document.write( "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. \n" ); document.write( "
Algebra.Com's Answer #134625 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
(5y-4)^-1/4 - 5(5y-4)^-4/3 =0
\n" ); document.write( " (5y-4)^(-3/12) - 5(5y-4)^(-16/12) = 0\r
\n" ); document.write( "\n" ); document.write( "----
\n" ); document.write( "[(5y-4)^(-16/12)][(5y-4)^(-12/12) - 5] = 0\r
\n" ); document.write( "\n" ); document.write( "[(5y-4)^(-16/12)][(5y-4)^-1 - 5] = 0\r
\n" ); document.write( "\n" ); document.write( "Solve for \"y\":\r
\n" ); document.write( "\n" ); document.write( "(5y-4)^(-16/12) = 0 or (5y-4)^(-1) = 5\r
\n" ); document.write( "\n" ); document.write( "y cannot equal 4/5 because the exponent is negative:
\n" ); document.write( "-----------
\n" ); document.write( "Solve (5y-4)^-1 = 5
\n" ); document.write( "Invert to get:
\n" ); document.write( "5y-4 = 1/5
\n" ); document.write( "5y = (1/5)+4
\n" ); document.write( "5y = 21/5
\n" ); document.write( "y = 21/25
\n" ); document.write( "=================
\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
\n" ); document.write( " \n" ); document.write( "
\n" );