Question

(y+6)^2/y^2-36 X 3y-18/2y +12

Answer 1

ok first u have to make factors of that y^2-36 in the denominator..

Answer 2

(y+6)(y-6)

Answer 3

ok good now u can cancel y+6 with the one in numerator..

Answer 4

so now it's (y+6)/(y-6)

Answer 5

yes.. on other side.. u can take 3 common from numerator and 2 from denominator.

Answer 6

I don't get how to do that part

Answer 7

take 3 common in numerator.
3(y-6)
and now denominator.
2(y+6)

Answer 8

now do you cancel more?

Answer 9

yes if possible

Answer 10

wouldn't it just be 3/2?

Answer 11

\[\frac{ y+6 }{ y-6 } * \frac{ 3( y-6)}{ 2(y+6) }\]

Answer 12

yes.

Answer 13

is that right? i feel like you can't cancel them if the coefficients are there in the second pa\rt

Answer 14

I realize that this has almost been solved but.
\[\frac{(y+6)^2}{y^2-36}\times \frac{(3y-18)}{(2y+12)}\]
\[\frac{(3y-18)(y+6)}{(y-6)(2y+12)}\]
\[\frac{3y^2+18y-18y}{2y^2+12y-12y}\]
\[\frac{3y^2}{2y^2}\]
Then, you can cancel, and you get your answer.

Answer 15

I of course skipped a step at the beginning, I didn't show my factoring.

Answer 16

so is it 3/2 or do you keep the y^2

Answer 17

There is a y^2 on top and bottom, so you could cancel them. The answer should be 3/2.

Answer 18

could you help me with a division one i'll ask

Answer 19

Sure!