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Subscripted Variables

  1. May 9, 2006 #1
    What would be the formula for solving problems like this?

    Solve for all variables.

    RETE = RWTW, RE = 200, RW = 250, 9 - TE = TW

    Answer in the form: (TE, TW)


    Seems simple, but not for me.
     
  2. jcsd
  3. May 9, 2006 #2
    RETE = RWTW, RE = 200, RW = 250, 9 - TE = TW

    Answer in the form: (TE, TW)


    RE(200) x TE(5) = 1000

    9-5=4(TW)

    RW(250) x TW(4) = 1000

    I enter the answer 5, 4


    But it says im wrong.


    -------------------------------

    And for another one

    Solve for the unknown variables.

    RETE + RCTC = 263, RE = 40, RC = 7, TC = TE + 4.

    Answer in the form: TE, TC

    The answer is 5, 9. (I'm correct)

    -----------------------------------------------

    I'm stuck on this one though

    NN + ND = 150

    5NN + 10ND = 450

    Answer in the form: (NN, ND)
     
    Last edited: May 9, 2006
  4. May 9, 2006 #3

    Pyrrhus

    User Avatar
    Homework Helper

    Ever heard of the reduction method?
     
  5. May 9, 2006 #4
    Not "reduction method". My teahcer doesn't teach, he expects us to know everything.

    Yet, i don't understand how i got the first problem wrong. I understand what i'm doing though, to an extent.
     
  6. May 10, 2006 #5

    VietDao29

    User Avatar
    Homework Helper

    Uhmm, you've solved the first problem correctly, the answer is (5, 4). However, tis problem should not be done by Trial and Error method. This problem is to solve the system of equations.
    The first equation is RE TE = RW TW
    So your first equation is 200 TE = 250 TW, or in other words, if you divide both sides by 50, you'll get 4 TE = 5 TW, or 4 TE - 5 TW = 0.
    Now the second equation is 9 - TE = TW or TE + TW = 9
    So we have a syatem of 2 linear equations, and we need to solve for 2 unknows, i.e TE, and TW:
    [tex]\left\{ \begin{array}{l} 4TE - 5TW = 0 \quad (1) \\ TE + TW = 9 \quad (2) \end{array} \right.[/tex]
    We first multiply 2 sides of the equation (2) by 4. Why? In the first equation, we have 4 TE, and we need 4 TE in the second equation, too.
    After multiplying, we will have:
    [tex]\Leftrightarrow \left\{ \begin{array}{l} 4TE - 5TW = 0 \quad (1) \\ 4 TE + 4 TW = 36 \quad (2) \end{array} \right.[/tex]
    Now, if we subtract 2 sides of equation (1) from 2 sides of equation (2), we'll have:
    9 TW = 36 => TW = 4
    Then we can either plug TW = 4 into equation (1), or (2), to solve for TE. Say we'll plug it in equation (1) 4 TE - 5 . 4 = 0 <=> 4TE = 20 <=> TE = 5.
    So the solution to the system is:
    [tex]\Leftrightarrow \left\{ \begin{array}{l} TE = 5 \\ TW = 4 \end{array} \right.[/tex]
    Can you get this? :)
    The third problem can be solved in the same manner. :)
     
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