Answer:
Neither of them
Step-by-step explanation:
[tex]\dfrac{-6y-4}{-8y-4} = \dfrac{2(-3y-2)}{4(-2y-1)} = \dfrac{-2(3y+2)}{-4(2y+1)} = \dfrac{2(3y+2)}{4(2y+1)}[/tex]
We can cancel this fraction and obtain:
[tex]\dfrac{3y+2}{2(2y+1)}[/tex]
So neither of them wrote an expression that is equivalent to the original expression. Felipe's expression is missing a 2 in the denominator.