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Simple Algebra Question

  1. Dec 12, 2004 #1
    Alright this is almost embarrassing! :blushing:

    I'm given

    (1) [tex]i=i_1+i_2[/tex]
    (2) [tex]i_1R_1+ir=E[/tex]
    (3) [tex]i_2R_2-i_1R_1=0[/tex]

    I'm being asked to show that [tex]i=\frac{R_1+R_2}{R_1R_2+r(R_1+R_2)}E[/tex]

    I'm doing this as a self-study program using a Dover book entitled Electromagnetic Fields and Waves by Validimir Rojanski. (page 6)

    Anyhow, I'm familiar with the electronics, I just don't seem to have the necessary algebra skills to solve this simple system of three equations.

    I've been trying substitutions like replacing:

    [tex]i_2=\frac{i_1R_1}{R_2}[/tex]

    And

    [tex]i_1=\frac{E-ir}{R_1}[/tex]

    And then trying to write [tex]i=i_1+i_2[/tex] all in terms of [tex]i_1[/tex]

    But that just seems to lead to redundant identities!

    Do I need to add some equations together or something first?

    This can't be this difficult. Is there some trick involved here that I'm not thinking about? I guess I need to brush up on basic algebra.
     
  2. jcsd
  3. Dec 12, 2004 #2

    Pyrrhus

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    Try adding equations 2 and 3, and the express in the resultant equation i2 in terms of i and then take equation 2 and express i1 in terms of i, now substitute in equation 1, and get the terms with i variable together to factor the i.
     
  4. Dec 12, 2004 #3
    Ok. That worked! :approve:

    But what's the deal with adding equations together?

    I'm real good at manipulating things around and making replacements, but I'm not sure I follow the reasoning behind adding equations together.

    I have some first year college algebra books here somewhere maybe they explain this in the section on solving simultaneous equations?

    And to think that I just passed a course on linear algebra too! Argh!!!

    How do people like me pass these courses when we obviously don't know what the heck we're doing???

    I swear a person could get a degree in the maths without having a clue!
     
  5. Dec 13, 2004 #4

    Pyrrhus

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    It's the metódo de reducción (in spanish), reduction method in english? :rofl:

    Heh, don't say that people tend to forget :smile: .
     
  6. Dec 13, 2004 #5

    cepheid

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    I don't see why not:
    Say you have two equations. The first

    [tex] LHS_1 = RHS_1 [/tex]

    [tex] LHS_2 = RHS_2 [/tex]

    The equations each state (and this is the key) that whatever is on the Left-Hand Side of the equation is equal to whatever is on the Right-Hand Side. So when you add them together, the sum on the new LHS must be equal to the sum on the new RHS...you're adding together equivalent terms from each equation on each side. Don't believe me? Well, if LHS1 = RHS1, then they are interchangeable...so if you makes some interchanges...you get from this:

    LHS1 + LHS2 = RHS1 + RHS2

    to EITHER this:

    LHS1 + LHS2 = LHS1 + LHS2 ->(replaced right-hand sides with their corresponding left-hand sides, to which they are equal)

    OR this:

    RHS1 + RHS2 = RHS1 + RHS2 ->(replaced left-hand sides with their corresponding right-hand sides, to which they are equal)

    These statements are clearly true.

    I'm just trying to show that there should be no reason why we can't add equations. It's not intended as a formal proof...I'm not sure that one is even necessary.
     
  7. Dec 13, 2004 #6
    I understand why we can do it. What I don't understand is what we gain from having done it?

    I'm really confused today. I just did the very same problem the way that I originally tackled it and it worked this time! I must have made some errors last night.

    In other words,... I just went back to what I did in my original post and got the correct answer without having to add any equations together. So while we can add equations together, it's not really necessariy to get the solution.

    In other words, if we add equations 2 and 3 together we get:

    [tex]i_2=\frac{E-ir}{R_2}[/tex]

    But we can also just rearrange equation 3 like so:

    [tex]i_2=\frac{i_1R_1}{R_2}[/tex]

    Then rearrange equation 2 like so:

    [tex]i_1=\frac{E-ir}{R_1}[/tex]

    Then substituing this in for [tex]i_1[/tex] in the preceding equation we get:

    [tex]i_2=\frac{E-ir}{R_2}[/tex]

    Which is the same thing we got by adding the equations together.

    So it appears that the methods of substitution and the methods of adding equations together lead to the same results.

    It seems to me though, that in some situations it's necessary to add equations together because making substitutions only leads to nothing more than simple identities.

    I never really understood why adding equations together is absolutely necessary in some cases, and why in others it's just an alternative method.

    Or maybe I'm confused about that. Maybe the subtitution will always work?

    I think I need to go out and smell the roses for a while. I've been doing too much mathematics lately! :yuck:
     
  8. Dec 13, 2004 #7

    matt grime

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    "So it appears that the methods of substitution and the methods of adding equations together lead to the same results."

    Yep, obviously, and with the obvious provisos about cases such as taking square roots and dividing by zero.

    It is which ever is easiest that you should do. The problems always arise because you've made a simple arithmetic mistake (especially when you change from one sheet of paper to another).

    Adding equations together is generally better as it's simpler, especially if the is an obvious cancellation.

