NationStates Jolt Archive

Yet again, I have another math problem that requires the assistance of others. If you wouldn't mind taking several minutes to point me in the right direction, that is.

The goal is to try to set the two sides equal to each other by solving them indepedently. Numbers can't be plugged in either. I tried taking the natural log of either of the sides, but I still am unsure of where to go from there.

http://img87.echo.cx/my.php?image=equation9ov.png

What variable are you solving for?

What is the digit in front of the x?

I'm at a loss as to what you need to do. Why can't you just subtract the C/2 from both sides, and the other variable to both sides, and go from there?

ego contemno mathematics! I took basic math in college, and I thought that was hard. :(

The end result is not an actual value for a single variable but two equal expressions. The actual question is to show that the relationship holds by computing both sides seperately.

The variable in front of the "x" is a "-r" and an "r" respectively for both sides.

C/2 can't be subtracted from both sides because the sides of the equation are to be solved seperately.

Maths! Seriously this in inexecusable American language malformation. It is plural! Do you say mathamatic? No of course you dont. You say mathamatics. In the same way it is Maths, not MATH!

Yet again, I have another math problem that requires the assistance of others. If you wouldn't mind taking several minutes to point me in the right direction, that is.

The goal is to try to set the two sides equal to each other by solving them indepedently. Numbers can't be plugged in either. I tried taking the natural log of either of the sides, but I still am unsure of where to go from there.

http://img87.echo.cx/my.php?image=equation9ov.png

Can you manually type the equation in as part of a post? that link won't open up.

As you displayed, the 2 sides are already set equal to each other. So what exactly are you trying to solve for? Are you having to find the value of a certain variable? Or are you having to show proof that the statement is true? What are the directions that accompanied this problem?

And actually the signs are different so you couldn't just subtract c over 2 from both sides.

EDIT: I see you answered my question as I was posting it. That clears things up a bit.

Yes, yes. Yet you still understood what I meant. Either way, you fail to help me.

I'm not sure of the point of this equation. C and r can be any value you'd like and the equation would still hold.

1)cancel all the C's

2)add 1/2 to each side

3)multiply the 1 by (1+exp(rx))/(1+exp(rx)) and simplify

4)multiply everything by (1+exp(rx)) and simplify

and you get exp(rx)=exp(rx)

C/2 can't be subtracted from both sides because the sides of the equation are to be solved seperately.
You can't "solve" anything if there isn't an equation

The symmetry property mentioned in Excercise 2(b)* can be expressed more generally for the graph of the logistic function

f(x) = c / 1+e^-rx

by saying that

f(x) - c/2 = c/2 - f(-x)

for all x. Show that this relationship holds by computing both sides seperately.

*I don't think this part is necessary to solve the problem, but if it happens to be, let me know and I'll provide you with the info.

What about "The Fall of Math" by 65 Days of Static :p

I'm not sure of the point of this equation. C and r can be any value you'd like and the equation would still hold.

1)cancel all the C's

2)add 1/2 to each side

3)multiply the 1 by (1+exp(rx))/(1+exp(rx)) and simplify

4)multiply everything by (1+exp(rx)) and simplify

and you get exp(rx)=exp(rx)

Yes, but I believe it's trying to get us to prove it through evaluting the individual expressions and getting two identical ones rather than actually plugging in numbers.

Yes, but I believe it's trying to get us to prove it through evaluting the individual expressions and getting two identical ones rather than actually plugging in numbers.

Well i didnt plug in the numbers. I just proved both sides are equal. Thats what you wanted isnt it?

Well i didnt plug in the numbers. I just proved both sides are equal. Thats what you wanted isnt it?

Yes, sorry. So,

1. Cancel all the C's; so the problem should look like this?

1/1+e^(-rx) - 1/2 = 1/2 - 1/1+e^(rx)

2. Add 1/2 to each side. Looks like this?

1/1+e^(-rx) = 1 - 1/1+e^(rx)

3. Multiply 1 by (1+exp(rx))/(1+exp(rx)) yields...

1/1+e^(-rx) = e^(rx)/1+e^(rx)

And this is where I lost you. Most likely through my stupidity. But,

4. Multiplying everything by (1+exp(rx)) gets,

e^(rx) = 1+e^(rx)/1+e^(-rx)

?

