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    solve these equations simultaneously
    x+2y=6
    4y^2-5x^2=36

    and i have no idea how to do this (can any 1 describe the method and soluton for me please )
    Find the values of A B and C
    3x^3-16x-1 = Ax(x-1)(x+2) + B(x-1)^2 + C(4x^2+3)
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    x+2y=6
    4y^2-5x^2=36

    x=6-2y

    substitute in to 4y^2-5x^2=36

    4y^2-5(6-2y)(6-2y)=36

    multiply out the brackets
    collect like terms on the left hand side
    do a quadratic, you will have two values for Y
    substitute back into
    x=6-2y
    for each value of Y you will get a X value

    you will have the answer to be
    x= ?, y = ? or x=??,y=??

    sry i didn't expand brackets for you, i have to just finish my media cwk off before rush hour 2 is on
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    Yeah, as above. Heres the working.

    4y^2 - 5(6-2y)(6-2y) - 36 = 0

    4y^2 - 5(36-24y+4y^2) - 36 = 0

    4y^2 - 180 + 120y - 20y^2 - 36 = 0

    4y^2 - 20y^2 + 120y - 180 - 36 = 0

    -16y^2 + 120y - 216 = 0

    Factor of 8

    -2y^2 + 15y - 27 = 0

    Using the forumla;

    Discriminant = b^2 - 4ac

    15^2 - 4 x -2 x - 27 = 9

    Y = -15 + 3 / -4 or -15 - 3 / -4

    Y = 3 or 4.5

    Sub into x = 6 -2y

    When Y = 3

    x = 6 - 6
    x = 0

    When Y = 4.5

    x = 6 - 9
    x = -3

    Answers x = -3, y = 4.5 or x = 0, y = 3
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    how do you do the second Q mentioned please?
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    3x^3-16x-1=Ax(x-1)(x+2) + B(x-1)^2 + C(4x^2+3)

    Expanding the brackets on the R. H. S. gives:

    Ax((x^2)+x-2)+B((x^2)-2x+1)+C(4(x^2)+3)

    =>A(x^3)+A(x^2)-2Ax+B(x^2)-2Bx+B+4C(x^2)+3C

    The coefficients, on the R. H. S. of:

    (x^3):A

    (x^2):A+B+4C

    (x):-2A-2B

    Constants on the R. H. S.:

    (c):B+3C

    Now equating these coefficients to what is on the L. H. S.:

    A=3 (i)

    A+B+4C=0 => B+4C=-3 (ii)

    -2A-2B=16 => -6-2B=-16 (iii)

    B+3C=-1 (iv)

    Soving (iii):

    2B=10 => B=5

    Substituting into (ii):

    5+4C=-3

    Therefore:

    4C=-8

    => C=-2

    A=3, B=5, C=-2


    Newton.
 
 
 
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