A bunch of cunts, mostly in the Australian sense. Except that one guy.

My professor

I just got off the phone with him.

He recommended a Schaum's book and some Spotted Cow beer for my algebra woes.
Permalink Send private email sharkfish 
September 29th, 2006 6:23pm
http://www.newglarusbrewing.com/

I'm thinking Fat Squirrel sounds good.
Permalink Send private email sharkfish 
September 29th, 2006 6:25pm
You're in the right location for these too:

http://www.gooseisland.com/
Permalink Send private email bon vivant 
September 29th, 2006 6:41pm
"New Glarus beer is available only in Wisconsin. There are only so many hours in the day to make beer and we can only keep up with the local demand."

Pffffftt!!!
Permalink Send private email bon vivant 
September 29th, 2006 6:46pm
Yeah, he's a cheesehead.
Permalink Send private email sharkfish 
September 29th, 2006 7:03pm
Schaum's rules.  Those are awesome.  Got me through several classes.
Permalink Send private email Aaron F Stanton 
September 29th, 2006 7:31pm
Yeah, the problem with this Schaums though (College Algebra is what he recommended and what I bought) is that it beats you over the head with polynomials, which I find easy.

It is these that stump me: 

x-1/1-x - x+1/x-2  /  x+1/1+x  + x+2/x-1

They have one of these and they don't have the answer, fuckers. 

So I rode the train home going through, thinking I might just make it through all this, and all I got was a pounding on what I am already comfortable with.  Although I do now have some of the factoring patterns more solidified now. (a^2-b^2) = (a-b)(a+b) kinda stuff.

They don't have enough solutions for the hard ones.

So I'm back on the internet.  Fucking fuck.
Permalink Send private email sharkfish 
September 29th, 2006 9:05pm
Not sure you posted the problem correctly as (assuming some brackets) x+1/1+x cancels to one.

Even so, transform the top and the bottom expressions from a/b+c/d to (ad+cb)/bd and simplify.

Then you'll have something of form p/q / r/s to be rearranged as ps/qr and again simplified.

NB a,b,c,d and p,q,r,s could themselves be bracketed expressions.
Permalink trollop 
September 29th, 2006 10:55pm
Oh.

The actual problem is

(x+1/x-1  - x-1/x+1)  / (1/x+1 + 1/x-1)  = 2

(show the work to get it to equal 2)

They don't show the steps, and I'm baffled.

I've done most of the other problems, but this is the only one like it.
Permalink Send private email sharkfish 
September 29th, 2006 10:59pm
With extra brackets, it would be like this?

[(x+1)/(x-1) - (x-1)/(x+1)] / [(1/(x+1) + 1/(x-1)] = 2
Permalink Send private email bon vivant 
September 29th, 2006 11:04pm
yup.
Permalink Send private email sharkfish 
September 29th, 2006 11:06pm
OK, so you have a big fraction with complicated things above and below the line.

First step, suppose you multiply top and bottom by (x+1)? What happens then?
Permalink Send private email bon vivant 
September 29th, 2006 11:08pm
Shouldn't I be simplifying each half first?

Wouldn't that mean multiplying each part of the top half by (x+1)(x-1)?
Permalink Send private email sharkfish 
September 29th, 2006 11:14pm
Yes, it would. But an important part of simplifying is to get rid of division signs.
Permalink Send private email bon vivant 
September 29th, 2006 11:17pm
Hmm. multiply the top half by x+1 is looking more correct.

lemme write some of this..
Permalink Send private email sharkfish 
September 29th, 2006 11:17pm
Don't post your answer yet...
Permalink Send private email sharkfish 
September 29th, 2006 11:23pm
S'OK, I'm making food.
Permalink Send private email bon vivant 
September 29th, 2006 11:24pm
the top half simplifies to x/x
Permalink Send private email sharkfish 
September 29th, 2006 11:27pm
oops no.  it simplifies to 1 (same thing :))
Permalink Send private email sharkfish 
September 29th, 2006 11:28pm
Let's look at the working for multiplying by (x+1):

[(x+1)/(x-1) - (x-1)/(x+1)] / [(1/(x+1) + 1/(x-1)]

= [(x+1)^2/(x-1) - (x-1)] / [1 + (x+1)/(x-1)]

