Answer
$\displaystyle \frac{3y+2}{y-3}$
Work Step by Step
To subtract rational expressions with the same denominator,
subtract numerators and place the difference over the common denominator.
If possible, factor and simplify the result.
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Don't forget to place the numerators being subtracted inside parentheses.
$\displaystyle \frac{6y^{2}+y}{2y^{2}-9y+9}-\frac{2y+9}{2y^{2}-9y+9}-\frac{4y-3}{2y^{2}-9y+9}$ =
$ = \displaystyle \frac{6y^{2}+y-(2y+9)-(4y-3)}{2y^{2}-9y+9}$
$ =\displaystyle \frac{6y^{2}+y-2y-9-4y+3}{2y^{2}-9y+9}$
$ =\displaystyle \frac{6y^{2}-5y-6}{2y^{2}-9y+9}$
Factoring trinomials $ax^{2}+bx+c:$
Find factors of $ac$ whose sum is $b.$
If they exist, rewrite $bx$ and factor in pairs.
$6y^{2}-5y-6 =$ ... factors of -36 with sum -5 are -9 and +4
$=6y^{2}-9y+4y-6$
$=3y(2y-3)+2(2y-3)=(2y-3)(3y+2)$
$2y^{2}-9y+9=$ ... factors of $18$ with sum $-9$ are $-6$ and $-3$
$=2y^{2}-6y-3y+9$
$=2y(y-3)-3(y-3)=(y-3)(2y-3)$
$... =\displaystyle \frac{(2y-3)(3y+2)}{(y-3)(2y-3)}$
... reduce (divide the numerator and denominator with common factors)
$=\displaystyle \frac{3y+2}{y-3}$
