# Solving for C

1. Aug 30, 2013

### Wumbolog

$V=C-\frac{C-S}{L}N$
Solve for C
I am extremely frustrated and have made countless attempts at this.

I would really appreciate a step by step on this. Thanks.

2. Aug 30, 2013

### phyzguy

Try this:

(1) Get all terms with C in them on one side of the equation, and all terms without C on the other side. In other words:

Terms with C = Terms without C

(2) Use the distributive law to write the equation as:

C*(Something without C) = Something else without C

(3) Divide both sides by (Something without C) so you have:

C = (Something else without C) / (Something without C)

Hope this helps.

3. Aug 30, 2013

### Staff: Mentor

I see an equation comprising 3 terms (one on the left of the equals sign, and two on the right hand side). What does it look like if you multiply each of those 3 terms by L?

4. Aug 31, 2013

### Wumbolog

$V=C-\frac{C-S}{L}N$

$LV=LC-C-SN$

$\frac{LV}{C}=L-SN$

$C=\frac {L-SN}{LV}$

but apparently this is wrong...

the book gives $C=\frac{LV-SN}{L-S}$

Last edited: Aug 31, 2013
5. Aug 31, 2013

### Staff: Mentor

$$\frac {C-S}{L} N = \frac {(C-S)N}{L}$$

Do you see it now?

6. Aug 31, 2013

### lendav_rott

Here's what I get:

V = C - N(C-S)/L | * L
LV = LC - NC + NS
LC - NC = LV - NS
C = (LV - NS)/(L-N)

So I have almost the same answer as the book, but I don't even begin to comprehend how an S can ever wind up as part of the demoninator.

7. Aug 31, 2013

### Ray Vickson

You should have $LV = LC -CN + SN$. Do you see the two errors you made in line 2? Can you see why it is wrong to go from the line
$$LV = LC - C - SN$$ (which, by itself, is wrong) to
$$\frac{LV}{C}=L-SN ?$$ Would you say it is correct to go from $5\times 3 - 3 - 4\times 2$ to $3 \times (5 - 4 \times 2)?$ You are claiming these are the same!

8. Aug 31, 2013

### Wumbolog

Thanks guys I finally got it. And the denominator should have been L-N, my mistake.