Answer:
[tex]\dfrac{y^2+2y-13}{(y-4)(y+3)(y-3)}[/tex]
Step-by-step explanation:
Given the expression
[tex]\dfrac{y+7}{y^2-y-12}-\dfrac{2}{y^2-9}[/tex]
First, factor each denominator:
1. [tex]y^2-y-12=y^2-4y+3y-12=y(y-4)+3(y-4)=(y-4)(y+3)[/tex]
2. [tex]y^2-9=(y-3)(y+3)[/tex]
Now the expression is
[tex]\dfrac{y+7}{(y-4)(y+3)}-\dfrac{2}{(y-3)(y+3)}[/tex]
Now, multiply the first numerator by (y-3) and the second by (y-4) and write the result as a fraction with denominator [tex](y-4)(y+3)(y-3):[/tex]
[tex]\dfrac{(y+7)(y-3)-2(y-4)}{(y-4)(y+3)(y-3)}=\dfrac{y^2-3y+7y-21-2y+8}{(y-4)(y+3)(y-3)}=\dfrac{y^2+2y-13}{(y-4)(y+3)(y-3)}[/tex]
This fraction cannot be simplified more.
