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Mathematics 20 Online
OpenStudy (anonymous):

does this equation define y as a function of x: x^2y+y=1

OpenStudy (anonymous):

\[x^2y+y=1~~\iff~~y(x^2+1)=1~~\iff~~y=\frac{1}{x^2+1}\]

OpenStudy (anonymous):

so is that yes?

OpenStudy (anonymous):

Yep, if you can express \(y\) explicitly in terms of \(x\) only, then you've got yourself a function.

OpenStudy (anonymous):

okay. how did you get \[y(x^2+1)=1\]

OpenStudy (anonymous):

@SithsAndGiggles

OpenStudy (anonymous):

Factor out the \(y\).

OpenStudy (anonymous):

okay thank you!

OpenStudy (anonymous):

you're welcome

OpenStudy (anonymous):

what about \[x=y^3\]

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