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Homework Help: Basic Algebra Questions

  1. Mar 27, 2006 #1
    I have a few nagging questions that are preventing me from solving calculus problems.. Can someone give me a hand?


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    Question 1

    [tex] e^xy' = (4x+1)y^2 [/tex]

    From the equation above, is it possible to do this:

    [tex] \frac{y'}{y^2} = \frac{4x+1}{e^x} [/tex]

    Aren't you supposed to divide one at a time, like this?

    [tex] \frac{e^xy'}{y^2} = (4x+1) [/tex]

    How would I get the [tex] \inline e^x [/tex] to the other side?


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    Question 2

    How do you tell the difference between these two?

    [tex] 2 - 3 = -1 [/tex] and [tex] 2-3 = -6 [/tex] ?

    (One latter being (2)(-3) = - 6?)
     
  2. jcsd
  3. Mar 27, 2006 #2
    Q1. Yes, it is possible. How to get the [itex]e^x[/itex] to the other side? Just divide both sides by [itex]e^x[/itex]. Which is the same as your first statement.

    Q2. What do you mean? One DOES NOT write [itex]2\cdot(-3)[/itex] as [itex]2-3[/itex]!
     
  4. Mar 28, 2006 #3

    HallsofIvy

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    Science Advisor

    There is, in fact, no "law" against dividing by several things at the same time: you could go from exy'= (4x+ 1)y2 to
    [tex]\frac{y'}{y^2} = \frac{4x+1}{e^x}[/tex]
    by dividing both sides by exy2
    OR by first dividing by y2 and THEN dividing by ex. The math doesn't care.
     
  5. Mar 29, 2006 #4

    J77

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    I guess the dash represents differentiating w.r.t x...

    [tex]e^x \frac{dy}{dx}=(4x+1)y^2[/tex]

    [tex]\frac{1}{y^2}\frac{dy}{dx}=\frac{(4x+1)}{e^x}[/tex]

    [tex]\int_y{y^{-2}}dy=\int_x{(4x+1)e^{-x}dx[/tex]
     
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