Math

√(5y+1)-√(3y-5)=2, I need to solve algebraically, and state any restrictions on the values for the variables, which I can't remember how to do

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asked by Al
  1. since √x is not real for x < 0, we need

    5y+1 >= 0
    y >= -1/5

    3y-5 >= 0
    y >= 5/3

    So, we need y >= 5/3 for the domain

    Now proceed to solve:

    √(5y+1) - √(3y-5) = 2
    5y+1 - 2√[(5y+1)(3y-5)] + 3y-5 = 4
    8y - 8 = 2√[(5y+1)(3y-5)]
    (4y-4)^2 = (5y+1)(3y-5)
    16y^2 - 32y + 16 = 15y^2-22y-5
    y^2 - 10y + 21 = 0
    (y-3)(y-7) = 0

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    posted by Steve

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