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Math Help - Simultaneous equations

  1. #1
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    Simultaneous equations

    Solve the following pairs of simultaneous equations for x and y:
    (a + b)x + cy = bc
    (b + c)y + ax = -ab

    unsure how to start any help would be greatful!

    Also write s in terms of a in the following pairs of equations
    1) s = ah
    h = 2a + 1

    2) s = h^2 + ah
    h = 3a^2
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  2. #2
    Super Member bigwave's Avatar
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    Cool then

    solve simultaniously for x and y

    (a + b)x + cy = bc
    (b + c)y + ax = -ab
    Last edited by bigwave; January 6th 2010 at 10:06 PM. Reason: signs
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  3. #3
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    I have to solve for x and y not a and b though
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  4. #4
    Super Member bigwave's Avatar
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    Cool

    do you mean like this:
    {x = c (b-y)/(a+b),y = -a(x-b)/(b+c)}
    Last edited by bigwave; January 6th 2010 at 10:29 PM.
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  5. #5
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    Sorry but I don't know hat you have done there!
    The answer is:
    x = c
    y = -a
    I just don't know how to get there!
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  6. #6
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    Quote Originally Posted by scubasteve94 View Post
    1) s = ah
    h = 2a + 1
    Just replace h in the first equation with what h is in the second

    s = ah

    s = a(2a + 1)
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  7. #7
    Super Member bigwave's Avatar
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    the answers are seen in the x= and y= (post #4)
    but don't know how the rest cancels
    just hope another reply comes to rescue

    was wondering if your text had an example of a problem like this
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  8. #8
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    Pickslides: thank you


    No the text does not have an example
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  9. #9
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    Quote Originally Posted by scubasteve94 View Post
    Solve the following pairs of simultaneous equations for x and y:
    (a + b)x + cy = bc
    (b + c)y + ax = -ab

     (a + b)x + cy = bc...(1)

     (b + c)y + ax = -ab...(2)

    Working on (1) to make x the subject

     (a + b)x + cy = bc

     (a + b)x = bc-cy

     x = \frac{bc-cy}{a+b}

    Now sub this into (2) for x

     (b + c)y + ax = -ab

     (b + c)y + a\left(\frac{bc-cy}{a+b}\right) = -ab

    exapnding in the a and separately dividing the denominators

     (b + c)y + \left(\frac{abc}{a+b}\right) - \left(\frac{acy}{a+b}\right)= -ab

    Grouping the y terms together

     (b + c)y - \left(\frac{acy}{a+b}\right)+\left(\frac{abc}{a+b}  \right) = -ab

    Taking out y as a common factor

     y\left((b + c) - \left(\frac{ac}{a+b}\right)\right)+\left(\frac{abc  }{a+b}\right) = -ab

     y\left((b + c) - \left(\frac{ac}{a+b}\right)\right) = -ab-\left(\frac{abc}{a+b}\right)

    and finally dividing

     y = \frac{-ab-\left(\frac{abc}{a+b}\right)}{\left((b + c) - \left(\frac{ac}{a+b}\right)\right)}

    You can simplify further and sub back in to find x.
    Last edited by pickslides; January 7th 2010 at 12:50 AM. Reason: bad latex
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  10. #10
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    Thank you so much
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  11. #11
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    Quote Originally Posted by pickslides View Post
     (a + b)x + cy = bc...(1)

     (b + c)y + ax = -ab...(2)

    Working on (1) to make x the subject

     (a + b)x + cy = bc

     (a + b)x = bc-cy

     x = \frac{bc-cy}{a+b}

    Now sub this into (2) for x

     (b + c)y + ax = -ab

     (b + c)y + a\left(\frac{bc-cy}{a+b}\right) = -ab

    exapnding in the a and separately dividing the denominators

     (b + c)y + \left(\frac{abc}{a+b}\right) - \left(\frac{acy}{a+b}\right)= -ab

    Grouping the y terms together

     (b + c)y - \left(\frac{acy}{a+b}\right)+\left(\frac{abc}{a+b}  \right) = -ab

    Taking out y as a common factor

     y\left((b + c) - \left(\frac{ac}{a+b}\right)\right)+\left(\frac{abc  }{a+b}\right) = -ab

     y\left((b + c) - \left(\frac{ac}{a+b}\right)\right) = -ab-\left(\frac{abc}{a+b}\right)

    and finally dividing

     y = \frac{-ab-\left(\frac{abc}{a+b}\right)}{\left((b + c) - \left(\frac{ac}{a+b}\right)\right)}

    You can simplify further and sub back in to find x.
    I'd be inclined to use the elimination method:

    Multiply equation (1) by a and equation (2) by (a + b), then substract one equation from the other and make y the subject.

    Then, multiply equation (1) by (b + c) and equation (2) by c, then substract one equation from the other and make x the subject.
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