Is that what you got?
Permalink Send private email bon vivant 
September 29th, 2006 11:33pm
nope. I multipled the top half (that's as far as I've gotten)
like so

(x+1)(x-1)        (x+1)      (x-1)
___________  .  _____  -  _______

(x+1)(x-1)        (x-1)      (x+1)
Permalink Send private email sharkfish 
September 29th, 2006 11:44pm
the dot = multiply, and move things over to the right some so it looks right.
Permalink Send private email sharkfish 
September 29th, 2006 11:45pm
then

(x+1)(x-1)(x+1) - (x-1)
________________________

(x+1)(x-1)(x-1) - (x+1)


remember this is just the top half.

then

(x-1) (x+1) (x+1) - 1
_____________________
(x+1) (x-1) (x+1) - 1

then

crossing things out leaves -1/-1
Permalink Send private email sharkfish 
September 29th, 2006 11:49pm
"top half" means everything on the left of the */* below

[(x+1)/(x-1) - (x-1)/(x+1)] */* [(1/(x+1) + 1/(x-1)]
Permalink Send private email sharkfish 
September 29th, 2006 11:50pm
Hmmm. I'm not really following. You seem to be trying to say that

a/b - c/d = (a-c) / (b-d)

That's not right. (That is to say, the right hand side is not equal to the left hand side).
Permalink Send private email bon vivant 
September 29th, 2006 11:53pm
I have no idea.

I don't have a pedagogy here to follow.  They don't have any problems like it, so I really haven't a clue.
Permalink Send private email sharkfish 
September 29th, 2006 11:56pm
there is something here that seems to mean that


1/x + 1/y = y+x/xy

so then

my top half should be

(x+1) + (x-1) / (x-1)(x+1)
Permalink Send private email sharkfish 
September 29th, 2006 11:59pm
There's nothing new in this problem, really. You just need to apply the things you already know one step at a time.

Let's look at it this way.

You have a big fraction, like this:

[(x+1)/(x-1) - (x-1)/(x+1)]
-------------------------------------
[(1/(x+1) + 1/(x-1)]

Let's call it

A
--
B

Now to simplify it, you would like to get rid of B.

My suggestion is to start by multiplying by (x+1)/(x+1) like this:

(x+1)A
---------- = ?
(x+1)B

Do you see?
Permalink Send private email bon vivant 
September 30th, 2006 12:02am
nope.

the rules say to use  lcd (lowest common denominator).  To me that would me multiply by (x+1)(x-1)

So no, I don't see.  What rule are you applying?  Or is this intuition?
Permalink Send private email sharkfish 
September 30th, 2006 12:03am
The effect of your demo there is multiplying by one. But why pick x+1 as opposed to x-1?
Permalink Send private email sharkfish 
September 30th, 2006 12:04am
Yes, multiplying top and bottom of the big fraction by (x+1)(x-1) is the right step. But let's do it in two small steps to keep the algebra more manageable. First (x+1), then (x-1).
Permalink Send private email bon vivant 
September 30th, 2006 12:05am
Sigh. I don't have any examples so this flow is really difficult.

I'm going to have to search on the internet.  I only have so many hours before I have to turn in my work.
Permalink Send private email sharkfish 
September 30th, 2006 12:12am
I could give you the answer, but I'm trying to lead you through it.

Would you like part of the working?
Permalink Send private email bon vivant 
September 30th, 2006 12:16am
I'm looking at this page for the hint I need, example 4

http://www.themathpage.com/alg/add-algebraic-fractions.htm

so if like you say, my lcm is (x+1)(x-1), then that's what I'm going to have to do in a pressure situation.  I'm not going to have some instinct to tell me to use x+1, no matter how hard the algebra turns out to be.

That's all I'm saying. I appreciate your help, and I apologize if I don't seem very patient. I'm a bad student and a lot of teachers find me irritating. So it is me, I realize.
Permalink Send private email sharkfish 
September 30th, 2006 12:20am
Here, I'm going to try to explain.

This is what we are starting with:

[(x+1)/(x-1) - (x-1)/(x+1)]
---------------------------------
  [(1/(x+1) + 1/(x-1)]

Notice how it has four division signs. And our goal in simplifying is to remove division signs.