And this is where I lost you. Most likely through my stupidity. But,

4. Multiplying everything by (1+exp(rx)) gets,

e^(rx) = 1+e^(rx)/1+e^(-rx)

?
No as far as I can tell you're right. I got to the same point and all your math so far is correct. However I think his last step is off because the exponents do not have the same signs. One is positive and the other is negative. In addition you cannot just take out the 1's since they are being added to the e^rx and e^-rx respectively. San haiti, if you come back into the forum (now that it's working again) could you please explain how you worked out your final step mentioned earlier? Thanks

Nova Roma, I haven't figured out the last part yet, sorry. But I'll post again if I can work it out.

Thanks for the help!

Maths! Seriously this in inexecusable American language malformation. It is plural! Do you say mathamatic? No of course you dont. You say mathamatics. In the same way it is Maths, not MATH!Gaudere's Law?

That's a good one, Mycon. I had to go look it up, but I like it!

Gaudere's Law: Any post made to point out a Quick Facts about: spelling
Forming words with letters according to the principles underlying accepted usagespelling or Quick Facts about: grammar
Studies of the formation of basic linguistic unitsgrammar error will invariably contain a spelling or grammar error.

I'm sorry, but you can't cancel the Cs. Doing so changes the equation. You could try dringing out a C but then you would end up with a (e^-rx)/C.

f(x) = c / 1+e^-rx

f(x) - c/2 = c/2 - f(-x)

c/1 + e^-rx -c/2 = c/2 + e^-rx

c/2 - (c/1 + e^rx) = -(c/2 + e^rx)

I don't know if that makes any sense. It doesn't to me.
That's probably wrong, but still, you can't cancel the Cs without changing the problem.


c/1 + e^x != c/c + e^x

c/1 + e^x = c(1 + (e^x)/c)

I'm sorry, but you can't cancel the Cs. Doing so changes the equation. You could try dringing out a C but then you would end up with a
Cancelling out the C is correct and vastly simplifies the problem.


http://img246.echo.cx/img246/181/untitled5wn.jpg


I included the boxed sidenote in case you don't understand the progression from equation 7 to equation 8.

Now, that makes sense. My mistake was order of operations.

However, didn't the problem say to solve both sides seperatly?

I would sugget taking the derivative or the integral, but the dirivatives of the two sides are vastly different. They graph as identical functions and identical functions should have identical tangents. Strange.

Edit: My mistake They do have identical diravitives, I just left out a dirivative of the inside in one.
Just take the dirivative of both side and work both of out as partial fractions.

You could also do this...

Remember that e^(-rx) = 1/e^(rx)

If you make the switch then the c/(1+e^(-rx)) on the left becomes c/(1+1/e^(rx))

Since 1 = e^(rx)/e^(rx) that then is the same as c/((e^(rx)+1)/e^(rx)) which becomes c * e^(rx)/(1+e^(rx))


So your original equation is really this: c*e^(rx)/(1+e^(rx)) - c/2 = c/2 - c/(1+e^(rx))

then just use algebra

Cancelling out the C is correct and vastly simplifies the problem.


http://img246.echo.cx/img246/181/untitled5wn.jpg


I included the boxed sidenote in case you don't understand the progression from equation 7 to equation 8.

I don't think he's allowed to get rid of the constant. The only reason his teacher would have put it there was so that he'd to arrive at c=c. If teacher's a dick he'll mark it wrong first line if the c's go, even though that's a perfectly justified thing to do.

