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Math Help - Solve for x

  1. #1
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    Solve for x

    Solve the following for x:
    a) b/(x-a) = 2b/(x+a)
    b) (p-q*x)/t + p = (q*x-t)/p
    With b) I did this:
    (p-q*x)/t - (q*x-t)/p = -p
    (p(p-q*x)-t(q*x-t))/pt = -p
    I don't even know if thats right. But I got stuck and couldn't go any further. I just need someone to help me through the steps. Thanks.
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  2. #2
    Master Of Puppets
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    Hi there,

    Quote Originally Posted by grooverandshaker View Post
    Solve the following for x:
    a) b/(x-a) = 2b/(x+a)
    \frac{b}{x-a} = \frac{2b}{x+a}

    Cross multiply gives,

    b(x+a) = 2b(x-a)

    Now expand each side, what do you get.
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  3. #3
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    Thanks so much. I didn't even think to cross multiply.
    I ended up getting x=3a. Hopefully thats right.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by grooverandshaker View Post
    Solve the following for x:
    b) (p-q*x)/t + p = (q*x-t)/p
    With b) I did this:
    (p-q*x)/t - (q*x-t)/p = -p
    (p(p-q*x)-t(q*x-t))/pt = -p
    I don't even know if thats right. But I got stuck and couldn't go any further. I just need someone to help me through the steps. Thanks.
    No need to get fancy with your starting point. You have a common denominator of pt, so
    (p-q*x)/t + p = (q*x-t)/p

    (pt)(p-qx)/t + (pt)p = (pt)(qx-t)/p

    (p)(p-qx) + tp^2 = (t)(qx-t)

    p^2 - pqx + tp^2 = qtx - t^2

    Can you take it from here?

    -Dan
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  5. #5
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    Thanks so much. I didn't think I was on the right track.
    I can take it from there.
    Thanks again everyone.
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