Results 1 to 2 of 2

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

  1. #1
    Newbie
    Joined
    Aug 2017
    From
    USA
    Posts
    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,345
    Thanks
    2860

    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: Nov 20th 2012, 04:50 AM
  2. Integration? Looks simple enough? Literally
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Apr 25th 2010, 10:23 PM
  3. [SOLVED] no idea for this(hard for me )!
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Apr 2nd 2010, 01:49 AM
  4. [SOLVED] No idea about this. (sine/cosine)
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: Mar 10th 2009, 07:40 PM
  5. [SOLVED] No idea where to start!
    Posted in the Algebra Forum
    Replies: 11
    Last Post: Jul 25th 2008, 09:44 AM

/mathhelpforum @mathhelpforum