Question

Question can someone perform this operation? simplify the expression: numerator (6r^2-7r-3) divided by denominator (r^2-1) divided by numerator (4r^2-12r+9) divided by denominator (r^2-1) multiply by numerator (2r^2-r-3) divided by denominator (3r^2-2r-1)

Answer 1

Can I just give the answer out? Too much typing.

Answer 2

Yes

Answer 3

Your question is worded a little bit strangely...
When you say numerator and then the denominator, do you mean each of those are one set of terms?
aka
(6r^2 - 7r - 3) / (r^2 - 1)
------------------------- <--divided by
(4r^2 - 12r + 9) / (r^2 - 1)
-------------------------
(2r^2 - r - 3) / (3r^2 - 2r - 1)

Answer 4

Yes like numerator 12 divided by denominator 2=6

Answer 5

Can you write this out using grouping symbols to make the question easier to read?

Answer 6

Yes all of the numerators are numerator but in sets like you have up top there.

Answer 7

For example:

4/2/2 can be taken several ways:

(4/2)/2 = 2/2 = 1
4/(2/2) = 4/1 = 4

As you can see; 4 does not equal 1 so the way the problem is set up is important

Answer 8

Sorry, going to start it now.

Answer 9

(3r + 1)^ 2 * (r^2 - 1)

Answer 10

If you define a way to set the problem up like using "//" to be the determining division between top and bottom... that would be helpful:

example:


4//2/2 = 4

4/2//2 = 1

Answer 11

(r+1)/(r-1)

Answer 12

6r^2-7r-3/r^2-1 / 4r^2-12r+9/r^2-1 times 2r^2-r-3/3r^2-2r-1

Answer 13

Beens that looks like the right solution, does anyone have more input?

Answer 14

(2r-3)(3r+1)
--------
(r+1)(r-1) (2r-3)(r+1)
-------------- times --------- = ?
(2r-3)(2r-3) (r-1)(3r+1)
----------
(r+1)(r-1)

Answer 15

If so, I get:

(3r+1)(2r-3)(r+1) (r+1)
--------------- = ----- same as Beens
(2r-3)(r-1)(3r+1) (r-1)