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Posted by **Neil** on Monday, December 17, 2012 at 3:38pm.

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

- Math For College Readiness -
**Steve**, Monday, December 17, 2012 at 3:49pmI'm sure there are some missing parentheses. Making a guess, I'd say you meant

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

putting all over a common denominator, we have

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

((2y^2-11y+15) - (y^2-y-2) + (-4y+7))/(2y^2-9y+10)

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

= (y-2)(y-12)/[(y-2)(2y-5)]

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

- Math For College Readiness -
**Neil**, Monday, December 17, 2012 at 3:54pmThank you very much. Yeah, I should've added the parenthesis to make it look familiar.

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