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Homework Help: Systems Of Equations Word Problems

  1. Jul 5, 2010 #1
    The problem statement, all variables and given/known data
    1.
    A small corporation borrowed $1,500,000 to expand it's product line. Some of the money was borrowed at 8%, some at 9%, and some at 12%. How much was borrowed at each rate if the annual interest was $133,000 and the amount borrowed at 8% was 4 times the amount borrowed at 12%

    2.
    attachment.php?attachmentid=26851&stc=1&d=1278366366.jpg



    2. Relevant equations



    3. The attempt at a solution

    1. I know you could write the total money borrowed and interest equations as

    a+b+c=1,500,000
    .08a+.09b+.12c=133,000

    However I dont know how to write the equation where the amount 'a' borrowed is 4 times the amount 'c' borrowed.

    2.
    A.
    I'm not even sure how I'd start to write a system of equations for this.

    Something like this?
    x3+x4-x5+x6+x7+500=600
    x1+x2-x3-x4+x5+600=500
    Am I on the right track? What would you do with the constant at the end subtract it from the RHS of the equation?
     

    Attached Files:

  2. jcsd
  3. Jul 5, 2010 #2

    Mark44

    Staff: Mentor

    You're going to want to slap your forehead.
    a = 4c is the third equation.
    It doesn't look like you're on the right track here at all. Look again at what they said about water flowing into a junction being equal to the water flowing out of the same junction.

    For each of the six junctions write an equation that represents the flows. For example in the upper left junction, 600 = x1 + x2. Do a similar analysis on each of the junctions. BTW, you ought to assign numbers to the junctions, with possibly the upper left one being junction 1, the upper middle one being junction 2, and so on. It doesn't matter much how you label them, but it might help you be more organized.

    For part a, you should get 6 equations in 7 unknowns, so you won't be able to get a unique solution. The b and c parts should allow you to get unique solutions.
     
  4. Jul 5, 2010 #3
    1. Haha. Don't know how that slipped me.

    2.
    Ok, I think I understand this but, for the upper left one wouldn't it be 600=x1+x3 because the 2 outputs of the upper left junction are x1 and x3?
     
  5. Jul 5, 2010 #4

    Mark44

    Staff: Mentor

    Right. I was thinking x3, but must have hit the wrong key.
     
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