Question

Find the restrictions for the following complex fraction.

Answer 1

\[\frac{\frac{8x^2y}{x+1}}{\frac{6xy^2}{x+1}}\]

This complex fraction simplifies to \[\frac{4x}{3y}\]

I only need help finding the restrictions. I'm not sure how to do it with 2 variables.

Answer 2

HI!!

Answer 3

hello :)

Answer 4

from your completed answer, and the fact that you cannot divide by zero, it is pretty clear that \(y\) cannot be zero right?

Answer 5

That is correct.

Answer 6

but there is another restriction because at the beginning you had another denomiator

Answer 7

\(x \neq -1\)?

Answer 8

yeah that too

Answer 9

Okay haha

Answer 10

WAIT

Answer 11

4x has no restrictions

Answer 12

only y cannot be 0

Answer 13

lol no

Answer 14

b/c you cannot divide by 0
but you CAN divide 0 by something

Answer 15

I know haha we were talking about the initial complex fraction ;)

Answer 16

ahhh

Answer 17

stiill let me check

Answer 18

hmm maybe yeah

Answer 19

thank god i have a real man to come in and check my work
i was worried that i was competent to explain on my own

Answer 20

haha

Answer 21

Do we have any other restrictions?

Answer 22

idn ask @AlexandervonHumboldt2

Answer 23

Okay, thank you so much! :)

Answer 24

\[\color\magenta\heartsuit\]

Answer 25

@misty1212 is right
x cannot be -1 or 0 and y cannot be 0

Answer 26

When is @misty1212 wrong? :P
Don't challenge her :)

Answer 27

:D

Answer 28

\[\frac{\frac{8x^2y}{x+1}}{\frac{6xy^2}{x+1}} \\ =\frac{8x^2y}{x+1} \cdot \frac{x+1}{6xy^2}\]

Answer 29

the bottom factors are (x+1) and 6 and x and y^2

Answer 30

we know 6 can never be 0 so we don't worry about that guy

Answer 31

That's what I did ;) I forgot to put \(x \neq 0\) on here after I write it down....

Answer 32

wrote* sorry

Answer 33

Shoot....I think I just did the wrong question :/

Answer 34

we care about the denominator of the bottom fraction but I decided we didn't need to care about x+1 being 0 twice
for example if we had:
\[\frac{\frac{8x^2y}{x+1}}{\frac{6xy^2}{x+2}} \\ =\frac{8x^2y}{x+1} \cdot \frac{x+2}{6xy^2}
\]since x+2 was on the bottom of a fraction I would also care about that
so I would care wehn x+2=0 and when x+1=0 and when x=0 and y=0 for this example I just made

Answer 35

lololol

Answer 36

Okay...thanks @freckles :)

Answer 37

Yep, I did do the wrong one :'( now I have to re-write a whole lot of math :/

Answer 38

It's a good thing I noticed this before I finished :/