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22 April, 09:09

The system of equations

/[/frac{xy}{x + y} = 1, / quad / frac{xz}{x + z} = 2, / quad / frac{yz}{y + z} = 3/]has exactly one solution. What is $z$ in this solution?

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  1. 22 April, 10:44
    0
    z = - 12

    Step-by-step explanation:

    The given system of equations is:

    xy / (x + y) = 1 ... (1)

    xz / (x + z) = 2 ... (2)

    yz / (y + z) = 3 ... (3)

    From (1) : x + y = xy

    => y = xy - x

    y = x (y - 1)

    x = y / (y - 1) ... (4)

    From (2) : 2 (x + z) = xz

    => 2x + 2z = xz

    2x = xz - 2z

    2x = z (x - 2)

    z = 2x / (x - 2) ... (5)

    From (3) : 3 (y + z) = yz

    => 3y + 3z = yz

    3y = yz - 3z

    3y = z (y - 3)

    z = 3y / (y - 3) ... (6)

    Comparing (5) and (6)

    2x / (x - 2) = 3y / (y - 3)

    2x (y - 3) = 3y (x - 2)

    2xy - 6x = 3xy - 6y

    6 (y - x) = xy ... (7)

    But from (1) : xy = x + y

    Using this in (7), we have

    6 (y - x) = x + y

    6y - y - 6x - x = 0

    5y - 7x = 0

    5y = 7x

    x = 5y/7 ... (8)

    Using this in (4)

    5y/7 = y / (y - 1)

    1 / (y - 1) = 5/7

    (y - 1) = 7/5

    y = 1 + 7/5

    y = 12/5 ... (9)

    Using this in (8)

    x = 5 (12/5) / 7 = 12/7 ... (10)

    Using (10) in (5)

    z = 2x / (x - 2)

    z = 2 (12/7) : (12/7 - 2)

    = 24/7 : - 2/7

    = 24/7 * (-7/2)

    = - 24/2 = - 12

    z = - 12.
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