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Math Help - need an explanation

  1. #1
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    need an explanation

    hi, i need to know sometheing guys

    i have an equation here, which reads: x = 2/(y-3)^2 - 1
    which then, according to the answer goes to: x = 2/y^2 - 6y + 8

    can anyone tell me how the 6y and 8 in the second equation come about, because i don't understand where they come from??

    thank you,
    ryry
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by diablo View Post
    hi, i need to know sometheing guys

    i have an equation here, which reads: x = 2/(y-3)^2 - 1
    which then, according to the answer goes to: x = 2/y^2 - 6y + 8

    can anyone tell me how the 6y and 8 in the second equation come about, because i don't understand where they come from??

    thank you,
    ryry
    Apparently your equation is:
    x = 2/[(y-3)^2 - 1]
    because that's the only thing that makes sense with what you are saying.

    Look at just the denominator:
    (y - 3)^2 - 1

    I'm going to multiply out the (y - 3)^2:
    (y - 3)^2 = (y - 3)(y - 3) = y(y - 3) - 3(y - 3) = y^2 - 3y - 3y + 9 = y^2 - 6y + 9

    So
    (y - 3)^2 - 1 = y^2 - 6y + 9 - 1 = y^2 - 6y + 8
    as advertised.

    -Dan
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  3. #3
    Senior Member OReilly's Avatar
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    Quote Originally Posted by diablo View Post
    hi, i need to know sometheing guys

    i have an equation here, which reads: x = 2/(y-3)^2 - 1
    which then, according to the answer goes to: x = 2/y^2 - 6y + 8

    can anyone tell me how the 6y and 8 in the second equation come about, because i don't understand where they come from??

    thank you,
    ryry
    When you expand (y-3)^2 - 1 you get( y^2-6y+9) - 1 = y^2 - 6y + 8.

    This can also calculate as (y-3)^2 - 1^2 = (y-3-1)(y-3+1) = (y-4)(y-2) = y^2-2y-4y+8 = y^2 - 6y + 8
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  4. #4
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    wow, thats real good stuff guys,

    i understand now, and i have a couple more equations if you're not too bothered. i'm not super when it comes to algebra, but i'm learning...

    1. its the same equation, but earlier, there is a transition from √2/x+1 = y - 3 to 2/x +1 = (y-3)^2. i understand squaring y-3 but i thought you had to square both sides to remove the square root.

    2. again, what process do you go through in your head, when going from 2/x = (y-3)^2 to x = 2/(y-3)^2

    3. and the process going from B = x/2∂ x R to B x R = x/2∂

    thanks a lot people. your help is appreciated.
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  5. #5
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    Next time make a new thread.
    Quote Originally Posted by diablo View Post
    1. its the same equation, but earlier, there is a transition from √2/x+1 = y - 3 to 2/x +1 = (y-3)^2. i understand squaring y-3 but i thought you had to square both sides to remove the square root.
    sqrt(2/(x+1))=y-3
    Then,
    2/(x+1)=(y-3)^2
    Correct.
    But if you had,
    sqrt(2/x)+1=y-3
    Then you cannot square.
    Subtract 1 from both sides,
    sqrt(2/x)=y-4
    Then attempt to square,
    2/x=(y-4)^2
    2. again, what process do you go through in your head, when going from 2/x = (y-3)^2 to x = 2/(y-3)^2
    Reciprocal of both sides.
    Then multiply by two to remove the 2 from denominator.

    3. and the process going from B = x/2∂ x R to B x R = x/2∂
    Multiply both sides by R, then the R's cancel on the right hand side.
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  6. #6
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    thank you, that was lots of help. is there any websites you know of which practice Qs and As for algebra???
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