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Rewriting again

  1. Oct 4, 2006 #1
    I know you all are probably sick of me posting these kind of "easy" problems, but rewriting equations is one of those things that I just seem to struggle with. So that being said, can I get help with [tex]2A=h(b_{1}+b_{2})[/tex]? I need to solve for [tex]b_{1}[/tex]. Maybe if someone just gave me the first step, I could get the rest ???
    Last edited: Oct 4, 2006
  2. jcsd
  3. Oct 4, 2006 #2
    [tex] 2A=h(b_{1}+b_{2}) [/tex].
    [tex] 2A = hb_{1}+hb_{2} [/tex].
    [tex] 2A-hb_{2} = hb_{1} [/tex].

    So how would you solve for [tex] b_{1} [/tex]?
  4. Oct 4, 2006 #3
    Wouldn't you divide both sides by h?
  5. Oct 4, 2006 #4
    yes you would.
  6. Oct 4, 2006 #5
    So the final answer would be [tex]\frac{2A-b_{2}}{h}=b_{1}[/tex] ?
  7. Oct 4, 2006 #6
    no it would be [tex] b_{1} = \frac{2A-hb_{2}}{h} [/tex] or [tex] b_{1} = \frac{2A}{h} - b_{2} [/tex]
    Last edited: Oct 4, 2006
  8. Oct 4, 2006 #7
    Oh you're right, I just forgot the 2nd h in my equation.

    And thanks for all your help. I think I'm starting to get this more, so hopefully this will be my last rewriting problem!
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