How do you solve #(1+2y)/(y-4 )= (4y^2+5y)/(2y^2-7y-4)#?

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Answer 1 · 2015-05-29T21:37:31

#(1+2y)/(y−4)=(4y^2+5y)/(2y^2−7y−4)#

#(4y^2+5y)/color(red)(2y^2-8y+y-4)#

#(4y^2+5y)/[color(red)(2y(y-4)+1(y-4)]#

#=(4y^2+5y)/[(1+2y)(y-4)]#

#(1+2y)/color(red)cancel(y−4)=(4y^2+5y)/[(1+2y)color(red)cancel((y-4)]#

#color(red)[(1+2y)^2]=4y^2+5y#

By expansion: #->(a+b)^2=a^2+2ab+b^2#

#color(red)[1+4y+cancel(4y^2)]=cancel(4y^2)+5y#

ANSWER
#y=1#

Answer 2 · 2015-05-29T21:37:33

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