Re-arrange in terms of P:

$\displaystyle P = x + SPT / 100$

Thank you.

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- Feb 15th 2011, 10:51 AMromeRe-arrange
Re-arrange in terms of P:

$\displaystyle P = x + SPT / 100$

Thank you. - Feb 15th 2011, 11:14 AMAckbeet
What are you solving for? P?

- Feb 15th 2011, 11:29 AMrome
- Feb 15th 2011, 11:30 AMAckbeet
Ok, so what's the usual first step? Or what have you done so far?

- Feb 15th 2011, 11:35 AMrome
$\displaystyle P - SPT/100 = x$

P.S. I don't need a worked solution- but I do need to re-arranged pretty quickly (it's not school work). - Feb 15th 2011, 11:37 AMAckbeet
That's correct. What's next?

- Feb 15th 2011, 11:41 AMrome
I don't know. That's why I'm here.

- Feb 15th 2011, 11:43 AMAckbeet
- Feb 15th 2011, 11:46 AMrome
Like this?

$\displaystyle P - P(ST/100) = x$ - Feb 15th 2011, 11:47 AMAckbeet
Well, that's certainly not incorrect, but can't you factor even more?

- Feb 15th 2011, 11:49 AMrome
How do you mean? Surely P-P=0 ?

- Feb 15th 2011, 11:52 AMAckbeet
No, I was thinking more like this:

$\displaystyle P - P(ST/100) = x$

$\displaystyle P(1- ST/100) = x.$

That's because of the distributive law. Do you see how that's done? - Feb 15th 2011, 11:53 AMrome
Edit: posted this before I saw your post

Hold on, like this:

$\displaystyle P - PST(0.01) = x$

? - Feb 15th 2011, 11:58 AMrome
- Feb 15th 2011, 12:00 PMAckbeet
Can you finish? What condition do you think you'll need to mention as a caveat?