If you just pull the c/2's to the right and the rest to the left you get:
c*[e^(rx)/(1+e^(rx)) + 1/(1+e^(rx))] = c

then c*[(e^(rx)+1)/(1+e^(rx))] = c

which is c*[1]=c so c=c

If you just pull the c/2's to the right and the rest to the left you get:
c*[e^(rx)/(1+e^(rx)) + 1/(1+e^(rx))] = c

then c*[(e^(rx)+1)/(1+e^(rx))] = c

which is c*[1]=c so c=c

True, but Nova Roma's problem instructions said specifically to solve the sides separately to prove that they are in fact equal. So she would not be allowed to move any parts across the equal symbol. (But in any other situation, I agree on it being a great way to simplify.)

Maths! Seriously this in inexecusable American language malformation. It is plural! Do you say mathamatic? No of course you dont. You say mathamatics. In the same way it is Maths, not MATH!

Listen, you crumpet-sucking jackass, IT IS A BLOODY BACK-FORMED ABBREVIATION! You don't need to pluralize the damn thing. MATH is short for MATHEMATICS because it's a back-formation and it's easier to SAY. Putting the "S" on the end makes you a pretentious, snobby little bastard who doesn't realize that complaining about US English's abbreviations when you've got a whole lexicon filled with a load of useless bloody "U"s is pretty stupid. Why add another phoneme to make the it harder to pronounce? Oh, because it makes me fees superior and aristocratic. Well good for you, Lord Tosh.

"Pre-calculus" ends in "S", but everyone says "pre-calc." Why? 'Cause abbreviationas and back-formations are supposed to be more convenient. I know it's equally petty for me to jump down the throat of someone being petty, but this kind of thing sets my teeth on edge. Next time, just realize you're on an international site and LET IT GO!

This thread confuses me...but I believe the poster is trying to prove an identity. Thus it is allowed to simplify the given equation using standard algebra. If you come up with an obvious equality in the end (such as 1 = 1), you have proven the identity. Conversely, if you come up with a contradiction (such as 1 = 0), the identity is false.

Since we have come up with something similar to 1 = 1, I believe the problem is solved...

[my 2 cents]

No as far as I can tell you're right. I got to the same point and all your math so far is correct. However I think his last step is off because the exponents do not have the same signs. One is positive and the other is negative. In addition you cannot just take out the 1's since they are being added to the e^rx and e^-rx respectively. San haiti, if you come back into the forum (now that it's working again) could you please explain how you worked out your final step mentioned earlier? Thanks

Nova Roma, I haven't figured out the last part yet, sorry. But I'll post again if I can work it out.

Yeah, stupid NS crashed last night before i was able to post the correction. It should be:

4)invert both sides

5)multiply both sides by exp(rx) and simplify

and you get exp(rx)=exp(rx)

I shouldn't try to do math at 1AM

Just add e^-rx and subtract e^-rx from the first side numerator after taking out the Cs

1/(1+e^-rx) - 1/2 = (1+e^-rx - e^-rx)/(1+e^-rx) -1/2

= 1 - 1/[e^rx(1+ e^-rx)] - 1/2 = 1/2 - 1/(e^rx + 1)

1/2 - 1/(1+ e^rx) = 1/2 - 1/(1+e^rx)

Listen, you crumpet-sucking jackass....
Intangelon: Official Warning - Flaming. You seem to be new, here, so I'll inform you that accumulating enough official warnings gets you banned from the forums (or even gets your nation deleted fi the offense is serious enough).

Neo Cannen: Stay on-topic. I trust that I am clear.

--The Modified Democratic States of Cogitation
"Think about it for a moment."
NationStates Game Moderator

--snip--
Neo Cannen: Stay on-topic. I trust that I am clear.



I will if you will, pal.

I certainly hope that YOU got the same warning -- because your post (about how "mathematics" should be abbreviated when the thread was asking for help on an equation) was decidedly off-topic as well. I'd hate to think you were getting preferential treatment in a game nominally dedicated to some kind of egalitarian ethos.

By the way, "I trust that I am clear"? What are you, a part-time actor who only plays headmasters in films shot at British prep schools?

I will if you will, pal.
I'm not sure what you mean or who you're talking to, here.

