
Focal length, $f_1 = +10$ cm
Using the lens formula, $\dfrac{1}{v_1} - \dfrac{1}{u_1} = \dfrac{1}{f_1}$
$\Rightarrow \dfrac{1}{v_1} - \dfrac{1}{-30} = \dfrac{1}{10}$
$\Rightarrow \dfrac{1}{v_1} + \dfrac{1}{30} = \dfrac{1}{10}$
$\Rightarrow \dfrac{1}{v_1} = \dfrac{1}{10} - \dfrac{1}{30} = \dfrac{3 - 1}{30} = \dfrac{2}{30} = \dfrac{1}{15}$
$\Rightarrow v_1 = +15$ cm
The image $I_1$ formed by $L_1$ acts as a virtual object for the second lens $L_2$.

For the second lens $L_2$ (concave lens):
Distance between the lenses, $d = 3$ cm
Object distance for $L_2$, $u_2 = v_1 - d = 15 - 3 = +12$ cm
Focal length, $f_2 = -10$ cm
Using the lens formula, $\dfrac{1}{v_2} - \dfrac{1}{u_2} = \dfrac{1}{f_2}$
$\Rightarrow \dfrac{1}{v_2} - \dfrac{1}{12} = \dfrac{1}{-10}$
$\Rightarrow \dfrac{1}{v_2} = \dfrac{1}{12} - \dfrac{1}{10} = \dfrac{5 - 6}{60} = -\dfrac{1}{60}$
$\Rightarrow v_2 = -60$ cm

The negative sign indicates that the final image is formed $60$ cm to the left of the concave lens.

Answer: $60$ cm to the left of the concave lens