# Making a pronumeral the subject

• Jan 14th 2010, 01:25 AM
JKDste
Making a pronumeral the subject
Can someone please solve Rp=R1 R2/R1+R2 to make R the subject!

And for a bunus C=N-n/2p to make p the subject.

Thanks so much!!
• Jan 14th 2010, 01:40 AM
Prove It
Quote:

Originally Posted by JKDste
Can someone please solve Rp=R1 R2/R1+R2 to make R the subject!

And for a bunus C=N-n/2p to make p the subject.

Thanks so much!!

Please either use brackets in the correct places, or use LaTeX.

Is the first one

$\displaystyle R_p = \frac{R_1R_2}{R_1 + R_2}$?

Is the second

$\displaystyle C = \frac{N - n}{2p}$?
• Jan 14th 2010, 02:00 AM
JKDste
Yeah thats it! Sorry I didn't know how to type it out like that!

Thanks so much!
• Jan 14th 2010, 02:10 AM
Prove It
Your first question doesn't make sense, since you don't have $\displaystyle R$ as a variable.

Are you trying to make $\displaystyle R_1$ or $\displaystyle R_2$ the subject?

Q.2) $\displaystyle C = \frac{N - n}{2p}$

$\displaystyle 2Cp = N - n$

$\displaystyle p = \frac{N - n}{2C}$.
• Jan 14th 2010, 02:27 AM
JKDste
Quote:

Originally Posted by Prove It
Your first question doesn't make sense, since you don't have $\displaystyle R$ as a variable.

Are you trying to make $\displaystyle R_1$ or $\displaystyle R_2$ the subject?

Q.2) $\displaystyle C = \frac{N - n}{2p}$

$\displaystyle 2Cp = N - n$

$\displaystyle p = \frac{N - n}{2C}$.

Thanks Prove it, the subject for the other one was R1
• Jan 14th 2010, 02:32 AM
Prove It
$\displaystyle R_p = \frac{R_1R_2}{R_1 + R_2}$

$\displaystyle R_p(R_1 + R_2) = R_1R_2$

$\displaystyle R_1R_p + R_2R_p = R_1R_2$

$\displaystyle R_1R_p - R_1R_2 = -R_2R_p$

$\displaystyle R_1(R_p - R_2) = -R_2R_p$

$\displaystyle R_1 = -\frac{R_2R_p}{R_p - R_2}$

$\displaystyle R_1 = \frac{R_2R_p}{R_2 - R_p}$.