# Re-arrange

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• February 15th 2011, 10:51 AM
rome
Re-arrange
Re-arrange in terms of P:

$P = x + SPT / 100$

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

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

Yes.
• February 15th 2011, 11:30 AM
Ackbeet
Ok, so what's the usual first step? Or what have you done so far?
• February 15th 2011, 11:35 AM
rome
$P - SPT/100 = x$

P.S. I don't need a worked solution- but I do need to re-arranged pretty quickly (it's not school work).
• February 15th 2011, 11:37 AM
Ackbeet
That's correct. What's next?
• February 15th 2011, 11:41 AM
rome
I don't know. That's why I'm here.
• February 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?
• February 15th 2011, 11:46 AM
rome
Like this?

$P - P(ST/100) = x$
• February 15th 2011, 11:47 AM
Ackbeet
Well, that's certainly not incorrect, but can't you factor even more?
• February 15th 2011, 11:49 AM
rome
How do you mean? Surely P-P=0 ?
• February 15th 2011, 11:52 AM
Ackbeet
No, I was thinking more like this:

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

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

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

Hold on, like this:

$P - PST(0.01) = x$

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

Originally Posted by Ackbeet
No, I was thinking more like this:

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

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

That's because of the distributive law. Do you see how that's done?

I think so. Thank you.
• February 15th 2011, 12:00 PM
Ackbeet
Can you finish? What condition do you think you'll need to mention as a caveat?
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last