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

[carlburns]

I am feeling like an idiot. I know I should know this, but I am at a loss for how to solve this 1/q + 1/p = 1/f for f

This is my actual problem and I have several similar in tonights homework that I am at a loss for. I know once I see the process I am gonna be like DUH! But I just started college again after 20 years gone and my head is starting to hurt looking at the problems. I just can't figure out how to isolate f when it is a denominator for some reason. I do not need you to solve my actual problem, but if you could show me the process for isolating the variable in a problem similar to this it would much appreciated!

Thanks

 

[carlburns]

Thought process

If 1/f is equal to 1/1*f can I just cancel the 1's? That doesn't make sense to me because then I would have 0f or do I times each side by 10? But then I dont know if f is 10... bah, I am super confused.

 

[carlburns]

maybe

if I multiply both sides by pqf would I have f=pq??

 

[carlburns]

I think I got it

1/q * p/p * f/f + 1/p * q/q * f/f = 1/f *p/p * q/q

for

pf/pqf + qf/pqf = pq/pqf

then multiply each side by pqf for

pf + qf = pq

then

f(p+q) = pq

then

f = pq/p+q

can anyone tell me if this is close?

 

[carlburns]

And if so...

for I= NE/NR + R for N

I = N(E/R+R)

I(E/R+R)=N

Yes or have I screwed it all up once I get here?

 

[JeffM]

if I multiply both sides by pqf would I have f=pq?? NO

Well, it's reasonable to start by multiplying both sides of the equation by pqf because doing so gets rid of the fractions, but you have to do it correctly.

\(\displaystyle \dfrac{1}{q} + \dfrac{1}{p} = \dfrac{1}{f} \implies pqf\left(\dfrac{1}{q} + \dfrac{1}{p}\right) = pqf * \dfrac{1}{f} \implies pf + qf = pq.\)

Now what?

 

[JeffM]

for I= NE/NR + R for N

I = N(E/R+R)

I(E/R+R)=N

Yes or have I screwed it all up once I get here?

I presume you mean

\(\displaystyle I = \dfrac{NE}{NE + R}.\) Is that correct? What you wrote means \(\displaystyle I = \dfrac{NE}{N} * R + R.\)

 

[carlburns]

yes

I = ne/nr+r ... sorry

 

[JeffM]

1/q * p/p * f/f + 1/p * q/q * f/f = 1/f *p/p * q/q

for

pf/pqf + qf/pqf = pq/pqf

then multiply each side by pqf for

pf + qf = pq

then

f(p+q) = pq

then

f = pq/p+q

can anyone tell me if this is close?

Well this is close. You made the work a whole lot more complicated than it needs to be, and then you wrote

\(\displaystyle f = \dfrac{pq}{p} + q = q + q = 2q\), which is wrong. You meant \(\displaystyle f = \dfrac{pq}{p + q},\) which is correct.

 

[JeffM]

I = ne/nr+r ... sorry

That is what you wrote the first time. It clarifies nothing. Do you know PEMDAS?

I gave you two options; you did not tell me which you meant.

 

[carlburns]

Yes

the first one was correct... maybe it looks different to then what I am typing I mean:

I = NE over NR+R ...

 

[carlburns]

faq

maybe there is a frequently asked questions that better explains how to write out the problem correctly using keyboard?

 

[JeffM]

the first one was correct... maybe it looks different to then what I am typing I mean:

I = NE over NR+R ...

That should be written I = NE / (NR + R). You need to review PEMDAS.

\(\displaystyle I = \dfrac{NE}{NR + R} \implies I * (NR + R) = \dfrac{NE}{NR + R} * (NR + R) \implies INR + IR = NE \implies \)

\(\displaystyle NE - INR = IR \implies N(E - IR) = IR \implies N = \dfrac{IR}{E - IR}.\)

 

[carlburns]

ok

ty... and I completely understand how what I was writing was looking confusing when I looked at it a second time... thank you.

 

[stapel]

maybe there is a frequently asked questions that better explains how to write out the problem correctly using keyboard?

This article may help with typing math clearly.