Magnifying factors of multiple lenses (closely packed) Well the following question I really struggle to find an answer

A thin lens creates an image of an object with magnification -3. A
  second identical lens is positioned very close to the first lens such
  that the distance between the two lenses can be neglected. What is the
  magnification with this two lens system assuming that the object
  distance has not changed?

At first the solution seemed very simple, name the distances between some virtual objects $d_0$, $d_1$ & $d_2$. 
Then the total magnifying factor can be found using:
$$M_t = -\frac{d_2}{d_0}$$
$$M = -3 = -\frac{d_1}{d_0}$$
$$M = -3 = -\frac{d_2}{-d_1}$$
(last equation has an extra minus sign as the image is a virtual location, the image $d_1$ is behind the lens, the distance of this virtual image to the lens is $d_1$ given that the lenses are "positioned very close".
Now the problem is: I can't solve this, 4 unknowns and only 3 equations. Also going by hand and stating simply "well the magnification factor is the combination of both lenses so $M_t = M * M = 9$ results in a "wrong answer".
 A: There are several ways to solve this. The simplest is to Google for the formula for the focal length of two lenses in contact, but let's instead do the problem step by step. This shows the situation with just one lens.

We know th magnification is -3, so we know that $v = -3u$ so the lens equation tells us (I'm taking +ve to be to the left i.e. $u$ is positive and $v$ is negative):
$$ \frac{1}{u} - \frac{1}{v} = \frac{1}{f} $$
and since $v = -3u = -3d$ we have $f = 3d/4$.
Now use the image from the first lens as the object for the second i.e. $u_2 = -3d$ and we know $f = 3d/4$, then we have:
$$ -\frac{1}{3d} - \frac{1}{v_2} = \frac{4}{3d} $$
and a quick rearrangement gives $v_2 = -3d/5$ and the magnification is $M = -3/5$.
The alternative is to Google for the focal length of two lenses in contact, and amongst the hits will be this page telling you that the combined focal length is given by:
$$ F = \frac{f_1 f_2}{f_1 + f_2} $$
or when the two lenses are identical:
$$ F = \frac{f^2}{2f}  = \frac{f}{2} $$
We've already worked out that $f = 3d/4$, so $F = 3d/8$, and using the lens equation we get:
$$ \frac{1}{d} - \frac{1}{v} = \frac{8}{3d} $$
and solving for $v$ gives $v = -3d/5$, just as we got before.
