How do you evaluate #\frac { 2y ^ { 2} + 9y - 5} { y ^ { 2} - 25} \times \frac { y - 5 } { 2y ^ { 2} - y }#?

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Answer 1 · 2017-10-19T01:17:35

#1/y#

If you factor everything you can in the problem, then much of it cancels out.

#((y+5)(2y - 1))/((y+5)(y-5)) * (y-5)/(y(2y -1)#

#(cancel(y+5)cancel(2y-1))/(cancel(y+5)cancel(y-5)) * cancel(y-5)/(y(cancel(2y-1))#

(which I suspect was the reason for giving you this problem, so you would learn to look for cancellations like this)

Leaving you with #1/y#

GOOD LUCK

Answer 2 · 2017-10-19T01:27:03

1/y

#2y^2+9y-5 = 2y^2 -y+10y -5 = y(2y-1) +5(2y-1) = (2 y-1)(y+5)#

#y^2-25 = (y+5)(y-5)#

#2y^2-y = y(2y-1)#

#:.((2y^2+9y-5)/(y2-25))*((y-5)/(2y^2-y)) #

#= cancel(color(red)((2y-1)(y+5)(y-5)))/(cancel(color(red)((y+5)(y-5)(2y-1)))y) = 1/y#