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Homework Help: Solving for C

  1. Aug 30, 2013 #1
    Solve for C
    I am extremely frustrated and have made countless attempts at this.

    I would really appreciate a step by step on this. Thanks.
  2. jcsd
  3. Aug 30, 2013 #2


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    Science Advisor

    Try this:

    (1) Get all terms with C in them on one side of the equation, and all terms without C on the other side. In other words:

    Terms with C = Terms without C

    (2) Use the distributive law to write the equation as:

    C*(Something without C) = Something else without C

    (3) Divide both sides by (Something without C) so you have:

    C = (Something else without C) / (Something without C)

    Hope this helps.
  4. Aug 30, 2013 #3


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    Staff: Mentor

    I see an equation comprising 3 terms (one on the left of the equals sign, and two on the right hand side). What does it look like if you multiply each of those 3 terms by L?
  5. Aug 31, 2013 #4



    ##C=\frac {L-SN}{LV}##

    but apparently this is wrong...

    the book gives ##C=\frac{LV-SN}{L-S}##
    Last edited: Aug 31, 2013
  6. Aug 31, 2013 #5


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    Staff: Mentor

    [tex]\frac {C-S}{L} N = \frac {(C-S)N}{L}[/tex]

    Do you see it now?
  7. Aug 31, 2013 #6
    Here's what I get:

    V = C - N(C-S)/L | * L
    LV = LC - NC + NS
    LC - NC = LV - NS
    C = (LV - NS)/(L-N)

    So I have almost the same answer as the book, but I don't even begin to comprehend how an S can ever wind up as part of the demoninator.
  8. Aug 31, 2013 #7

    Ray Vickson

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    Homework Helper

    You should have ##LV = LC -CN + SN##. Do you see the two errors you made in line 2? Can you see why it is wrong to go from the line
    [tex] LV = LC - C - SN[/tex] (which, by itself, is wrong) to
    [tex] \frac{LV}{C}=L-SN ?[/tex] Would you say it is correct to go from ##5\times 3 - 3 - 4\times 2## to ##3 \times (5 - 4 \times 2)?## You are claiming these are the same!
  9. Aug 31, 2013 #8
    Thanks guys I finally got it. And the denominator should have been L-N, my mistake.
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