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Math Help - algebra Qs

  1. #1
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    algebra Qs

    hi people, im working on rearranging formulae at the mo, and im tuck on two questions. you couldnt give me a hand could you???

    1. make y the subject:
    2/y = x-3y/2y

    2. make x the subject:
    4a = b-3c/c

    could you take me through the steps to do these. im sure its all very logical!!

    ryan xx
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by diablo View Post
    1. make y the subject:
    2/y = x-3y/2y
    You mean "solve for y"? Also, please use parenthesis. This could mean
    2/y = x - (3y)/(2y)
    2/y = (x - 3y)/(2y)
    and a few other things.

    I will assume your equation is:
    \frac{2}{y} = \frac{x - 2y}{2y} Multiply both sides by 2y:

    (2y) \frac{2}{y} = (2y)\frac{x - 2y}{2y}

    2 \cdot 2 = x - 2y Add 2y to both sides:

    2y + 4 = x - 2y + 2y

    2y + 4 = x Add -4 to both sides:

    2y + 4 - 4 = x - 4

    2y = x - 4 Divide both sides by 2:

    \frac{2y}{2} = \frac{x - 4}{2}

    y = \frac{x-4}{2}

    (I should mention that we can't let x = 4 in this expression, since that would make y = 0 and the original expression forbids this.)

    Quote Originally Posted by diablo View Post
    2. make x the subject:
    4a = b-3c/c
    This equation has no x in it!!

    -Dan
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  3. #3
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    thanks for the help topsquark. one thing i dont understand is the first step you did, you multiplied both sides by 2y and on the left side, (2y)2/y became 2 . 2 what does this mean??

    sorry about the 2nd equation. it was 'make c the subject'. if u got a mo, could u help me out with that.

    top quality stuff,
    ryan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by diablo View Post
    thanks for the help topsquark. one thing i dont understand is the first step you did, you multiplied both sides by 2y and on the left side, (2y)2/y became 2 . 2 what does this mean??
    That step in detail then:

    (2y) \frac{2}{y} = \frac{(2y) \cdot 2}{y} = \frac{4y}{y} = 4

    -Dan
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by diablo View Post
    2. make c the subject:
    4a = b-3c/c
    Again, I must make an assumption as to what you mean. Is this
    4a = (b - 3c)/c ?

    Actually it is very much like the other one.

    4a = \frac{b - 3c}{c} Multiply both sides by c:

    c \cdot 4a = c \cdot \frac{b - 3c}{c}

    4ac = b - 3c Add 3c to both sides:

    4ac + 3c = b - 3c + 3c

    4ac + 3c = b New trick. We're going to factor the common "c" from the RHS:

    (4a+3)c = b Now divide both sides by 4a + 3:

    \frac{(4a+3)c}{4a+3} = \frac{b}{4a+3}

    c = \frac{b}{4a+3}

    (For this one we ought to mention that we can't have b = 0, or a = -3/4 either.)

    -Dan
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  6. #6
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    good stuff, thanks a lot for all the help!!!
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