Dividing a rational expression

[silverdragon316]

y^2-16 divided by y-4
3y-9 divided by y-3

(y+4)(y-4)(y-3)
3(y-3)(y-4)


When you divide it's the same a s multiplying by the reciprical right?
then factor.
then cancel out. I came up with.
y+4
3

 

[stapel]

y^2-16 divided by y-4
3y-9 divided by y-3

You might want to check your posts, or "Preview" them before posting. If you'd done so, you'd have noticed that your attempt at space-delineated multi-line formatting hadn't been successful.

Does your expression (above) mean either of the following?

. . . . .[(y^2 - 16) / (y - 4)] / [(3y - 9) / (y - 3)]

. . . . .[(y^2 - 16) / (3y - 9)] / [(y - 4) / (y - 3)]

Or something else?

When you divide it's the same a s multiplying by the reciprical right?

Yes.

then factor. then cancel out.

I'm sorry, but I don't know what this means...?

Please reply with clarification. Thank you.

Eliz.

 

[soroban]

Hello, silverdragon316!

If the problem is: \(\displaystyle \L\:\frac{\frac{y^2\,-\,16}{y\,-\,4} }{\frac{3y\,-\,9}{y\,-\,3}}\;\) your answer is correct.


We have: \(\displaystyle \L\:\frac{\sout{(y\,-\,4)}(y\,+\,4)}{\sout{y\,-\,4}}\,\cdot\,\frac{\sout{y\,-\,3}}{3\sout{(y\,-\,3)}} \;=\;\frac{y\,+\,4}{3}\)

 

[silverdragon316]

Hello, silverdragon316!

If the problem is: \(\displaystyle \L\:\frac{\frac{y^2\,-\,16}{y\,-\,4} }{\frac{3y\,-\,9}{y\,-\,3}}\;\) your answer is correct.


We have: \(\displaystyle \L\:\frac{\sout{(y\,-\,4)}(y\,+\,4)}{\sout{y\,-\,4}}\,\cdot\,\frac{\sout{y\,-\,3}}{3\sout{(y\,-\,3)}} \;=\;\frac{y\,+\,4}{3}\)

yes, thank you