# Thread: This is literally for my job, and I have no idea of it can even be solved

1. ## This is literally for my job, and I have no idea of it can even be solved

I've managed to get the problem down into a simple equation, but I'm not sure if it's even possible top solve. I don't think I have enough information to do it. I need to solve whatever "A" is in order to find "X" and "Y".
A - B = X, C - A = Y. I always know B and C, but not A, X, or Y. C is always a larger number than B. We usually have A, but I need to depend on the shift before me to get me that number, or get someone with access to past numbers to get it for me. It frequently ends up that I'm the only person on duty, and can't get the number for A. I'm wondering if there even is a way to get it, or if I'll just have to keep leaving blank paperwork forever.
I've used some example numbers to make it more easy for me to look at, but it's not helping. I'll leave it here if it helps.

A - 50 = X, 100 - A = Y

Can this even be solved for A?

2. ## Re: This is literally for my job, and I have no idea of it can even be solved

You have two equations with three "unknowns", A, X, and Y. That is not enough information. There are literally an infinite number of sets (A, X, Y) that satisfy those equations. What you can do is solve for two variables in terms of the third. Here we have X= 50- A, Y= 100- A.
We cannot "solve" this for A. We could choose any value we want for A and calculate X and Y using those equations. For example, if we choose A= 0, we have X= 50 Y= 100. That obviously satisfies 50= 50- 0 and 100= 100- 0. But if we choose A= 30, we have X= 50- 30= 20, Y= 100- 30= 70. 20= 50- 30 and 70= 100- 30 so that is also a valid solution.