Re-arrange

**rome** · February 15th 2011, 10:51 AM

Re-arrange in terms of P:

$P = x + SPT / 100$

Thank you.

**Ackbeet** · February 15th 2011, 11:14 AM

What are you solving for? P?

**rome** · February 15th 2011, 11:29 AM

Originally Posted by Ackbeet

What are you solving for? P?

Yes.

**Ackbeet** · February 15th 2011, 11:30 AM

Ok, so what's the usual first step? Or what have you done so far?

**rome** · February 15th 2011, 11:35 AM

$P - SPT/100 = x$

P.S. I don't need a worked solution- but I do need to re-arranged pretty quickly (it's not school work).

**Ackbeet** · February 15th 2011, 11:37 AM

That's correct. What's next?

**rome** · February 15th 2011, 11:41 AM

I don't know. That's why I'm here.

**Ackbeet** · February 15th 2011, 11:43 AM

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?

**rome** · February 15th 2011, 11:46 AM

Like this?

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

**Ackbeet** · February 15th 2011, 11:47 AM

Well, that's certainly not incorrect, but can't you factor even more?

**rome** · February 15th 2011, 11:49 AM

How do you mean? Surely P-P=0 ?

**Ackbeet** · February 15th 2011, 11:52 AM

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?

**rome** · February 15th 2011, 11:53 AM

Edit: posted this before I saw your post

Hold on, like this:

$P - PST(0.01) = x$

?

**rome** · February 15th 2011, 11:58 AM

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.

**Ackbeet** · February 15th 2011, 12:00 PM

Can you finish? What condition do you think you'll need to mention as a caveat?

Thread: Re-arrange