rational expressions: (6p-18/9p)/(3p-9/p^2+2p)

(6p-18/9p)/(3p-9/p^2+2p)
[6(p-3)/9p]*[p(p+2)/3(p-3)]
(2/9p)*[p(p+2)/1]

[2p(p+2)]/9p

HELP! is this correct?

Re: rational expressions

jane said:

(6p-18/9p)/(3p-9/p^2+2p)
[6(p-3)/9p]*[p(p+2)/3(p-3)]
(2/9p)*[p(p+2)/1]

[2p(p+2)]/9p

HELP! is this correct?

Click to expand...

I cannot figure out what you are writing!!

Please use parenthesis as I had suggested in:

http://www.freemathhelp.com/forum/viewt ... ght=#95859

I did!!! Its fractions.

Fraction one is (6p-18) over 9p
that fraction is divided by 3p-9 over p squared +2p
then you have to factor it....

jane said:

I did!!! Its fractions.
Fraction one is (6p-18) over 9p
that fraction is divided by 3p-9 over p squared +2p
then you have to factor it....

Click to expand...

BUT you posted this: (6p-18/9p)/(3p-9/p^2+2p) ; that's NOT the same thing!

PROPERLY shown is this way:
[(6p - 18) / (9p)] / [(3p - 9) / (p^2+2p)]

The [(6p - 18) / (9p)] simplifies to [(2p - 6) / (3p)]

Remember that (a/b) / (c/d) = (a/b) * (d/c); so:

[(2p - 6) / (3p)] * [(p^2+2p) / (3p - 9)]

[(2p - 6)(p^2 + 2p)] / [3p(3p - 9)]

That's all the typing I'm doing!
Carry on: complete the multiplications, simplify and you'll end up with a nice 2(p + 2) / 9

Thanks for the help.