SOLUTION: I'm working on quadratic equations Equations and inequalities of one variable, in higher degree polynomial equations. Equation (5y-4)^-1/4

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Question 179694: I'm working on quadratic equations Equations and inequalities of one variable, in higher degree polynomial equations. Equation (5y-4)^-1/4 - 5(5y-4)^-4/3 =0. I understand that I must factor out (5y-4)^-4/3. I understand "factor out" as meaning to remove this from the equation. However, this results in (5y-4)^-4/3 ((5y-4)-4)=0, which I do not understand. Can someone explain what took place.
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
(5y-4)^-1/4 - 5(5y-4)^-4/3 =0
(5y-4)^(-3/12) - 5(5y-4)^(-16/12) = 0
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[(5y-4)^(-16/12)][(5y-4)^(-12/12) - 5] = 0
[(5y-4)^(-16/12)][(5y-4)^-1 - 5] = 0
Solve for "y":
(5y-4)^(-16/12) = 0 or (5y-4)^(-1) = 5
y cannot equal 4/5 because the exponent is negative:
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Solve (5y-4)^-1 = 5
Invert to get:
5y-4 = 1/5
5y = (1/5)+4
5y = 21/5
y = 21/25
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Cheers,
Stan H.

SOLUTION: I'm working on quadratic equations Equations and inequalities of one variable, in higher degree polynomial equations. Equation (5y-4)^-1/4 - 5(5y-4)^-4/3 =0. I understand that I m