Ask your own question, for FREE!
Mathematics 20 Online
OpenStudy (anonymous):

What is the simplified form of the quantity 15 x y squared over x squared minus 5x plus 6 all over the quantity 5 x squared y over 2x squared minus 7x plus 3

OpenStudy (anonymous):

@driftracer305

OpenStudy (anonymous):

I honestly have no clue how to start this one off.

OpenStudy (driftracer305):

k hang on....let me simply it for u

OpenStudy (anonymous):

ok :)

OpenStudy (driftracer305):

k u'll have this \[(15xy^2 (2x^2-7x+6)) \div ((x^2-5x+6) (5x^2y)\]

OpenStudy (anonymous):

Ok, so from here what do i do?

OpenStudy (driftracer305):

hang on....its a long calculation....thats y the other guy ran off

OpenStudy (anonymous):

ohh, Im sorry :(

OpenStudy (anonymous):

well i have my pen and paper ready! haha

OpenStudy (driftracer305):

do u knw how to multiply the exponents above??

OpenStudy (anonymous):

I dont think so :/

OpenStudy (anonymous):

if you kinda guided me i think i could catch on

OpenStudy (driftracer305):

can i see your options??

OpenStudy (driftracer305):

coz i got \[6xy^2-21xy+18y \div x^3-5x^2+6x\]

OpenStudy (anonymous):

ergg yeah thats not an option, one sec let me get them

OpenStudy (anonymous):

OpenStudy (anonymous):

OpenStudy (anonymous):

OpenStudy (anonymous):

OpenStudy (anonymous):

Ok those are the options :)

OpenStudy (driftracer305):

alright hang one 1 sec

OpenStudy (anonymous):

ok :)

OpenStudy (driftracer305):

r u sure u typed the question as it is??

OpenStudy (anonymous):

what is the simplified form of

OpenStudy (driftracer305):

coz it should be \[(3y(2x^2-7x+3)) / (x(x^2-5x+6))\]

OpenStudy (driftracer305):

ohh ok wait...i see my fault

OpenStudy (anonymous):

hmm, we can always move on to another question? its ok :)

OpenStudy (driftracer305):

yea...its 6xy-3y all over x (x-2)

OpenStudy (driftracer305):

so the last one

OpenStudy (driftracer305):

yeah 4.

OpenStudy (anonymous):

Yay! thanks :D

OpenStudy (driftracer305):

np.. ;)

OpenStudy (anonymous):

Youre the best help ever :)

OpenStudy (anonymous):

@driftracer305

OpenStudy (driftracer305):

xd.......glad to help

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!