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Algebra 15 Online
OpenStudy (anonymous):

Divide: 8xy^2 - 14y^2/2y

OpenStudy (anonymous):

4xy^2-7y^2 4x-7 4xy-7y 4y-7x

OpenStudy (anonymous):

\[\frac{ 8xy^{2} - 14y^{2} }{ 2y }\] Right?

OpenStudy (mosaic):

(8xy^2 - 14y^2) / 2y = 8xy^2 / 2y - 14y^2 / 2y = ?

OpenStudy (anonymous):

Yup @jcpd910

OpenStudy (anonymous):

How did you do that @jcpd910

OpenStudy (anonymous):

I used the equation button.

OpenStudy (anonymous):

do you get that?

OpenStudy (anonymous):

Yeah, I'm solving it.

OpenStudy (anonymous):

Okay, so first... \[\frac{ 8xy^{2} }{ 2y } - \frac{ 14y^{2} }{ 2y }\]

OpenStudy (anonymous):

Would it be b

OpenStudy (anonymous):

No.

OpenStudy (anonymous):

Ii'm getting 4xy - 2y

OpenStudy (anonymous):

@phi

OpenStudy (anonymous):

I just called in backup.

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

@kirbykirby do you get this

OpenStudy (kirbykirby):

14/2 = 7

OpenStudy (anonymous):

Im getting B

OpenStudy (anonymous):

4x-7

OpenStudy (kirbykirby):

not quite, \[ \frac{y^2}{y}=y^{2-1}=y\]

OpenStudy (kirbykirby):

The y doesn't completely disappear

OpenStudy (anonymous):

Ah, I used 4 not 14, I got the answer now.

OpenStudy (anonymous):

[4xy^2-7y^2\]

OpenStudy (anonymous):

kirby can explain it better.

OpenStudy (anonymous):

So it has to be either A or C

OpenStudy (kirbykirby):

Well @jcpd910 got most of it down already :) Just have to be careful when dividing out the y You have a y^2 in the top and y in the bottom when you divide that out, you do: \(\large \frac{y^2}{y}=y^{2-1}=y\), so a y is still there

OpenStudy (kirbykirby):

only y is left, not y^2

OpenStudy (anonymous):

Would anything still be squared

OpenStudy (anonymous):

Ok so c

OpenStudy (kirbykirby):

yes

OpenStudy (anonymous):

Thanks

OpenStudy (kirbykirby):

=]

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