SOLUTION: This problem was on my practice test, and I'm confused as to how to do it.. {{{I = nE/(nr+R)}}} For "n" I don't even know where to begin... Thanks in advance!

Algebra ->  Test -> SOLUTION: This problem was on my practice test, and I'm confused as to how to do it.. {{{I = nE/(nr+R)}}} For "n" I don't even know where to begin... Thanks in advance!      Log On


   

Question 13436: This problem was on my practice test, and I'm confused as to how to do it..

I+=+nE%2F%28nr%2BR%29 For "n"
I don't even know where to begin...
Thanks in advance!
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
I+=+nE%2F%28nr%2BR%29

First of all congratulations for writing the equation in correct algebra.com format!!

Next, let's clear the fraction by multiplying both sides of the equation by (nr+R):
I+=+nE%2F%28nr%2BR%29
I%2A%28nr%2BR%29++=+%28nE%2F%28nr%2BR%29%29+%2A+%28nr%2BR%29

The (nr+R) divides out on the right side:
I%28nr%2BR%29+=+nE
Inr+%2BIR+=+nE

Get all the n terms to the right side by subtracting Inr from each side:
Inr+-+Inr++%2BIR+=+nE-Inr+

Factor out the n to get the n in one place:
IR+=+n%28E-Ir%29+

Divide both sides by (E-Ir):
%28IR%29%2F%28E-Ir%29+=+%28n%28E-Ir%29%29%2F%28E-Ir%29+

%28IR%29%2F%28E-Ir%29+=+n+

R^2 at SCC