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Math Help - Values for N

  1. #1
    Junior Member Dragon's Avatar
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    Values for N

    What are the values of n for which the system 3x+ny=3,2x-4y=1 has a solution x<0,y<0
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Dragon View Post
    What are the values of n for which the system 3x+ny=3,2x-4y=1 has a solution x<0,y<0
    3x + ny = 3
    2x - 4y = 1

    Solve the bottom equation for y:
    y = (1/2)x - (1/4)

    and insert this into the top equation:
    3x + n[(1/2)x - (1/4)] = 3

    Now solve for x:
    (3 + n/2)x - n/4 = 3

    (6 + n)/2 * x = n/4 + 3

    x = 2*(n/4 + 3)/(n + 6)

    x = (n/2 + 6)/(n + 6)

    To simplify this a bit, multiply the numerator and denominator by 2:
    x = (n + 12)/(2n + 12)

    So
    y = (1/2)[(n + 12)/(2n + 12)] - (1/4)

    y = (n + 12)/(4n + 24) - (1/4)

    y = (4n + 12 - (4n + 24))/[4(4n + 24)]

    y = 9/(4n + 24)

    So
    x = (n + 12)/(2n + 12)
    y = 9/(4n + 24)

    For what values of n are both x and y negative? Well, both the numerators and denominators have to have the opposite sign of each other. But since the "9" in the y equation is always positive, we require that 4n + 24 < 0, or n < -6.

    Now, for -12 < n < -6 the numerator of x is positive and the denominator negative, so x is also negative here. But for n < -12 the numerator of x is negative, so x is positive. Thus

    -12 < n < -6

    -Dan

    -Dan
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