Is 2-(((2*x^2)/y^2)) the same as ((2-(2*x^2))/y^2)?

I meant something like this:

      $${\mathtt{0}}$$ $${\mathtt{25}}$$     

$${\mathtt{5}}$$$${\sqrt{{\mathtt{125}}}}$$

    - $${\mathtt{10}}$$ |

       __ ↓

         $${\mathtt{25}}$$

       - $${\mathtt{25}}$$

         ___

          $${\mathtt{0}}$$ $${\mathtt{0}}$$

For explanation of long division, not calculation.

No, they are not the same.

Here is why:

(Assuming you know the value of y and x and solve.)

In the first one, everything is subtracted from two at the end.

In the second one, since 2 is inside parenthesis lower than y^2, the problem ends with everything being divided by y^2.

$$\\Is\;\; 2-(((2*x^2)/y^2))$ the same as $((2-(2*x^2))/y^2)\\\\
1st=2-(\frac{2x^2}{y^2})=\frac{2y^2-2x^2}{y^2}\\\\
2nd=\frac{2-2x^2}{y^2}$$

So you can see that they are different.

Melody how do you do the long division bracket with Math Formula or LaTeX Formula?

What long division?

Math formual is easier to use than LaTex  but it does not do as many things.

If you want a fraction or a poser written properly just open Math formual and write it in there.

ex

3/4*6^2

I'll enter is in [Math Formula] just as it is ( and press enter and ok)

$${\frac{{\mathtt{3}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{{\mathtt{6}}}^{{\mathtt{2}}}$$       and this is the output

---------------------------------------------------

Now if I want the same thing using LaTex I would enter

\frac{3}{4}\times 6^2

and this would be the result

$$\frac{3}{4}\times 6^2$$

----------------------------------------------------

It would be great is you learn bot hthese tools.

Familiarise yourself with Math Formula first because it is easy.

No, Yours is probably the BEST long division that i have seen presented on the forum (or other forums for that matter)

I have done algebraic division using LaTex.  It was alright but not supurb.  

This was one of my attempts at algebraic division using LaTex

http://web2.0calc.com/questions/latex#r49