# Re-arrange

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• Feb 15th 2011, 10:51 AM
rome
Re-arrange
Re-arrange in terms of P:

\$\displaystyle P = x + SPT / 100\$

Thank you.
• Feb 15th 2011, 11:14 AM
Ackbeet
What are you solving for? P?
• Feb 15th 2011, 11:29 AM
rome
Quote:

Originally Posted by Ackbeet
What are you solving for? P?

Yes.
• Feb 15th 2011, 11:30 AM
Ackbeet
Ok, so what's the usual first step? Or what have you done so far?
• Feb 15th 2011, 11:35 AM
rome
\$\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 AM
Ackbeet
That's correct. What's next?
• Feb 15th 2011, 11:41 AM
rome
I don't know. That's why I'm here.
• Feb 15th 2011, 11:43 AM
Ackbeet
Quote:

Originally Posted by rome
I don't know. That's why I'm here.

That's fine. That's why we're here, as well: to help you get unstuck. Try factoring the P out. What do you get next?
• Feb 15th 2011, 11:46 AM
rome
Like this?

\$\displaystyle P - P(ST/100) = x\$
• Feb 15th 2011, 11:47 AM
Ackbeet
Well, that's certainly not incorrect, but can't you factor even more?
• Feb 15th 2011, 11:49 AM
rome
How do you mean? Surely P-P=0 ?
• Feb 15th 2011, 11:52 AM
Ackbeet
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 AM
rome
Edit: posted this before I saw your post

Hold on, like this:

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

?
• Feb 15th 2011, 11:58 AM
rome
Quote:

Originally Posted by Ackbeet
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?

I think so. Thank you.
• Feb 15th 2011, 12:00 PM
Ackbeet
Can you finish? What condition do you think you'll need to mention as a caveat?
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