Answer
$\displaystyle \frac{2y-3}{3y-2}$
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.
$\displaystyle \frac{3y^{2}-2}{3y^{2}+10y-8}-\frac{y+10}{3y^{2}+10y-8}-\frac{y^{2}-6y}{3y^{2}+10y-8}$ =
$ = \displaystyle \frac{3y^{2}-2-(y+10)-(y^{2}-6y)}{3y^{2}+10y-8}$
$ =\displaystyle \frac{3y^{2}-2-y-10-y^{2}+6y}{3y^{2}+10y-8}$
$ =\displaystyle \frac{2y^{2}+5y-12}{3y^{2}+10y-8}$
Factoring trinomials $ax^{2}+bx+c:$
Find factors of $ac$ whose sum is $b.$
If they exist, rewrite $bx$ and factor in pairs.
$2y^{2}+5y-12 =$ ... factors of $-24$ with sum $5$ are $+8$ and $-3$
$=2y^{2}+8y-3y-12$
$=2y(y+4)-3(y+4)=(y+4)(2y-3)$
$3y^{2}+10y-8=$ ... factors of $-24$ with sum $10$ are $+12$ and $-2$
$=3y^{2}+12y-2y-8$
$=3y(y+4)-2(y+4)=(y+4)(3y-2)$
$... =\displaystyle \frac{(y+4)(2y-3)}{(y+4)(3y-2)}$
... reduce (divide the numerator and denominator with common factors)
$=\displaystyle \frac{2y-3}{3y-2}$
