For problems like these, you must start by factoring everything.
#3y^2 + 18y + 5# can be factored as:
#3(y^2 + 6y + 5) = 3(y + 5)(y + 1)#
#6y + 6# can be factored as:
#6(y + 1)#
#y - 5# is as simplified as it can be.
#y^2 - 25# can be factored as #(y + 5)(y - 5)#
Putting all of this back together:
#(3(y + 5)(y + 1))/(6(y + 1)) xx (y - 5)/((y + 5)(y - 5))#
See what you can cancel out, using the property #a/a = 1#:
#(cancel((3))cancel((y + 5))cancel((y + 1)))/(cancel((6))^2cancel((y + 1))) xx (cancel(y - 5))/(cancel((y + 5))cancel((y - 5)))#
We are left with:
#1/2#
Now, before stating the final answer, we need to determine any restrictions on the variable. These will occur when the denominator equals #0#. Hence, they can be found by setting the denominator to #0# and solving for #x#.
#6y + 6 = 0" and "y^2 - 25 = 0#
#y = -1 " and "y = +-5#
Thus, #y != -1, 5, -5#
Hopefully this helps!
