Question

1/f = 1/p + 1/q
solve for p.

Answer 1

So we have:
\[\frac{1}{f}-\frac{1}{q}=\frac{1}{p}
\]Multiplying by \(pf\):
\[
p-f=\frac{pf}{q}
\]Dividing by \(pf\):
\[
\frac{p-f}{pf}=\frac{1}{q}
\]And taking the reciprocal:
\[
\frac{pf}{p-f}=q
\]And we are done.

Answer 2

\[{1 \over f} = {1 \over p }+{1 \over q} \]\[\implies {1 \over f} = {q + p \over pq}\]\[\implies {pq \over f} = { q + p} \]\[\implies pq = f(p + q)\]\[\implies pq = fp + fq\]\[\implies pq - fq= fp\]\[\implies {q(p - f) \over f} = p \]

Answer 3

@LolWolf Unfortunately, all that was done for \(q\). :P

Answer 4

Oh, I solved for the wrong variable, but you can switch them, equally XD

Answer 5

Haha, yeah, @ParthKohli

Answer 6

Oy, be careful @ParthKohli , you still have a remaining \(p\) in your final equation.

Answer 7

Oh... what the heck

Answer 8

@emb27 I solved for the wrong variable, but you can switch them, equally, they are the same thing.

Answer 9

\[\implies {q(p - f) \over f} ={pf \over f}\]Equating numerators:\[\implies pq - pf = pf\]\[\implies pq = 2pf\]\[\implies p = {2pf \over q}\]

Answer 10

Is that correct? I need to sleep!

Answer 11

I don't think it is, because we can do:
\[
\frac{1}{f}-\frac{1}{q}=\frac{1}{p}
\](Won't solve for the wrong variable this time!) To equal:
\[
\frac{f-q}{fq}=\frac{1}{p}
\]So:
\[
\frac{1}{\frac{f-q}{fq}}=\frac{1}{\frac{1}{p}}=p=\frac{fq}{f-q}
\]And you still have a remaining \(p\) on the other side XD

Answer 12

Honestly, I am so confused. Which is the correct answer to 1/f = 1/p + 1/q. Solve for p.

Answer 13

The correct answer is:
\[
p=-\frac{fq}{f-q}
\]

Answer 14

Is the answer a positive or negative, both are shown?

Answer 15

Oh, dang, I forgot to put a negative sign in my answer? What is wrong with me, to-day? Lack of sleep, I guess... yes, it's negative. I've messed up on the explanation, but the final case is correct.