I certainly hope that YOU got the same warning -- because your post (about how "mathematics" should be abbreviated when the thread was asking for help on an equation) was decidedly off-topic as well. I'd hate to think you were getting preferential treatment in a game nominally dedicated to some kind of egalitarian ethos.
It sounds like you're talking to "Neo Cannen", but I'm going to respond to this, anyway.

"Neo Cannen"s offense was mild; not bad enough to issue an official warning over (for an isolated offense), but bad enough that he has to be told to get back in line. If he persisted, then I'd slap him with an official warning for topic hijacking. The action I took against him was proportional to his offense.

You called him a "crumpet-sucking jackass". Direct attacks against other NationStates players* are not allowed and this is severe enough for an official warning; the fact that "Neo Cannen" started it is largely irrelevant because retaliation is not a justification for violating NationStates rules. Thus, the action I took against you was proportional to your offense.

* ...outside of a roleplay context, which this isn't....

By the way, "I trust that I am clear"? What are you, a part-time actor who only plays headmasters in films shot at British prep schools?
I assume that you're now talking to me.

Comments deliberately intended to provoke insults in return are called "flamebait". It is not allowed on NationStates. Flamebaiting a NationStates Moderator is particularly unwise.

"Think about it for a moment."

--The Modified Democratic States of Cogitation
NationStates Game Moderator

Thanks for the help. I've decided to go with Antebellum South's method, despite the fact that it included both sides of the equation. However, as no one else in the class even attempted the problem, I'm sure it won't matter how I got it now, as long as I did.

Thanks for the help fellows. I never thought of setting the equation equal to one. I guess it just never dawned upon me.

Point taken.

Point taken.

In a few words "Dont screw with the mods"

Okay, I have another problem. This time, I got the answer, but I don't think I followed the parameters.

Jorge Martinez is making a business trip by car. After driving half the total distance, he finds he has averaged only 20 mi/h, because of numerous traffic tie-ups. What must be his average speed for the second half of the trip if he is to average 40 mi/h for the entire trip? Answer this question using the following method.

a. Let d represent the distance Jorge has traveled so far, and let r represent his average speed for the remainder of the trip. Write a rational function, in terms of d and r, that gives the total time Jorge's trip will take.

: I got ( d / 20 ) + ( d / r ) = t ; however, it didn't specify t as a possible variable, so I'm not sure if I can use this function.

b. Write a rational expression, in terms of d and r, that gives his average speed for the entire trip.

: I got ( 20 + r ) / 2 ; again, I'm not sure if this is possible because I didn't use d in the expression.

c. Using the expression you wrote in part (b), write an equation expressing the fact that this average speed for the entire trip is 40 mi/h. Solve this equation for r if you can. If you cannot, explain why not.

: Using my previous expression, you get ( 20 + r ) / 2 = 40 and r = 60 mi/h. This troubles me because the entire worksheet is supposed to be extremely challenging and it is dealing with complex rational equations. Yet I circumvented all that and still arrived at the same answer with a much simpler equation. Of course, my teacher is a bitch and will most likely take points from me for not following directions. My question is, is there a way to write these equations using both d and r?

: I got ( d / 20 ) + ( d / r ) = t ; however, it didn't specify t as a possible variable, so I'm not sure if I can use this function.You're using t as a dependant variable- in y=x^2, y isn't considered a variable because it depends upon the others, thus making it (effectively) a single-variable function. Thus, you are correct.

: I got ( 20 + r ) / 2 ; again, I'm not sure if this is possible because I didn't use d in the expression.Distance=rate*time. In your above answer, you state this as "time= Distance/rate."
Your answer is a rate, so you need rate=Distance/time. Since you don't have a time-variable, invent one by labelling the first & second distances D[1] and D[2] (D-sub-one... does this board have any subscript codes?) and using the above D/R/T equations. D[1] & D[2] are unecessary, as they're equal, but since you're adding unecessary variables anyway,
Ask again if you can't figure it from this.



