# 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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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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