Math For College Readiness
posted by Neil .
y - 3 / y - 2 - y + 1 / 2y - 5 + -4y + 7 / 2y^2 - 9y + 10
I'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)
Thank you very much. Yeah, I should've added the parenthesis to make it look familiar.