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Homework Help: Algebra problem

  1. Jan 18, 2006 #1
    I have an algebra problem which is giving me headaches.

    Lt = St
    Lt = Kt+1 - Kt
    St = s * Yt
    Kt = k * Yt

    K0 = 200, s = 0.3, k = 2

    Prove that Kt+1 = 1.15 * K

    Thanks.
     
  2. jcsd
  3. Jan 18, 2006 #2
    Now I'm not sure if this is correct, but I started off by subsituting in the known values.

    Lt = St
    Lt = Kt+1 - Kt
    St = 0.3 * Yt
    Kt = 2 * Yt

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

    Lt = St
    Lt = Kt+1 - Kt
    St = 0.3 * (0.5 * Kt)
    0.5 * Kt = Yt
     
  4. Jan 18, 2006 #3

    HallsofIvy

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    Science Advisor

    Since the very first equations says Lt= St, you can simplify slightly by writing that third equation as Lt= 0.3Yt. You now have 3 equations:
    Lt= Kt+1- Kt
    Lt= s Yt
    Kt= k Yt

    Also, You say you want to prove "Kt+1= 1.15*K" which makes no sense. I imagine it must be either "Kt+1= 1.15*k" or "Kt+1= 1.15*Kt". I suspect it is the latter.

    Since, however, you want Kt+1= something, I suggest rewriting that first equation as Kt+1= Lt + Kt (add Kt to both sides). Now, it looks to me like you just need to replace that Lt with something!

    You know that Lt= sYt. do you see that Yt= (1/k) Kt? That should make it clear!
     
    Last edited by a moderator: Jan 18, 2006
  5. Jan 18, 2006 #4
    The sheet I was given says "Prove that Kt+1 = 1.15 x K" with the K in upper case.
     
  6. Jan 18, 2006 #5

    VietDao29

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    Homework Helper

    I think that's a little typo. The sheet should read:
    Kt + 1 = 1.15 Kt.
    ---------------
    You can follow HallsofIvy's post, and you'll soon arrive at the answer,
    or you may also want to try a slightly different way:
    St = Lt
    <=> s Yt = Kt + 1 - Kt.
    What's Kt + 1? What's Kt?
    Can you go from here?
     
  7. Jan 18, 2006 #6
    I have also been able to confirm that the equation:

    Kt+1 = 1.15 * K

    should read:

    Kt+1 = 1.15 * Kt

    From the help that HallsofIvy and VietDao29 provided, you should come to the answer.

    I came to the answer a different way, but want to find out if you can use their help before I make another post.
     
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