Work Proof... Need help! I'm having trouble figuring out a proof at work, and to spare the business jargon I put it in terms of variables. Could someone help?
Prove: 
$$\frac{A-B+Y}{1-C} \cdot C +\frac{A-B+X}{1-D} \cdot D = \frac{A-B}{1-C-D} \cdot(C+D)$$
where:
$$X=\frac{A-B+Y}{1-C} \cdot C$$
$$Y=\frac{A-B+X}{1-D} \cdot D$$

From the condition we get
$(1-C)X-CY=(A-B)C$ and $-DX+(1-D)Y=(A-B)D$ or
$$(1-C)(1-D)X-C(1-D)Y=(A-B)(1-D)C$$ and $$-DCX+C(1-D)Y=(A-B)CD$$ and after adding we get $((1-C)(1-D)-DC)X=(A-B)C$, which gives
$X=\frac{(A-B)C}{1-C-D}$ and $Y=\frac{(A-B)D}{1-C-D}$.
Thus, $$\frac{A-B+Y}{1-C} \cdot C +\frac{A-B+X}{1-D} \cdot D=\frac{A-B+\frac{(A-B)D}{1-C-D}}{1-C} \cdot C +\frac{A-B+\frac{(A-B)C}{1-C-D}}{1-D} \cdot D=$$
$$=(A-B)\left(\frac{1+\frac{D}{1-C-D}}{1-C} \cdot C +\frac{1+\frac{C}{1-C-D}}{1-D} \cdot D\right)=\frac{(A-B)(C+D)}{1-C-D}$$
