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Rearranging eqn

  1. Mar 1, 2008 #1
    a1 + a3 = ( b2*(b1+b3) ) / (b1 + b2 + b3)

    a1 + a2 = ( b3*(b1+b2) ) / (b1 + b2 + b3)

    a2 + a3 = ( b1*(b2+b3) ) / (b1 + b2 + b3)

    those equations are given

    i need to show that

    a1 = (b2 * b3) / (b1 + b2 + b3)

    i just need some help starting this off i think i need to solve for a1 and then substitute but what ever i do i am always left with a2 or an a3 in my answer any help is appreciated
  2. jcsd
  3. Mar 1, 2008 #2


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    Subtract the third from the second, and add the resulting eq to the first, and multiply the result with 1/2
    Last edited: Mar 1, 2008
  4. Mar 1, 2008 #3


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    Try subtracting the third equation from the second. This will leave you with two equations with left hand sides (a1+a3) and (a1-a3), which you may see how to solve for a1.
  5. Mar 1, 2008 #4
    thanks for the advice im going to try it now
  6. Mar 1, 2008 #5
    i am just unclear on why you multiply by 1/2
  7. Mar 1, 2008 #6


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    Because prior to thar move, you'll have an equation with 2a1 on your left-hand side.
    Multiplying the equation with 1/2 eliminates that factor of 2 from the left hand side, leaving a1 on that side.
  8. Mar 1, 2008 #7
    ok now i got 2a1 = 2b2*b3 / b1 + b2 + b3

    multiply by 1/2 and i get the answer i understand how i did this. but what i am asking is every time i do a problem like this do i multiply by 1/2 at the end or it just worked out that way this time? thanks for the help i havent done these in two years so i am a bit rusty
  9. Mar 1, 2008 #8
    It depends on how many equations there are and what those equations are. Just make sure when you add equations, you add both sides. What you will then get is a new equation that is true; if it ends up that you have 2a1, then divide both sides by 2 (multiply by 1/2), if you have 3a1, then divide both sides by 3, etc.
  10. Mar 1, 2008 #9
    i just want to get more practice with this if i wanted to show :

    b1 = (a1*a2 + a2*a3 + a3*a1 ) / a1

    would i solve all three eqns for

    b1 + b3 eqn 1

    b1 +b2 eqn 2

    b2 + b3 eqn 3

    instead of having a1 + a3 and so forth and then you the same approach i used for my first problem?
  11. Mar 1, 2008 #10
    If that's how the equations look, sure. It doesn't matter what the letter is--it's just a place holder and could be called a, b, c, d, *, $ or whatever.
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