: Using my previous expression, you get ( 20 + r ) / 2 = 40 and r = 60 mi/h. This troubles me because the entire worksheet is supposed to be extremely challenging and it is dealing with complex rational equations.Well, if it were stated differently, then it'd be impossible (unless he can exceed C and go back in time, or teleport instantaneously), as he'd have to double distance travelled without increasing the time taken at all. However, stated as it is, there might be a legitimate answer, depending on interpretation.

In a few words "Dont screw with the mods"
Yes, I explained that to him, already.

Don't... provoke the situation further.

--The Modified Democratic States of Cogitation
NationStates Game Moderator

...

Well, if it were stated differently, then it'd be impossible (unless he can exceed C and go back in time, or teleport instantaneously), as he'd have to double distance travelled without increasing the time taken at all. However, stated as it is, there might be a legitimate answer, depending on interpretation.
I have to concur with The Mycon, here. Usually, when one talks about average speeds, one is talking about time-weighted averages. 60 mi/hr would be correct if you were using distance-weighted averages (which I find unlikely).

Jorge has so far traveled d for 20 mi/hr. This has required (d / 20) hours of time. He wants to travel 2 d for 40 mi/hr; this requires (2 d / 40) hours = (d / 20) hours of time. Thus, after traveling distance d for 20 miles per hour, Jorge has already completely expended the time needed to travel 2 d for an average of 40 mi/hr. Thus, his goal is impossible.

--The Democratic States of Cogitation
"Think about it for a moment."
Founder and Delegate of The Realm of Ambrosia

Sorry Nova, thought you were a guy. I have been confronted by my own stereotyping :(.


PS: Neo Cannen, you made an entire argument out of spelling and grammar while repeatedly failing to spell mathematics correctly.

PS: Neo Cannen, you made an entire argument out of spelling and grammar while repeatedly failing to spell mathematics correctly.The next person {to complain about the spelling of mathematics or its abbreviation} or {to complain about someone else complaining about the spelling of mathematics or its abbreviation} gets an official warning for topic hijacking.

--The Modified Democratic States of Cogitation

...

: I got ( d / 20 ) + ( d / r ) = t ; however, it didn't specify t as a possible variable, so I'm not sure if I can use this function.Correct. This is the total time that the total trip will require.

: I got ( 20 + r ) / 2 ; again, I'm not sure if this is possible because I didn't use d in the expression.
Average rate is distance/time, or 2d / t in your case. Can you figure it out from that?

--The Democratic States of Cogitation

Just out of sheer curiosity, what kind of class do you enrol in that forces you to solve equatios like this? I made it through my high school's calculus course because I was told it'd look impressive on an application for college. Unfortunately, I was put on that math track before I discovered that I wanted to be a musician and a music teacher. Point is, though I took calculus, I promptly forgot it all -- even the imaginary numbers.

It's impressive to see, well, a symposium, practically, on one equation. What are you all doing that such high-level stuff becomes your daily meat and milk?

Just out of sheer curiosity, what kind of class do you enrol in that forces you to solve equatios like this? I made it through my high school's calculus course because I was told it'd look impressive on an application for college. Unfortunately, I was put on that math track before I discovered that I wanted to be a musician and a music teacher. Point is, though I took calculus, I promptly forgot it all -- even the imaginary numbers.

It's impressive to see, well, a symposium, practically, on one equation. What are you all doing that such high-level stuff becomes your daily meat and milk?I'm also curious as to what he's doing in class, but I wouldn't call this "high-level". It simply looks like a very strict study of algebra.

--The Democratic States of Cogitation

Distance=rate*time. In your above answer, you state this as "time= Distance/rate."
Your answer is a rate, so you need rate=Distance/time. Since you don't have a time-variable, invent one by labelling the first & second distances D[1] and D[2] (D-sub-one... does this board have any subscript codes?) and using the above D/R/T equations. D[1] & D[2] are unecessary, as they're equal, but since you're adding unecessary variables anyway,
Ask again if you can't figure it from this.


