I’m a little worried that the OP doesn’t seem to know how to add fractions. Just because you have letters instead of numbers doesn’t mean the same basic principles apply…
I can add fractions and the thing is driving me nuts. v can’t equal u because then they would each have to equal f, which means that you would be dividing by zero in the second expression. When I use Ultrafilter’s method I wind up with an extremely unwieldy expression. Maybe that’s the only thing to do, but it strikes me that there should be a simpler way. I’m going to think about this for a bit.
It’s up to the instructor to determine what level of assistance is appropriate for students to seek on homework. Some teachers will let their students consult one another when doing homework, but not anyone else. Some will let them consult any book they want, but not any other person. Some will let them consult anyone, and some will let them consult no one. If the point of the homework is to help the students learn how to do the problem, then it makes sense to let them seek help, although the students should be seeking to understand how to solve problems such as the one they were assigned, rather than just finding a solution they can copy. On the other hand, if the point of the homework is to test how well the students understand the material, it doesn’t make sense to let them seek help, since this gives an inaccurate gauge of what they are able to do on their own.
But often, the point of homework is both to help the students learn to solve the problems and to gauge how much they’ve learned so far. In that case, something in between not allowing them to seek any help and allowing them to seek as much help as they want is appropriate. In any case, if a student is unclear on to what degree the teacher will let them collaborate with others, they should just ask the teacher that. It’s not cheating if you have permission.
(For what it’s worth, I’ve taught some intro-level college courses, and my policy was generally that I didn’t care what help students received on homework, so long as they understand it. Meaning, (1) if I ask them to explain their solution to me, they are expected to be able to do so, (2) if they are given a similar problem on a test or quiz, they should be able to recall how to solve it. For me, homework was mainly a teaching tool, and quizzes and tests were for testing their understanding.)
At any rate, dare_devil007_'s homework was probably already due, and since there are non-high-schoolers asking about it to, here’s a basic sketch of how to solve that problem.
[spoiler]One basic approach to solving the problem is to solve for f in the first equation, substitute this into the second equation, and simplify. Keep everything as factored as possible, and cancel any factors appearing in both the numerator and the denominator of the same fraction. Add fractions by finding a common denominator.
Basically, it goes like this (with some assumptions and some steps omitted.)
1/u + 1/v = 2/f
=> f = 2/(1/u+1/v) = 2uv/(u+v)
Thus, 1/(f-u) + 1/(f-v) = 1/(2uv/(u+v) - u) + 1/(2uv/(u+v) - v)
= (u+v)/(u(v-u)) + (u+v)/(v(u-v))
= ((uv + v[sup]2[/sup]) - (u[sup]2[/sup] + uv)) / (uv(v-u))
= (v[sup]2[/sup] - u[sup]2[/sup]) / (uv(v-u))
= (v+u)(v-u)/(uv(v-u))
= (v+u)/uv
= 2/f
Contingent on certain restrictions on the values of u and v (to avoid having any denominator equal to zero.)[/spoiler]
OK, but my question then is - what restrictions could be placed on u and v such that it would result in a valid equation where (u != v)? I can’t see any case were u would NOT equal v, (and therefore equalling f as well), at least when dealing with real numbers - is there something I’m missing?
As the Strinka noted, all the denominators have to be the same in order for the numerators to be added together such that 1+1 equals 2.
Only if you’re restricting the values, particularly f, to the set of integers. But what if f is not restricted to the integers?
For example, if u is 4 and v is 6, then 1/4 + 1/6 = 10/24. Put this in the form of 2/f, and f = 4.8
2 / 4.8 = .4166666 (repeating).
Now, if you plug in 4 and 6 for u and v to the other expression, you will end up with the same result (try it on a calculator): 2 / f
Well, thanks for all the tips.
And to clarify: I did ask all the other students in my math class, but most of them are stuck on the same question, so they didn’t offer much help. I only asked because I had no one else to ask and since it was weekend homework, I wasn’t able to ask my teacher.
But, today, I saw my teacher during lunch and I asked him for some help and he pretty much said the same things (make a common denominator, equate “f” from the first equation and substitute).
If this thread was inappropriate, then I apologize and mods will close it if they feel they must.
Pretty much.
Try
u = 2, v = 6, f = 3 (all integers, real, and unique)
or
u = 6, v = 12, f = 8 (all integers, real, and unique)
People have different learning styles. Practically an entire field of psychology is dedicated to studying it. Perhaps math professors get so used to dealing with emotionless, unresponsive math problems that it rubs off on their relationship with their students. In that case, why shouldn’t they consult the Internet, or better yet (as I do) hire a tutor. This strict regimen of analytical professor + symbolic representation that didn’t work the first time probably has a small subset of students that it works for in any sort of a way that can be constitued as learning.
Yes, I have done that. But I was never on the verge of tears; I considered it an interesting challenge.
For what it’s worth, I found it easier to work with x := 1/u, y:= 1/v, and g := 1/f instead of u, v, and f; but that’s a matter of personal taste.
I assure you I’m not. Try teaching math to kids in a culture that insists that mathematics is on the one hand impossibly difficult and on the other hand next to useless. After some exposure to the sense of entitlement this engenders, you’ll be a hardass too.
There are already cryptographic protocols to play fair poker. Surely one can be adapted.
Basically, Alice, Bob, Carol, and Ted agree on an encoding of each card and on a commutative cryptosystem. Alice encrypts all the cards with her key, then Bob with his, then Carol with hers, then Ted with his. Then they divvy up the encrypted cards and pass them around, each person taking of his or her encryption until the last person to decrypt each set of cards is their holder.
DAMMIT, wrong thread. okay, trying again:
Try disinterested students who want a mark more than comprehension.
There is no way that any substantial answer can be given in this forum other than to give away the answer. If the student doesn’t already know to solve for one variable and plug in, then he has much larger problems than this one exercise. If the student know what to do, but doesn’t know how to do it, a message board can’t help other than giving the answer.
Look, if this were a student in an actual class of mine the answer would be different, but then I’d know far more context on what the student knows, or at least what has been taught. Teaching requires much more of a relationship than anyone coming to a message board (or asking a parent’s friend at a party) can ask for.