Algebra question

[Div]

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 

[4D]

Where's "a" ?

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

* I expect the answer to be X=new handbag

[jimmy_onions]

What job would this help to get? 

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)

[manc red]

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...

[Don Rogers Shop]

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

Pah

[reeves4england]

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...

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

[blah blah]

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...

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

[blah blah]

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

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

[sheepshagger]

what fucking stupid alien language are you all talking

[Ben Wah Balls]

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