So from here I established this new expression:

(2d) / (d/20) + (d/r)

2d representing the total distance, as d represents half of the total distance. And the denominator representing t time which we said was equal to the (d/20) + (d/r) expression.

From there I established the equation:

(2d) / (d/20) + (d/r) = 40

Simplified the complex fraction through LCD to get:

(2d) / (rd + 20d)/20r = 40

Reciprocal multiplication:

2d * ( 20r / rd + 20d) = 40

40rd / rd + 20d = 40

Multiplied top by bottom to get:

40rd = 40rd + 800d

800d = 0

d != 0 (0 distance? :rolleyes: )

Is this all correct? If so, what is the reason you cannot solve for r? The fact that you can't isolate it without producing an erroneous (is that the right terminology?) equation?

By the way, not sure what made some of you think that I was a girl... But I'm a guy.

High-level? If only. To answer your questions, I am in an Honors Algebra II class; generally, the problems are much simpler and more numerous.

However, as my grade is too close to a B for my comfort (93.3) I've decided to take all the extra credit opportunities that I can. Some of them include essays (as you've seen my e essay) and others include these "Challenge" problems.

From my experience, however, they are more tedious and, as Cogitation said, strict on restrictions than they are challenging.

So from here I established this new expression:

(2d) / [ (d/20) + (d/r) ]

2d representing the total distance, as d represents half of the total distance. And the denominator representing t time which we said was equal to the (d/20) + (d/r) expression.

From there I established the equation:

(2d) / [ (d/20) + (d/r) ] = 40

Simplified the complex fraction through LCD to get:

(2d) / [ (rd + 20d)/20r ] = 40

Reciprocal multiplication:

2d * [ 20r / ( rd + 20d ) ] = 40

40rd / ( rd + 20d ) = 40

Multiplied top by bottom to get:

40rd = 40rd + 800d

800d = 0

d != 0 (0 distance? :rolleyes: )

Is this all correct? If so, what is the reason you cannot solve for r? The fact that you can't isolate it without producing an erroneous (is that the right terminology?) equation?

By the way, not sure what made some of you think that I was a girl... But I'm a guy.
[Corrected or parenthesis.]

Please remember your parenthesis and order-of-operations.

d=0 Hmmm... I hadn't thought of that.

Yes, d=0 is a solution, but it's a trivial solution that we're not interested in.

Let's back up to this equation:
(2d) / [ (d/20) + (d/r) ] = 40

Notice that the ds cancel on the left-hand side.
2 / [ (1/20) + (1/r) ] = 40

Multiply both sides by [ (1/20) + (1/r) ]
2 = 40 * [ (1/20) + (1/r) ]

Distribute the right-hand side.
2 = (40/20) + (40/r)
2 = 2 + (40/r)

Subtract 2 from both sides
0 = 40/r

You are therefore looking for a value of r such that 40 divided by r is 0. r would have to be infinity; there is no real value of r that does this. Alternately, you can try to multiply both sides by r and divide both side by zero to get:
r = 40/0

...which is undefined.

Again, this is explained by saying that if Jorge wanted to averge 40 mi/hr over a distance of 2d, then he already exhausted all of his time for that goal after traveling 1d for 20 mi/hr.

--The Democratic States of Cogitation
"Think about it for a moment."

Interesting, thanks. However, I'm still unsure what you mean by,

Again, this is explained by saying that if Jorge wanted to averge 40 mi/hr over a distance of 2d, then he already exhausted all of his time for that goal after traveling 1d for 20 mi/hr.

Interesting, thanks. However, I'm still unsure what you mean by,


He means that if Jorge wanted to travel 40 miles in 1 hour then he is out of luck.

However, when computing the average rate of change, 160m/4h = 40m/1h

Okay, I have another problem. This time, I got the answer, but I don't think I followed the parameters.


: I got ( d / 20 ) + ( d / r ) = t ; however, it didn't specify t as a possible variable, so I'm not sure if I can use this function.


Read the problem again. It asks for a function that represents the total TIME.

