Getting Composition of a Mixture of Colors I wanted to validate a thought on blending of colours digitally.

Imagine I have two colours, $C_1$ $(R_1,G_1,B_1)$ and $C_2$ $(R_2,G_2,B_2)$. Now, I create colour $C3$ as follows:
$C_3 = \alpha C_1 + (1-\alpha) C_2$

Where $\alpha$ is a value between $0$ and $1$.
So, yes, basically, I am trying to put one object over another with a transparency associated with it. 

What I want to understand is that, if I have $C_3$, $\alpha$ and $C_1$ in the above equation, will I get an accurate value of $C_2$?

Math suggests that I should, but I just want to confirm that I am not missing anything. 
Also, can I keep a really high but $&lt; 1$ value for $\alpha$ and still accurately get $C_2$?
You actually have 3 independent equations.
If one of them is solvable, all three are.  

Now, the equation is given by:

$$ z = \alpha x + \left( 1 - \alpha \right) y $$

Assume $ \alpha, x, z $ are known, then:

$$ y = \frac{z - \alpha x}{1 - \alpha} $$

For any value of $ \alpha \neq 1 $ the solution is valid.
Moreover, if $ alpha = 1 $ then $ z = x $ and $ y $ is irrelevant.  

Looking at the other case, restoring $ x $, yields:

$$ x = \frac{z - \left( 1- \alpha \right) y}{\alpha} $$

For any value of $ \alpha \neq 0 $ the solution is valid.
Moreover, if $ alpha = 0 $ then $ z = y $ and $ x $ is irrelevant.  

As you wrote, in order to have a Linear Interpolation the parameter $ \alpha $ must be in the range $ \left[ 0, 1 \right] $.