I just got off the phone with him.

He recommended a Schaum's book and some Spotted Cow beer for my algebra woes.

http://www.newglarusbrewing.com/

I'm thinking Fat Squirrel sounds good.

You're in the right location for these too:

http://www.gooseisland.com/

"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!!!

Yeah, he's a cheesehead.

Schaum's rules.  Those are awesome.  Got me through several classes.

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.

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.

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.

With extra brackets, it would be like this?

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

yup.

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?

Shouldn't I be simplifying each half first?

Wouldn't that mean multiplying each part of the top half by (x+1)(x-1)?

Yes, it would. But an important part of simplifying is to get rid of division signs.

Hmm. multiply the top half by x+1 is looking more correct.

lemme write some of this..

Don't post your answer yet...

S'OK, I'm making food.

the top half simplifies to x/x

oops no.  it simplifies to 1 (same thing :))

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?

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)

the dot = multiply, and move things over to the right some so it looks right.

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

"top half" means everything on the left of the */* below

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

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).

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.

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)

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?

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?

The effect of your demo there is multiplying by one. But why pick x+1 as opposed to x-1?

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).

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.

I could give you the answer, but I'm trying to lead you through it.

Would you like part of the working?

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.

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?

I can't count.

There were five, not four, division signs to start with and three afterwards.

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.

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.

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.

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.

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.

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.

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.