We had a few solutions sent in for this challenge. Simran from Beauchamp College England wrote;

D = Danesh
D-4=12
D=12+4
D=16
M = Meg
M+8=12
To 'undo' minus 8
M=12-8
M=4

C = Chris
C/2=12
To 'undo' times 2
C=12*2
C=24

Ore from Sir Joseph Williamsons Mathematical School wrote

In all these problems you must do the opposite of what has been done to the
original number to 12.
For Danesh 12+4=16
For Meg 12-8=4
And Chris 12*2=24

From Martha at Langtons Junior School

You need to do the inverse (opposite)operation. For instance Chris halved
his starting number and finished with twelve. The inverse of halving is
doubling so you double twelve so Chris's starting number is 24. it is the
same for the other sums.

Right at the end of the month we received this contribution from Daniel at Moss Hall Primary School in England in which he describes very thoroughly and with excellent mathematical skills what he did.

Let d = Danesh's starting number, m = Meg's starting number, and c = Chris' starting number. We use letters to represent their starting numbers since we do not know their values yet. These are called unknowns/variables in an equation.
For example, $x + 4 = 5$ . $x = 1$. We have 'solved' the equation and found the value of the unknown/variable.

If Danesh started with d, then took away 4 from it, and finished with $12$, we can write this as the expression:
$d - 4 = 12$ (he started with d, took away $4$, which made it equal to $12$).
This is a linear equation and can be solved by adding 4 to both sides:
$d - 4 + 4 = 12 + 4$
$d = 16$
Danesh started with $16$, subtracted $4$, and finished with $12$.

Meg started with m and added $8$ to it. She ended up with $12$. This translates to:
$m + 8 = 12$
We want to get rid of the $8$ and leave m by itself so we know it's value. To do this, we have to minus $8$ from both sides of the equation. Since one side'equals' the other side in any equation, what we do to one side has to be done to the other side so that both sides remain equal.
$m + 8 = 12$
$m + 8 - 8 = 12 - 8$
$m = 4$
Meg started with $4$, added $8$, and ended up with $12$.

Finally, to find Chris' starting number, we use the same techniques as above. You should have noticed the rough outline of what we should do by now. First, we construct an equation relating the variables and the outcome. In this case, the outcome of the mathematical operations is $12$.
Since Chris 'halved' his starting number to get $12$, we simply right $c/2 =12$. Dividing by two is the same as multiplying by $1/2$, because c, is technically $c/1$. Any integer (whole number) is in fact a fraction, but we just right it without the fraction to make life easy for our selves. For
example, $5$, is actually $5/1$. $(5/1=5)$ but we just write $5$. In this case, c is actually $c/1$, and halving is multiplying by $1/2$, so $c/1 \times 1/2 = c / 2 = 12$ (we are told they all finish with $12$). To solve this, we multiply both side by $2$. Doing this gets rid of the $2$ on the left side, because $c/2 \times 2/1 = 2c/2 = c$. And the right side ends up as $12 \times 2 = 24$. Remember, an equation needs both sides to remain equal, what we do to one side has to be
done to the other, otherwise the equations will not remain equal most of the time. (Try and find something that leaves both sides the same when it is applied only to one side! Experiment :D)

We conclude from the working out above, that the numbers the three people
started with are:
$d = 16$
$m = 4$
$c = 24$

A good idea when you have finished solving an equation is to substitute your value back into the equation, for example for Chris' equation, we got $24$, so to check if our answer is right, we put it in place of c (since $c =24$): $c/2 = 12$ substituting $24$, we get: $24/2 = 12$ , $12 = 12$