1
$\begingroup$

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:

$$X=\frac{A-B+Y}{1-C} \cdot C$$

$$Y=\frac{A-B+X}{1-D} \cdot D$$

  • 0
    Suggestion: replace $A-B$ throughout with, say, $U$. That'll simplify a little bit. Also write $(1 - C - D)$ as $(1 - (C+D))$, so that the symmetries are more evident. – 2017-01-28
  • 0
    Umm so $X$ is defined in terms of $Y$ and $Y$ is defined in terms of $X$? – 2017-01-28

2 Answers 2

7

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, $$\frac{A-B+Y}{1-C} \cdot C +\frac{A-B+X}{1-D} \cdot D=\frac{A-B+\frac{(A-B)D}{1-C-D}}{1-C} \cdot C +\frac{A-B+\frac{(A-B)C}{1-C-D}}{1-D} \cdot D=$$ $$=(A-B)\left(\frac{1+\frac{D}{1-C-D}}{1-C} \cdot C +\frac{1+\frac{C}{1-C-D}}{1-D} \cdot D\right)=\frac{(A-B)(C+D)}{1-C-D}$$

  • 0
    How did you derive the initial X and Y condition in your solution? Thank you so much for helping out. – 2017-01-28
  • 0
    @J S I fixed my post for you. – 2017-01-28
  • 0
    Thank you so much, this has been bugging me all day!! I'm an actuary and this was essentially proving the logic our system was using (coded way back in the day) is calculating charges correctly. Again, thank you! – 2017-01-28
  • 0
    I guess the symbollic manipulation works out, but can these equations really represent anything since X is defined in terms of Y and Y is defined in terms of X? – 2017-01-28
  • 0
    @Ovi, great question. The answer is that this shows the equation works whenever you have $X$ and $Y$ that are related in that way. It does not, as you may be thinking, have to actually "make sense". For example, if we were told $X=Y+1$ and $Y=X+1$, we could probably prove things like the OP, and the proofs would be perfectly reasonable, but they might be vacuous since it's not clear if the hypothesis can be met. – 2017-01-28
  • 0
    @Pat Devlin I thought about it, but I still don't see how we can release it. – 2017-01-28
  • 0
    @PatDevlin Thanks for the response, it makes sense. @ Michael Rozenberg what do you mean "release it"? – 2017-01-28
  • 0
    @Ovi I meant to make it. – 2017-01-28
  • 0
    The two equations are of the form $X = aY + b$ and $Y = cX + d$. So we just need say $X= a(cX +d) +b$, which has a unique solution iff $ac \neq 1$. Similarly, we can solve for $Y$ iff $ac \neq 1$. In the original langauge, this requires $CD \neq (C-1)(D-1)$, which follows since $C+D -1 \neq 0$ by assumption. – 2017-01-28
  • 0
    I'll explain further. we have A - B, which represents a transfer amount subject to a charge. We have X and y which represent two seperate charge amounts, and C and D represent charge percentages respectively. These formulas are representative of what Charge X and Y would be if we were to base our charge off of the transfer (A - B) less the charges ( X+Y ). So we're pretty much calculating charges based off of the amount the transfer would be after those charges. It's pretty circular logic, but yes in practice it is representative of something. – 2017-01-28
  • 0
    @PatDevlin Ah good catch, so there is really no problem after all. – 2017-01-28
2

In the future, you can prove this sort of thing by typing it into free websites like Wolframalpha.com

For instance, you're trying to prove something like $\alpha + \beta = \gamma$, so you could type into the webpage $\alpha + \beta - \gamma$. This will simplify to 0, which proves the equation you wanted.

You could also plug in lots and lots of points. If you do this literally randomly (use some random number generator to plug in numbers between say 0 and 1). If the numbers are random, then for many equations, you can actually prove that they always work simply because they work when you plug in some random numbers.

A third option that works for things like this would be to multiply everything in sight to get rid of all the denominators. Then just multiply out everything and things will just cancel like crazy. This is very time-consuming to do by hand however, and people are prone to mistakes. [This is what the webpage would do though]