Melody

Are you the question asker?

Yes the first three answers should have been discarded but did you bother to check the next 2 answers.

You still have not said that you have?  Are they right?  Are they wrong?

If you knew some things were correct then you should have said so.

Your response was very rude.

You can politely suggest to people that maybe they should check their answer by substitution without being rude about it.

Did you bother to check if the other answers are correct?

Or are you just here to criticize?

\(\frac{2x^2-3x}{x^2-x} + 5x -11 = \frac{3x^2+5x+2}{x^2-1}\\ \frac{2x^2-3x}{x(x-1)} + 5x -11 = \frac{3x^2+5x+2}{(x-1)(x+1)}\\ x\ne\pm1,\quad x\ne0\\ [\frac{2x^2-3x}{x(x-1)} + 5x -11]*[x(x-1)(x+1)] = [\frac{3x^2+5x+2}{(x-1)(x+1)}]*[x(x-1)(x+1)] \\ \frac{2x^2-3x}{x(x-1)} *[x(x-1)(x+1)]+ (5x -11)*[x(x-1)(x+1)] = [\frac{3x^2+5x+2}{1}]*[x] \\ (2x^2-3x)(x+1)+ (5x -11)(x(x^2-1))=x (3x^2+5x+2) \\ (2x-3)(x+1)+ (5x -11)(x^2-1)= (3x^2+5x+2) \\ (2x^2-x-3)+ (5x^3 -11x^2-5x+11)= (3x^2+5x+2) \\ 5x^3 -9x^2-6x+8= 3x^2+5x+2 \\ 5x^3 -12x^2-11x+6= 0 \\ (x-3)(5x-2)(x+1)=0\\ \text{x can't equal -1 so}\\ x=3 \;\;or\;\;x=0.4 \)

LaTex

\frac{2x^2-3x}{x^2-x} + 5x -11 = \frac{3x^2+5x+2}{x^2-1}\\
\frac{2x^2-3x}{x(x-1)} + 5x -11 = \frac{3x^2+5x+2}{(x-1)(x+1)}\\
x\ne\pm1,\quad x\ne0\\
[\frac{2x^2-3x}{x(x-1)} + 5x -11]*[x(x-1)(x+1)] = [\frac{3x^2+5x+2}{(x-1)(x+1)}]*[x(x-1)(x+1)] \\
\frac{2x^2-3x}{x(x-1)} *[x(x-1)(x+1)]+ (5x -11)*[x(x-1)(x+1)] = [\frac{3x^2+5x+2}{1}]*[x] \\
(2x^2-3x)(x+1)+ (5x -11)(x(x^2-1))=x (3x^2+5x+2) \\
(2x-3)(x+1)+ (5x -11)(x^2-1)= (3x^2+5x+2) \\
(2x^2-x-3)+ (5x^3 -11x^2-5x+11)= (3x^2+5x+2) \\
5x^3 -9x^2-6x+8= 3x^2+5x+2 \\
5x^3 -12x^2-11x+6= 0 \\
(x-3)(5x-2)(x+1)=0\\
\text{x can't equal -1  so}\\
x=3 \;\;or\;\;x=0.4

I'll walk you through my solution without giving you the final answer.

Let RR sit together

If they are in the 1st and 2nd position how many ways can the there frogs be seated?

If they are in the 2ndand 3rd position how many ways can the there frogs be seated?

If they are in the 3rd and 4th position how many ways can the there frogs be seated?

If they are in the 4th and 5th position how many ways can the there frogs be seated?

If they are in the 5th and 6th position how many ways can the there frogs be seated?

Let there be just one seat between the 2 read frogs

Go through the same process as above.  Look at each alternative.

Do the same for  2 seats between the two red frogs.

3 seats between

4 seats between

and lastly 5 seats between.

Now add the numbers all up. 

The number of ways the 6 people can sit is  6! = 720 ways

Let the couples be  AB, CD, and EF

The number of ways that AB can sit together is  5!*2 = 240 ways

The number of ways that CD can sit together is  5!*2 = 240 ways

The number of ways that EF can sit together is  5!*2 = 240 ways

The number is ways that they can all sit in pairs is  3!*2^3 = 48

           I have already added this in 3 times so I will have to subtract it twice.

How many ways can there be in 2 pairs only.  

          The odd pair out will have to be in in the 3rd and 6th chairs

                  or the 1st and 6th

                  or the 1st and 4th

         So I have   3 * 3C2 *2^3 = 72 ways

         I have added each of these in twice so I have to subtract once

So I get that the number of ways for at least 1 couple to be stated together is

3*240 - 2*48 - 72 = 552 ways

So        P(at least one couple sits together) = 552/720 = 23/30

If you get confirmation that this answer is either correct or incorrect, please let me know. 

Well,  I will let you.  

Hi Ginger,

I have just been reading through those old posts.

Thanks for the trip down memory lane  :))

Solveit tells me that U just means union.

I have never seen it used like this before so anything I write is pure guess work.

It seems to me that 

   \(\displaystyle \cup _{n=1}^{5}\;A_n = \{A_1, A_2, A_3,A_4, A_5\}\)

\(\displaystyle \cup _{k=1}^{n}\;A_n = \{A_1, A_2, A_3,A_4, A_5, ....., A_n\}\\~\\ (\displaystyle \cup _{k=1}^{n}\;A_n) = \{A_1, A_2, A_3,A_4, A_5, ....., A_n\}\\~\\ \displaystyle \cup _{k=1}^{5}(\displaystyle \cup _{k=1}^{n}\;A_n) = \{A_1\}\cup \{ A_1, A_2\} \cup \{ A_1, A_2, A_3\} \cup \{ A_1, A_2, A_3, A_4\}\cup \{ A_1, A_2, A_3, A_4,A_5\}\\~\\ \displaystyle \cup _{k=1}^{5}(\displaystyle \cup _{k=1}^{n}\;A_n) =\{ A_1, A_2, A_3, A_4,A_5\}\\~\\ \displaystyle \cup _{k=1}^{5}(\displaystyle \cup _{k=1}^{n}\;A_n) =\displaystyle \cup _{k=1}^{5}\;A_n\\~\\\)

LaTex:

\displaystyle \cup _{k=1}^{n}\;A_n = \{A_1, A_2, A_3,A_4, A_5, ....., A_n\}\\~\\
(\displaystyle \cup _{k=1}^{n}\;A_n) = \{A_1, A_2, A_3,A_4, A_5, ....., A_n\}\\~\\

\displaystyle \cup _{k=1}^{5}(\displaystyle \cup _{k=1}^{n}\;A_n) = \{A_1\}\cup \{ A_1, A_2\}  \cup \{ A_1, A_2, A_3\} \cup \{ A_1, A_2, A_3, A_4\}\cup \{ A_1, A_2, A_3, A_4,A_5\}\\~\\
\displaystyle \cup _{k=1}^{5}(\displaystyle \cup _{k=1}^{n}\;A_n) =\{ A_1, A_2, A_3, A_4,A_5\}\\~\\
\displaystyle \cup _{k=1}^{5}(\displaystyle \cup _{k=1}^{n}\;A_n) =\displaystyle \cup _{k=1}^{5}\;A_n\\~\\

Hi Solveit,

Perhaps someone can tell me what the U symbol is,  I have not seen it before.   

Thanks Grace  :)