Let's multiply by (x+1)/(x+1) and see what happens:

(x+1)[(x+1)/(x-1) - (x-1)/(x+1)]
---------------------------------
  (x+1)[(1/(x+1) + 1/(x-1)]

      [(x+1)^2/(x-1) - (x-1)]
= ---------------------------------
        [(1 + (x+1)/(x-1)]

Notice how now there are only three division signs, so we have made step in the right direction.

The reason why we choose (x+1) is because it will make the 1/(x+1) on the bottom of the fraction turn into 1. That isn't really intuition, it is examining the what is laid out before you and making a decision based on looking ahead towards your goal.

Now, what would be the next logical step after the one above?
Permalink Send private email bon vivant 
September 30th, 2006 12:28am
I can't count.

There were five, not four, division signs to start with and three afterwards.
Permalink Send private email bon vivant 
September 30th, 2006 12:33am
Oh my goodness.

look at example e, Problem 10:

http://www.themathpage.com/alg/add-algebraic-fractions-2.htm


If you do just that, you get 2 as a result.  Mo.Fo.
Permalink Send private email sharkfish 
September 30th, 2006 12:59am
Absolutely. Take the next step from the step I gave before and multiply by (x-1)/(x-1):

(x-1)[(x+1)^2/(x-1) - (x-1)]
---------------------------------
  (x-1)[(1 + (x+1)/(x-1)]


    [(x+1)^2 - (x-1)^2]
= ----------------------------
        [(x-1) + (x+1)]

It becomes just how they show it.
Permalink Send private email bon vivant 
September 30th, 2006 1:11am
Given that this stuff qualifies as arithmetic, there isn't really anything to understand, and I will happily use that site for shortcuts. 

I am going to start memorizing those as of this weekend.

:)

Thank you.
Permalink Send private email sharkfish 
September 30th, 2006 4:07am
Oh dear. It's not arithmetic you know, it's algebraic manipulation. There really are things to understand, and they are assumed knowledge for your classes. I think one of the reasons you are struggling is because you are weak in that area.

With some rules of algebra you can solve the problems on that web site without needing to memorize the solutions. Here, let me show you with a simple example.

First some rules:

#1:    a(b+c) = ab + ac

#2:    b/c = ab/ac        (as long as "a" is not zero)

#3:    a/c + b/c = (a + b)/c

Now let's apply the rules to add together two fractions a/b and a/c.

We start with:

    a/b + a/c

Apply rule #2:

    = ca/cb + a/c

Apply rule #2 again:

    = ca/cb + ba/bc

    = ac/bc + ab/bc

Apply rule #3:

    = (ac + ab)/bc

Apply rule #1:

    = a(c+b)/bc

Now we can apply no more rules so we are done. More complicated examples like the one in this thread will break down with application of these rules in just the same way.

Learning the rules is not hard, but learning how to apply the rules does need practice. There is no shortcut. You can try to memorize answers, but if you come across a problem you haven't memorized before you will be stuck.

I hope you can see the truth in this.
Permalink Send private email bon vivant 
September 30th, 2006 3:12pm
Yes, but I saw the rule way after this post began and saved myself some time.

You were going to tutor me the long way, like a math prof, on a subject that even the best mathematicians consider crap work.

The formula was straight-forward.  I appreciate having them.

Yes, you can derive them.  But that isn't really my aim just now.

The guy who's site I linked to here has a quote--something about don't listen to math teachers :)  He's right.  People make this stuff hard, they really do.  It was a simple algebra problem if you know the shortcut.
Permalink Send private email sharkfish 
September 30th, 2006 3:42pm
I considered the important part of math to be the part that proves the rules.  Most of them are easy enough looking proofs, but it still strikes me as the critical part.

Homework problems were like recognition exercises of where to apply these "magic rules"/formulas.  In High School, the textbooks seemed terrible at explaining the formulas, way too brief. I rarely got a sense of 'why can't I apply this rule, but only this rule?' - but especially in Geometry.
Permalink LinuxOrBust 
September 30th, 2006 4:28pm
expect someone said this, but I always found this useful :

(a/b) + (c/d) =

(ad/bd) + (bc/bd) =

(ad + bc)/bd


which i think was part of what you were after, though i didn't scan all this stuff in depth.
Permalink captain fucking obvious ($--) 
September 30th, 2006 8:44pm

This topic is archived. No further replies will be accepted.

Other topics: September, 2006 Other topics: September, 2006 Recent topics Recent topics