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Math Help - equation

  1. #1
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    equation

    hi i've been trying to solve a simultaneous equation and i'm sure i've gone wrong somewhere and would be grateful if someone could point out where:

    5x + 3y = 4
    5x^2 - 3y^2 = 8

    i started by making x the determinant in the first equation then subbing it into equation 2 like so

    x = \frac{4-3y}{5}(1)

    5(\frac{4-3y}{5})^2 - 3y^2 = 8

    then multiplied everthing by 5 to remove the fraction
    25(4 - 3y)^2 - 15y^2 = 40 then expanded the squared bracketed term:
    25(16 - 24y + 9y^2) - 15y^2 = 40 then carried on with
    400 - 600y + 225y^2 - 15y^2 = 40 then
    360 - 600y + 210y^2 = 0

    i really don't think this is right, could someone explain this to me please? thanks
    Last edited by mark; September 21st 2009 at 06:35 AM.
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  2. #2
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    Quote Originally Posted by mark View Post
    hi i've been trying to solve a simultaneous equation and i'm sure i've gone wrong somewhere and would be grateful if someone could point out where:

    5x + 3y = 4
    5x^2 - 3y^2 = 8

    i started by making x the determinant in the first equation then subbing it into equation 2 like so

    x = \frac{4-3y}{5}(1)

    {5}(\frac{4-3y}{5}) - 3y^2 = 8

    then multiplied everthing by 5 to remove the fraction
    25(4 - 3y)^2 - 15y^2 = 40 then expanded the squared bracketed term:
    25(16 - 24y + 9y^2) - 15y^2 = 40 then carried on with
    400 - 600y + 225y^2 - 15y^2 = 40 then
    360 - 600y + 210y^2 = 0
    \not5(\frac{4-3y}{\not5}) - 3y^2 = 8
    Next step. ({4-3y}) - 3y^2 = 8
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  3. #3
    Senior Member pacman's Avatar
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    , you forgot to SQUARE the parenthetical expression

    (4 - 3y)^2 - 15y^2 = 40, expand

    16 - 24y + 9y^2 - 15y^2 = 40, sum up

    16 - 24y - 6y^2 = 40, re-arrange

    6y^2 + 24y - 34 = 0, DBS by 2

    3y^2 + 12y - 17 = 0,

    the quadratic formula can bash this . . . .
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  4. #4
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    what does the symbol replacing the 5's mean? and what happened to the square? i don't understand...
    by the way i missed out the square symbol in the original post but then realised i had and added it in later, i still carried on the workings out though as if the square symbol was there

    the answer the book gives is x = 2 and y = -2
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  5. #5
    Senior Member pacman's Avatar
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    see the graph
    Attached Thumbnails Attached Thumbnails equation-asw.gif  
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  6. #6
    Senior Member pacman's Avatar
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    do you remember this?



    the 5 in the denominator is squared, but you only have 5 in the numerator,

    in order to eliminate that 25 in the denominator, you multiply 5 by 5

    5()

    25[(4 - 3y)^2]/25 - (5)(3y^2) = 5(8),

    (4 - 3y)^2 - 15y^2 = 40, i know you can finish this now
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  7. #7
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    i've only been doing maths for about 6 weeks or so now and i don't really know what that graph was supposed to mean. do you think you could explain it in a different way?
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  8. #8
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    i still don't really follow, you said "the 5 in the denominator is squared, but you only have 5 in the numerator", where is the 5 in the numerator? i thought the numerator was the number(s) above the line in a fraction?

    then you multiplied everything by 5, which i understand but divided part of it by 25 like this:

    25[(4 - 3y)^2]/25 - (5)(3y^2) = 5(8)

    the only thing i have trouble understand is dividing the first 25 by 25 again...
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  9. #9
    Senior Member pacman's Avatar
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    ,

    (4 - 3y)/5 is a squared quantity, it means (4 - 3y) is squared and 5 in the D is squared

    now look , in the perenthesis term it is precede by 5 right?

    then you have 5^2 in the D, what will you multiply with the N to eliminate 25 in the D?

    5 right?

    so, 5[5/5^2][4-3y]^2 - 15y^2 = 40, that red thing cancels
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