Of course t is a variable.

The proper form would be

t= f(d) = d/(20mph) + d/r

Math is simple but someones people who write word problems try to trick you with useless information.

Thanks for the help; 40 extra points will now be added into the averaging of my quiz scores!

However I have another problem.

Instructions: Find A and B in the following equation so that the two sides are equal for all values of x for which they both are defined.

[(A) / (x-2)] + [(B) / (x+3)] = (3x-11) / [ (x-2)(x+3) ]

As I'm too lazy to type out all my steps, I think I can manage by saying I got 3x-11 = 3x-11. I suppose I'm looking for a second opinion to see if I actually answered the question. As both sides of the equation are equal for all values of x, I assume that I'm right, but I just want to make sure.

No, you didn't. You're supposed to find values for A & B such that the LHS equals the RHS. It doesn't ask you to eliminate A & B. I'll give the first part and then give you 15 hours before I give more information (since I have no home internet connection, and my last final ends in 14).

Instructions: Find A and B in the following equation so that the two sides are equal for all values of x for which they both are defined.

[(A) / (x-2)] + [(B) / (x+3)] = (3x-11) / [ (x-2)(x+3) ]Cross multiply LHS-
A(x+3)+B(x-2)/[(x-2)(x-3]=3x-11/[(x-2)(x-3] -bottoms are equal, can be dropped (except to note that x=/= -3, 2).

shifting around factors gets you
(A+B)x+3A-2B=3x-11

Since x is an irreplacable unit, (A+B)=3, and (3A-2B)=-11

Simple algebra from here.

(A+B)=3, and (3A-2B)=-11Okay, 14 & 1/2 hours later, a big hint.

Think of A & B as two variables. You want to use those two equations above to eliminate one of them. Since I'm shutting down soon (maybe 2 hours/week free online, and I still gotta read my comics/the Straight Dope. E-mail can be handled from work), the rest is hidden.

Add the first equation [(A+B)=3] to the second equation [(3A-2B)=-11] twice.

Solve for A, plug this in to the first equation, and you have B.

I don't know why anyone would try to prove these 2 are equal by simplifying each one, but i'll give it a shot:

Left one:
multiply the first part with e^rx/e^rx and the second one with (1+e^rx)/(1+e^rx)
=> C/[2*(1+e^rx)]*[2*e^rx-1-e^rx] = C/[2*(1+e^rx)]*[e^rx-1]

Right one:
multiply the first part with (1+e^rx)/(1+e^rx), the second with 2/2
=>C/[2*(1+e^rx)]*[1+e^rx-2] = C/[2*(1+e^rx)]*[e^rx-1]

Thus they are equal.
I still find it more logical to do it the traditional way ...

Thanks for the help; 40 extra points will now be added into the averaging of my quiz scores!

However I have another problem.

Instructions: Find A and B in the following equation so that the two sides are equal for all values of x for which they both are defined.

[(A) / (x-2)] + [(B) / (x+3)] = (3x-11) / [ (x-2)(x+3) ]

As I'm too lazy to type out all my steps, I think I can manage by saying I got 3x-11 = 3x-11. I suppose I'm looking for a second opinion to see if I actually answered the question. As both sides of the equation are equal for all values of x, I assume that I'm right, but I just want to make sure.

splitting up a rational function i see.
here is my approach:
For finding A: multiply both parts of the equation with (x-2), then take the limit of both sides for x->2; On the left side, everything will become 0 except for the term with A; the right side will be A.
(note:if you have higher order poles in your function, you'll have to be a bit cunning and do the same trick, but with deratives; the exact rule you'll have to find out for yourself)

A = -1
B = 4

Edit: I see that mycon has allready posted the easier to find method (which is usually less efficient)

Here is a fun one:

2 trains start 200 km from eachother at a speed of 50 km/h and go towards eachother. A fly starts on one train at a speed of 75km/h. When it reaches the second train, it goes back and continues between both trains. What is the total distance the fly has travelled when it gets crushed by the trains?