Solving for a variable. I=nE/nr+R solve this for n.

Question

anonymous · Answer

As this stands, $n$ is cancelled in the equation.

unklerhaukus · Answer

$$I=\frac{nE}{nr+R}$$?

anonymous · Answer

That's what I was thinking. The problem is I have an answer sheet from the doctor herself. In her infinite wisdom, she says the answer is 

n= -IR/Ir-E

How do you explain that (wish I could, lol).

anonymous · Answer

Has Rhaukus correctly typed the equation?

anonymous · Answer

Yes he has.

unklerhaukus · Answer

$$I{nr+R}={nE}$$$$I{nr+R}-{nE}=0$$$$(Ir-E)n+R=0$$$$(Ir-E)n=-R$$$$n=\frac{-R}{Ir-E}$$

anonymous · Answer

Then $n$ does not cancel. Multiply by the denominator and separate $n$ to one side. Then divide.

anonymous · Answer

Rhaukus, you messed up on step one.

anonymous · Answer

Yeah it's kind of a curve ball. I appreciate you guys trying to help me out too, by the way.

anonymous · Answer

I agree that it is a curveball in the sense that it is not the typical situation of "find $x$".

Also, Rhaukus, your error was minor. You forgot to distribute all of the denominator onto $n$.

anonymous · Answer

onto $I$, I should say.

unklerhaukus · Answer

$$I=\frac{nE}{nr+R}$$$$I(nr+R)={nE}$$$$Inr+IR-nE=0$$$$n(Ir-E)+IR=0$$$$n=\frac{-IR}{Ir-E}$$

anonymous · Answer

Yup. Correct.

anonymous · Answer

Yeah, that's it. I will have to remember that one. Thanks.

anonymous · Answer

You can remove the negative.
$$n=\frac{-IR}{Ir-E}=\frac{IR}{E-Ir}$$
Also, Kash, you don't need to remember this. Just derive it on your own.

anonymous · Answer

Well, what I mean is I need to remember the steps you used. It didn't occur to me to zero out one side and start the problem from there. That was extremely helpful.

unklerhaukus · Answer

are these variables inan electric circuit ?

anonymous · Answer

Oh, yeah! I totally advocate taking little tips and tricks and keeping them in mind. :D

anonymous · Answer

That was flippin' awesome. lol. I didn't even know they had this kind of stuff online. Maybe I can get on board and try to help someone.

anonymous · Answer

I just recommend reading the code of conduct (or at least skimming it) here http://openstudy.com/code-of-conduct . Helping others is greatly encouraged! :D