Answer:
Step-by-step explanation:
Hello, please consider the following.
We take y different from -5, as dividing by 0 is not allowed and we can compute as below.
[tex]\dfrac{y^2+3y-10}{3y+15}=\dfrac{(y+5)(y-2)}{3y+15}\\\\=\dfrac{(y+5)(y-2)}{3(y+5)}\\\\\large \boxed{\sf \bf \ \ =\dfrac{y-2}{3} \ \ }[/tex]
Thank you
