#When dividing rational expressions; do you need to keep in count the initial denominator (DOMAIN)?

32 messages · Page 1 of 1 (latest)

proper edge
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There seems to be a debate on the website im learning math on, on weather what the professor did was right or not.
I had the right answer but, like other people I rewrote the domain using only the inverse fractions' denominator that was not included in the end result.
The professor included the numbers that used to be the denominator AND he used the new denominators who used to be in as nominators (before inversing the division into a multiplication).

How i view it, It kind of doesn't make sense but ill learn it anyway. I would think you'd need to use only one of the denominators since the nominator doesn't matter, so if you had a 0 in the beginning as a denominator, because it's a division of two fractions you could just inverse them anyway. But at the same time, i guess that's what you could do if it was a multiplication all the same, so there is a point to it.
Anyway, i'll let you guys tell me what it is.

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gilded reefBOT
devout briar
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flip the denominator expression and multiply that to the numerator... that's how to resolve fractions divided by fractions

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Looks like in your example they did this and then broke it down to cross out the common expressions... are you saying that didn't make sense?

proper edge
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no im only talking about it's domain

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i have to include the original domain AND new domain after inversing fractions

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sorry i wasnt clear in my title

gilded reefBOT
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When dividing rational expressions; do you need to keep in count the initial denominator (DOMAIN)?

devout briar
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Domain as all possible values of x in the expression right? It should boil down to the same thing either way... his final answer should still satisfy the condensed form and the starting form of the expression

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Why doesn't he have x cannot = -1?

proper edge
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he only includes them in their factored form

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that's why

devout briar
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Oh nm.... X Can be -1 or 6... cannot be -4

proper edge
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sorry i cluttered the screen

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yeah

devout briar
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Yeah I dunno where the other stuff is coming from at first glance... the point of condensing it down is to find the answer... anything causing zero 0 in the top is valid answer, anything causing zero in the bottom is not valid

proper edge
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his denominator of his first denominator before inversing them is (x+5) (x-6)

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yeah my question here was only that I wasn't sure if domain was supposed to be for both inverse and non inverse form or just one of them

devout briar
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but I see before he crossed out (x+5) in the top and bottom, that -5 is therefore an invalid answer, because it did exist in the bottom before he crossed it out

proper edge
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like I thought only 1 would suffice

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yeah

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his final domain is =/ -5, 6, -4, 4

devout briar
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Same with x=4 before he crossed that out, so it makes sense now what he did

proper edge
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my final domain is =/ -4,4

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so what did was correct then right?

devout briar
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X can be -1, 6.... X cannot be 4, -4, -5?

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Actually in the original expression, there's X-6 in the bottom of the bottom, so that's why he's saying X cannot be 6 as well

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Think it makes sense now... you good too?

proper edge
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Yeah i guess so

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just wanted to verify it!

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thank you!!!

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.close