I bet no one can solve this equation.

- anonymous

I bet no one can solve this equation.

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- anonymous

5x^-2y^10 over 2x^-1(-3x^-3y^-1)^-2

- myininaya

Because there is no equation to solve. :p

- myininaya

Do you mean to simplify the expression.

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## More answers

- anonymous

Yes, straight from an SAT study book

- DawnR

this is http://media.tumblr.com/tumblr_memy8mw86I1qc9tqw.gif

- myininaya

I'm going to phrase this using latex just to make sure I understand what you have written:
\[\frac{5x^{-2}y^{10}}{2x^{-1}(-3x^{-3}y^{-1})^{-2}}\]

- anonymous

@DawnR XD hahaha

- myininaya

It is very important to recall the law of exponents here.
I will write some useful ones here regarding this problem:
\[x^{-n}=\frac{1}{x^n}\]
\[\frac{1}{x^{-n}}=x^{n}\]
\[x^{1}=x\]
\[x^0=1 \text{ note: } x \neq 0\]
\[(a^rb^tc^s)n=a^{rn}b^{tn}c^{sn}\]

- myininaya

also
\[x^nx^m=x^{m+n} \text{ and } \frac{x^n}{x^m}=x^{n-m}\]

- myininaya

These are good lows to recall for this problem.

- myininaya

also that one law is a little broken let me rewrite it

- myininaya

\[(a^rb^tc^s)^n=a^{rn}b^{tn}c^{sn}\]

- myininaya

can you use this last law here on the problem anywhere that you can see?

- anonymous

I'm not asking this question because I need help, I am asking to see if someone for actually get it

- anonymous

Can*

- myininaya

|dw:1403479310371:dw|
I will give you a hint:
look in the circle below:

- myininaya

|dw:1403479439985:dw|

- myininaya

that looks exactly like that law on the left hand side

- myininaya

now use the right hand side to rewrite this thing in the circle

- myininaya

I do get it
And I can help you

- anonymous

I don't need help, I wanted to see if anyone can actually get it. Its not HW or anything, just a fun equation

- myininaya

Ok, well I do get it...

- anonymous

I never said you didn't...

- myininaya

I guess I don't understand what you mean then.

- anonymous

Ok solve the equation then

- myininaya

I will help you to do just that.

- myininaya

If you are posting this as a challenge, I think I will let someone else do it.

- myininaya

This problem is not challenging to me. I'm sorry.

- mathmale

Drummer: If you want feedback on the solution of this problem, please share with the rest of us the work that you yourself have done. if you're not willing to provide such feedback, then, frankly, it sounds as that you may be trying to get someone else to do your work for you...and that'd be true regardless of whether you're doing this problem for the challenge or for an assignment.

- anonymous

Do you want the website I got this from?? I just wanted to see if people could actual do it

- mathmale

Again, please try doing it yourself first. If you want feedback on your work after you've done that, fine. To paraphrase you, I just want to see whether YOU can actually solve this problem at hand, before I invest time and effort in this problem solving myself. Isn't that fair?

- anonymous

Again, this is just a friendly question. It does not come from a quiz, worksheet, or homework of any kind. I simply goggled "Very hard math equations" and copy and pasted one from a website to give a fun challenge to people.

- mathmale

Fine, Drummin. Take the challenge and do the problem yourself, as far as you can. Then I'd be happy to give you feedback on how I think we might finish solving the problem.
Remember that everyone here on OpenStudy is here voluntarily, and thus is free to help or not help, get involved or not get involved, as he or she wishes. I've made my wants clear: show what you have done and list any questions you may have, before I get involved.
Glad you're curious and motivated and eager to solve challenging questions. Still, I feel more obligated to help those here on OpenStudy who try their best to get started out in their problem solving and share what they have done with me up front.

- anonymous

I have no clue how to do this, I don't even know where to start

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