Ask Question
3 April, 05:22

The system of equations

(xy) / (x+y) = 1, (xz) / (x+z) = 2, (yz) / (y+z) = 3 has exactly one solution. What is $z$ in this solution?

+4
Answers (1)
  1. 3 April, 05:50
    0
    z = - 12

    Step-by-step explanation:

    Rewritting the equations, we have:

    x + y = xy (eq1)

    2x + 2z = xz (eq2)

    3y + 3z = yz (eq3)

    From the first equation:

    x = y / (y-1) (eq4)

    From the third equation:

    y = 3z / (z - 3) (eq5)

    Using the value of y from (eq5) in (eq4), we have:

    x = [3z / (z - 3) ] / [3z / (z - 3) - 1]

    x = [3z / (z - 3) ] / [ (3z - z + 3) / (z - 3) ]

    x = 3z / (2z + 3) (eq6)

    Using the value of x from (eq6) in (eq2), we have:

    6z / (2z + 3) + 2z = (3z / (2z + 3)) * z

    (6z + 4z^2 + 6z) / (2z + 3) = 3z^2 / (2z + 3)

    12z + 4z^2 = 3z^2

    z^2 = - 12z

    z = - 12
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “The system of equations (xy) / (x+y) = 1, (xz) / (x+z) = 2, (yz) / (y+z) = 3 has exactly one solution. What is $z$ in this solution? ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers