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:

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

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