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I helped a friend's son with his Algebra the other day and came a cropper on a question. I'm telling myself it's because i haven't done it for 15 years, but it seemed really basic stuff and it's literally given me sleepless nights.

Right, it's a simple case of rework it or whatever it is you call it, to make the equation X=

a^2/x = yz

So, both sides, multiplied by X.

a^2 = x(yz) and then it looks as though you should divide by yz, but this seems a bit illogical to me, as i see it as this mess:

a^2/yz = x(yz)/yz

Although i believe it should be as so, but it just seems too clean:

x = a^2/yz

So, please tell me why dividing x(yz)/yz is x. With numbers it's obvious, but maybe it's just me, i really do not like this answer. It just seems wrong to me and has put my precious little pedantic nose right out of joint. :)

What if you just multiple by a^2?

Doesn't that become:

x = yz(a^2)?

I haven't done algebra for fucking ages, and was shit at it then.....

No

x = a^2/yz .... so if you multiply by a^2 then you get...... x(a^2) = a^4/yz

Barry's question is why does x(yz)/yz = x

Try thinking of it this way: if y and z are two numbers, they just make another number, which we can call w. w=yz.

x(w)/w = x

x times w, then divided by w, = x

In the same way that 5 times 2, then divided by two, = 5

I know this is your very basic GCSE level linear equation, but after spending my evening learning and trying to remember redemption yield formulas my brain is now fucked after trying to work out that equation.

I better call it a night.

Youre right, all the way because when it becomes a^2 = (yz)x/yz the denominator and the numerator which are (yz) cancel each other out. It's like saying 2x/2, if you forget the x for a minute it leaves you with 2/2 which makes 1,
1x = x

I may be wrong, and I've had a couple of beers, but I'll have a go...

a^2/x = yz

a^2= xyz

a^2/yz = x

Edit:  Realised you've done it like that...I'll have another beer.

Tick , or well thats what I got anyhow.

What job would this help to get?  ???

Maths Teacher

I did A level maths (many years ago) what the hell is the ^ ?

to the power of

a^2 = a squared

And they say shit is getting easier.

^ wasn't "to the power of" when I was at school

You need to be able to do that stuff if you want ANY kind of qualification or job which relies on mathematical analysis really

Yeah, this was how i clarified it in my mind, switch it for numbers and all is well. It's probably because i haven't touched algebra for ages, but i just find x(yz) divided by yz too, i dunno, convenient. I think i think of x(yz) as it's own entity; a completely new variable, that then becomes indivisible by yz in any manner that brings simplification.

I'm almost tempted to fish out my old maths books, because it now seems to make perfect sense and i'm questioning why i was so puzzled by it. :)


Algebra teacher.

r4e said it how i would have put it if i tried to tell you.

You could think of it another way though (albeit, similar to r4e method)

Given xyz=0 (sorry, for any sort of algebraic maneuver you need two sides to an equation, 0 in this case so nothing gets done on the RHS).

To get x, we can take steps through y and z separately, so that xz=0 (by diving by y first, you know this is what you get), then dividing by z, you get x=0. What you've essentially done though, is multiplied xyz by 1/y and then by 1/z, and since multiplication is associative (thats the technical jargain), (xyz * 1/y)*1/z = xyz*(1/y * 1/z) = xyz*(1/yz) = xyz/(yz) = x.

I like maths, maths is fun fun fun  :yay:

Where's "a" ?

Actually, don't worry, JFW can answer this tomorrow.

* I expect the answer to be X=new handbag

As a structural engineer I use eqns like this pretty much every day.

Barry, think of it this way...

x(yz)/yz

is the same as

X times (yz)/(yz)

and yz/yz is 1 right?

so its bascially X times 1

so a^2 = X (times 1)

Just to add to the confusion. 

     a = x            [true for some a's and x's]
   a+a = a+x          [add a to both sides]
    2a = a+x          [a+a = 2a]
 2a-2x = a+x-2x       [subtract 2x from both sides]
2(a-x) = a+x-2x       [2a-2x = 2(a-x)]
2(a-x) = a-x          [x-2x = -x]
     2 = 1            [divide both sides by a-x]

Spot the error...

What's wrong with 4+8 etc why get the numbers involved

Pah

i think I've got it.

Not an error as such, but 2(a-x) = a-x because a-x=0 so you're saying 2(0) = 0 rather than 2=1

2(a-x) / (a-x) = (a-x), not 2 !

No, fuck it, I'm talking bollocks again. I used to like maths..... :doh:

what fucking stupid alien language are you all talking ???

The error is you're dividing by zero, since a -x = 0.