Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (jazzyfa30):

(6y2 -6/ 8y2 +8y)/(3y -3 / 4y2 + 4y)

OpenStudy (luigi0210):

Easy, flip and multiply

OpenStudy (jazzyfa30):

???????

OpenStudy (anonymous):

flip the second one and multiply

OpenStudy (jazzyfa30):

?????

OpenStudy (anonymous):

6y2-6/8y2+8y x 4y2+4y/3y-3

OpenStudy (luigi0210):

\[\frac{6y^2-6}{8y^2+8y}*\frac{4y^2+4y}{3y-3}\]

OpenStudy (anonymous):

u can cross out 8y2 and change it to 2 and cross out the 4y2 completely

OpenStudy (anonymous):

same with 8y and 4y

OpenStudy (anonymous):

u also cancel out the 6y2 and make it 2y because 6/3= 2 and the y2 took the y on the bottom.

OpenStudy (anonymous):

and so ur left with 2y-2/2+2: 2y-2/4 simplify that and u get 1/2y-1/2

OpenStudy (jazzyfa30):

ummmmm thats wrong

OpenStudy (luigi0210):

@nickirivera stop >.>

OpenStudy (jazzyfa30):

thats wrong

OpenStudy (jazzyfa30):

help please

OpenStudy (luigi0210):

Okay hold on

OpenStudy (luigi0210):

\[\frac{6y^2-6}{8y^2+8y}*\frac{4y^2+4y}{3y-3}=\frac{6(y^2-1)}{8(y^2+1)}*\frac{4y(y+1)}{3(y-1)}\]

OpenStudy (luigi0210):

Can you finish it?

OpenStudy (jazzyfa30):

ummmm i think the answer 1

OpenStudy (luigi0210):

\[\frac{6(y+1)}{8(y^2+1)}*\frac{4y+1}{3}\] That equals 1?

OpenStudy (luigi0210):

You were right, nice job Jazzy! :)

OpenStudy (jazzyfa30):

the answer was wrong

OpenStudy (luigi0210):

Darn I see where I messed up.

OpenStudy (luigi0210):

\[\frac{6(y+1)}{8(y^2+1)}*\frac{4y(y+1)}{3}\]

OpenStudy (luigi0210):

Now simplify: \[\frac{2(y+1)}{2(y^2+1)}*\frac{y(y+1)}{1}=\frac{y+1}{y^2+1}*y(y+1)\]

OpenStudy (jazzyfa30):

ummmmmm idk

OpenStudy (luigi0210):

@nickirivera Is my process right? I need your feedback please.

OpenStudy (anonymous):

ohhhhh, lol, oki, yea hes right.. *props @luigi*

OpenStudy (luigi0210):

Thanks Nicki!

OpenStudy (anonymous):

ur welcome

OpenStudy (jazzyfa30):

hello

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!