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Math Help - solving these sorry

  1. #1
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    solving these sorry

    just seen a post underneath nearly the same, maybe were doing the same course :P but anyway im having real trouble with these and if anyone could help it would be great ( oh and i dont know how to write squared and such so i copied it off that post too

    a) x(2) + 3x + 5y = 20
    x + 3y = 1

    b) 2x + y = 8
    4x(2) - 3y(2) = 1

    c) x + 2y = 5
    5x(2) + 4y(2) + 12x = 29

    d) x(2) + y(2) + 4x + 6y - 40 = 0
    x - y = 10

    thank you
    thank you
    thank you
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  2. #2
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    Hello, Kim!

    I'll do the last one . . . it seems to be the messiest one.
    I hope you can use my solution as a template for solving the others.


    d) .[1] .x + y + 4x + 6y - 40 .= .0
    . . .[2] .x - y .= .10

    We have a quadratic equation and a linear equation.

    Solve the linear equation for one of its variables.
    . . From [2], we have: .y .= .x - 10 .[3]

    Substitute into [1]:
    . . . . .x + (x - 10) + 4x + 6(x - 10) - 40 .= .0
    . .x + x - 20x + 100 + 4x + 6x - 60 - 40 .= .0
    . . . . . . . . . . . . . . . . . . . . . . .2x - 10x .= .0
    Factor: . . . . . . . . . . . . . . . . . .2x(x - 5) .= .0

    Solve the two equations:
    . . . 2x .= .0 . . . x .= .0
    . . x - 5 .= .0 . . x .= .5

    Substitute into [3]:
    . . x = 0: . y .= .0 - 10 . . y .= -10
    . . x = 5: . y .= .5 - 10 . . y .= .-5


    Therefore, the solutions are: .(0,-10) and (5,-5)

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  3. #3
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    Quote Originally Posted by Kim2425 View Post
    a) x^2 + 3x + 5y = 20
    x + 3y = 1
    Since most people avoid fractions where ever they can I will solve the bottom equation for x. (Always solve the linear equation first, if there is one.)
    x = 1 - 3y

    Insert this value of x into the top equation:
    (1 - 3y)^2 + 3(1 - 3y) + 5y = 20

    (1 - 6y + 9y^2) +(3 - 9y) + 5y = 20

    1 - 6y + 9y^2 + 3 - 9y + 5y = 20

    9y^2 - 10y - 16 = 0

    y = [-(-10) (+/-) sqrt{(-10)^2 - 4*9*(-16)}]/(2*9)

    y = [10 (+/-) sqrt{100 + 576}]/18

    y = [10 (+/-) sqrt{676}]/18

    y = [10 (+/-) 26]/18

    y = [10 + 26]/18 = 2
    or
    y = [10 - 26]/18 = -8/9

    (Which means the quadratic factors. Ah well, the good ol' quadratic formula always works, at least! )


    So anyway:
    x = 1 - 3y

    y = 2 ==> x = 1 - 3*2 = 1 - 6 = -5

    y = -8/9 ==> x = 1 - 3*(-8/9) = 1 + 8/3 = 11/3

    So the solutions are
    (x, y) = (11/3, -8/9)
    (x, y) = (-5, 2)

    -Dan
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