    "It seems to me though, that in some situations it's necessary to add equations together because making substitutions only leads to nothing more than simple identities."

    No more than naive substitution will.
     
    Last edited: Dec 13, 2004
  9. Dec 13, 2004 #8

    Hurkyl

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    As an interesting side note, any system of polynomial equations (over a field) that has isolated solutions can be solved by the method of adding (multiples) of equations to each other, until you eliminate all but one of the variables, at which point you can back-substitute to get the rest.

    Similar can be done for nonisolated solutions (i.e. the solution might be a line), but I don't precisely recall just what can be done, at least in a sense that can be communicated simply.
     
  10. Dec 15, 2004 #9
    Didn't warrant its own thread, but I can't for the life of me simplify this:

    m = 90 - sqrt(100+m^2) - sqrt((10/m*10)^2+100)

    I want to find m, so I know I need to somehow get all the m's to one side.
     
  11. Dec 15, 2004 #10

    matt grime

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    put the 90 on the left, square, rearrange to get any the resulting square root on its own and square again.
     
  12. Dec 15, 2004 #11
    I give up.
     
    Last edited: Dec 15, 2004
  13. Dec 15, 2004 #12

    matt grime

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    erm, i think you need to remember that (a+b+c)^2 IS NOT a^2+b^2+c^2

    I've not worked out the answer, and I do mind writing it down for you.
     
  14. Dec 15, 2004 #13
    I can't do it. Thanks for the help, though.
     
  15. Dec 15, 2004 #14

    uart

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    Did you manage to get rid of the square-root's at least? Truely if you follow Matts advice it should be straight foward to get rid of all the square roots and get it into the form of a polynomial in m.

    How far did you get with it. Why dont you post what you've done.
     
  16. Dec 15, 2004 #15
    Just goofing off, please check!!!

    Note: I hope I got the original problem copied correctly!

    [tex]m-90 = -\sqrt{100+m^2}-\sqrt{\frac{100}{100m^2}+100}[/tex]

    [tex]m^2-180m+90^2 = 100+m^2 +2\sqrt{100+m^2}\sqrt{\frac{100}{100m^2}+100}+ \frac{100}{100m^2}+100[/tex]

    [tex]-180m+90^2-200-\frac{100}{100m^2}= 2\sqrt{100+m^2}\sqrt{\frac{100}{100m^2}+100} [/tex]


    [tex]\left(-180m+90^2-200-\frac{100}{100m^2}\right)^2= 4\left(100+m^2\right)\left(\frac{100}{100m^2}+100\right) [/tex]

    How'd I do so far??? This isn't even my problem!!!

    I probably made some really stupid errors! :yuck:
     
  17. Dec 23, 2004 #16
    Hi NeutronStar.

    I followed the advice of Cyclovenom (12-12-2004) but could not get any result, just got stuck in this crap:

    -ir(R1+R2) + E(R1+R2)
    ---------------------- = i
    R1R2

    How did you get to the correct result. Would ya show me pls????
    Thanx in advance.
    All the best.
     
    Last edited: Dec 23, 2004
  18. Dec 24, 2004 #17
    Ok, so far you have,…

    [tex]i=\frac{E(R_1+R_2)-ir(R_1+R_2)}{ R_1R_2}[/tex]

    You just need to go a little further,…

    [tex]i=\frac{E(R_1+R_2)} { R_1R_2}-\frac{ir(R_1+R_2)}{ R_1R_2}[/tex]


    [tex]i+\frac{ir(R_1+R_2)}{ R_1R_2}=\frac{E(R_1+R_2)} { R_1R_2}[/tex]


    [tex]\frac { iR_1R_2}{R_1R_2}+\frac{ir(R_1+R_2)}{ R_1R_2}=\frac{E(R_1+R_2)} { R_1R_2}[/tex]


    [tex]\frac {iR_1R_2+ir(R_1+R_2)}{ R_1R_2}=\frac{E(R_1+R_2)} { R_1R_2}[/tex]


    [tex]i \left( \frac {R_1R_2+r(R_1+R_2)}{ R_1R_2}\right) =\frac{E(R_1+R_2)} { R_1R_2}[/tex]


    [tex]i=\frac{E(R_1+R_2)} {R_1R_2+r(R_1+R_2)} [/tex]

    I didn't bother with English explanations. I think I did enough steps that you should be able to follow the algebra. If there's any step you don't follow let me know.

    Hope I didn't make any typos! :yuck:

    There might actually be a more eloquent way to do this. I do algebra one tiny little step at a time. :wink:
     
  19. Dec 24, 2004 #18
    Rereading what I did I can see now that I could have multiplied through the whole equation with [itex]R_1R_2[/itex] early on and saved myself about three steps. :cry:

    I was focusing on cleaning up the LHS to isolate i that I wasn't paying any attention to what the RHS had to offer that would help. :rolleyes:

    Still got the right answer though. :biggrin:

    I don't do eloquent algebra. :yuck:
     
  20. Dec 24, 2004 #19
    Equation

    Hi NeutronStar, very kind from you. The steps are clear enough and self explanatory. Sure, any algebraic expression can be optimized more or less.

    Silly from me, I forgot to put the "i"s together on the LHS and then factorize.

    Thanx a lot for your help.
     
    Last edited: Dec 24, 2004
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