Question

(2y^2-7y-15)/(3y^2-8y-3) times (9y^2-1)/(4y^2-9) divided by (y^2+3y-10)/(2y^2-9y+9) help??

Answer 1

\[\frac{(2y^2-7y-15)}{(3y^2-8y-3)}.\frac{(9y^2-1)}{(4y^2-9)} \div \frac{(y^2+3y-10)}{(2y^2-9y+9)}\]
flip the last one over; factor it all; cancel the like terms top and bottom and simplify :)

Answer 2

i see a few difference of squares that factor nicely

Answer 3

Okay! Lets see what I get... hahaha

Answer 4

(2y+3)(y-5)(3y+1)(3y-1)(2y-3)(y-3)
-------------------------------
(y-3)(3y+1)(2y-3)(2y+3)(y-5)(y+2)


(3y-1)
----- is what I get
(y+2)

Answer 5

After factoring and cancelling I got.... (3y-1)(y-5)/(y+5)(y-2)
I think I might have messed up though in the middle :/

Answer 6

its alot to keep track of :) but if you take it step by step; and verify your factors; your answer should match mine :)

Answer 7

okay! Thanks :)

Answer 8

lol ... or mine will match yours :)

Answer 9

hahaha :)

Answer 10

(y^2+3y-10)

(y+5)(y-2) is right; yours wins lol

Answer 11

really?? :D

Answer 12

(2y^2-7y-15)

(y-5)(2y+3); yep; yours is good ;)

Answer 13

yay!! thank you!! :)