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    Make r subject

    P = 2/3Mr^2

    Make x subject

    c = (ax+c)/(x-2)

    these two are really making my mind boggle, on the second one I moved x-2 to the other side and got cx-2c = ax+c

    then i added the 2 c and got

    cx = ax+3c

    is that even right? how do I get x from that =?
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    (Original post by TimetoSucceed)
    Make r subject

    P = 2/3Mr^2
    For this one, start by making r^2 the subject, and then just take the square root of both sides.

    (Original post by TimetoSucceed)
    Make x subject

    c = (ax+c)/(x-2)

    these two are really making my mind boggle, on the second one I moved x-2 to the other side and got cx-2c = ax+c

    then i added the 2 c and got

    cx = ax+3c

    is that even right? how do I get x from that =?
    You're almost there, subtract ax from both sides (to take it over to the LHS) and then take x out of the LHS as a common factor. Then there's something you can do to finish it off.
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    (Original post by TimetoSucceed)
    Make r subject

    P = 2/3Mr^2

    Make x subject

    c = (ax+c)/(x-2)

    these two are really making my mind boggle, on the second one I moved x-2 to the other side and got cx-2c = ax+c

    then i added the 2 c and got

    cx = ax+3c

    is that even right? how do I get x from that =?
    For the first one, rearrange to get r^2 as the subject and square root.
    For the second one, you're on the right track. When you have cx=ax+3c, move the ax to the other side, take a common factor of x out and divide by c+a.
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    thanks a lot =] makes sense now
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    (Original post by Gemini92)
    For the first one, rearrange to get r^2 as the subject and square root.
    For the second one, you're on the right track. When you have cx=ax+3c, move the ax to the other side, take a common factor of x out and divide by c+a.
    dont you mean c-a?

    x(c-a) = 3c so I divide 3c by c-a right? I get

    x = 3c/(c-a)
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    (Original post by TimetoSucceed)
    dont you mean c-a?

    x(c-a) = 3c so I divide 3c by c-a right? I get

    x = 3c/(c-a)
    That's right, yup.
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    (Original post by TimetoSucceed)
    dont you mean c-a?

    x(c-a) = 3c so I divide 3c by c-a right? I get

    x = 3c/(c-a)
    Yes, I did mean c-a, sorry.
 
 
 
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