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-   -   How to set up equations out of word problems. (https://www.askmehelpdesk.com/showthread.php?t=135512)

  • Sep 30, 2007, 11:49 AM
    CellistJames
    How to set up equations out of word problems.
    I am having some trouble thinking of how to set up equations out of word problems most of them I can get but there were a few on my homework that I am not getting.

    Here is the Problem:
    A mixture of 12 liters of chemical A, 16 liters of chemical B and 26 liters of chemical C is required to kill a destructive crop insect. Commercial spray X contains 1, 2, and 2 parts respectively, of these chemicals. Commercial spray Y contains only chemical C. Commercial spray Z contains only chemicals A and B in equal amounts. How much of each type of commercial spray is needed to get the desired mixture?

    I need to set up three equations out of this to solve it because there are three variables I am trying to solve for.

    I get to this point:

    X = 1/5A + 2/5B + 2/5C
    Y = C
    Z = 1/2A + 1/2B

    but I can not see how to solve for the variables so I am thinking that these equations are wrong.

    So my question is:
    How can I figure out what is important and what isn't important in the word problem above?
  • Sep 30, 2007, 01:51 PM
    StuMegu
    The way I would look at this is:

    since you can add chemical C directly from Spray Y that doesn't matter yet.

    spray Z only includes A and B in equal amounts, so it doesn't help you get 12 of A and 16 of B so that's no good yet

    spray x is how you get the difference between a and b (4 litres difference), to do this you need 20 litres of spray x giving 4A, 8B and 8C

    you then add 16 litres of spray Z to bring A up to 12litres and B up to 16litres C remains at 8 litres so you add 18 litres of spray Y to finish.

    I'm not sure how you do the equasions for this or even if you can but that is my logic - hope it helps.
  • Sep 30, 2007, 02:20 PM
    galactus
    StuMegu did good by reasoning it out.

    Here's an algebraic way to set it up using three equations:



    Solving, we get:

    20 liters of X, 18 liters of Y, and 16 liters of Z

    Which equals 54 liters as needed.
  • Sep 30, 2007, 02:45 PM
    CellistJames
    hmm I am a bit lost in StuMegu's reasoning, seems like he/she is guessing, I know that I am suppose to use Gaussian Elimination to solve this. I am not sure how you, galactus, got those equations. If you could explain it might help.
  • Oct 1, 2007, 02:45 PM
    CellistJames
    Ok I got some help from my math teacher today. Basically, you know you need chemicals A, B, and C in the given amounts. And you know of mixtures that can give you those amounts.
    So...
    Solution X: 1/5A + 2/5B + 2/5C
    Solution Y: C
    Solution Z: 1/2A + 1/2B

    So which solutions have amounts of chemical A?
    X and Z
    in what amounts?
    1/5 and 1/2
    so..
    1/5X + 1/2Z = 12
    the same logic follows the remaining ones

    final equations:
    1/5X + 1/2Z =12
    2/5X + 1/2Z = 16
    2/5X + Y = 26

    Which are the same equations that galactus posted, just in fraction form and re-arranged.

    So this problem is fixed!
  • Oct 1, 2007, 03:12 PM
    katringette22
    Wow you all make me feel really stupied but that's okay... I'm sure ill need your help to figure out a couple of question ( grade 11 pre-cal) it's the same type of stuff but with quadreatics and stuff

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