How do you combine #(y-3)/(y-2)-(y+1)/(2y-5)+(-4y+7)/(2y^2-9y+10)#?

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Answer 1 · 2016-07-16T15:55:40

#(y-12)/(2y-5)#

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

you can obtain an equivalent expression with a single fraction:

#((y-3)(2y-5)-(y+1)(y-2)+(-4y+7))/(2y^2-9y+10)#

#=(2y^2-5y-6y+15-y^2+2y-y+2-4y+7)/(2y^2-9y+10)#

#=(y^2-14y+24)/(2y^2-9y+10)#

Then, since

#y^2-14y+24=(y-2)(y-12)#,

you can factor:

#(cancel(y-2)(y-12))/(cancel(y-2)(2y-5))#

so the final expression is:

#(y-12)/(2y-5)#