simplify the following a. 1/y^2-4 + 3/y^2+y-6 divided by 1/y^2-4 + 2/y^2+3y+2 b. 4a/2a^2-1 algebra

1/y^2-4 + 3/y^2+y-6 divided by 1/y^2-4 +
2/y^2+3y+2

1/(y^2-4) + 3/(y^2+y-6) divided by 1/(y^2-4) +
2/(y^2+3y+2)

First we will factor the
equations.

1/((y-2)(y+2) + 3/((y+3)(y-2)) divided by
1/((y-2)(y+2)) + 2/((y+2)(y+1))

Now find the LCD of both
sides

(y-2)(y+2)(y+3) and (y-2)(y+2)(y+1) are the LCD's,
now get the same denominator for both fractions on both
equations.

1/((y-2)(y+2))*(y+3)/(y+3) +
3/((y+3)(y-2))*(y+2)/(y+2) and

1/((y-2)(y+2))*(y+1)/(y+1)+2/((y+2)(y+1))*(y-2)/(y-2)

to
get

(y+3)/((y-2)(y+2)(y+3)) + 3(y+2)/((y-2)(y+2)(y+3))
and
(y+1)/((y-2)(y+2)(y+1)) +
2(y-2)/((y-2)(y+2)(y+1))

Now that we have the same
denominators we add the
numerators

((y+3)+3(y+2))/((y-2)(y+2)(y+3)) and
((y+1)+2(y-2))/((y-2)(y+2)(y+1))

And simplify to
get

(4y+9)/((y-2)(y+2)(y+3)) and
(3y-3)/((y-2)(y+2)(y+1))

Now we divide by changing the
division to a multiplication by the
reciprocal

(4y+9)/((y-2)(y+2)(y+3)) *
((y-2)(y+2)(y+1))/(3y-3)

Cancel y+2 and y-2 to
get

((4y+9)(y+1))/(3(y-1)(y+3))

And
our final answer
is

((4y+9)(y+1))/(3(y-1)(y+3))