What is a common way to simplify this by getting rid of x?

What is a common way to simplify this by getting rid of x?

(bx - c) / x

Thanks bros!

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youtube.com/watch?v=ToEFkUxFURg
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I'd say the most common way is by reading an elementary school algebra textbook.

It's quicker to ask here than to sift through a book to the right page. It takes a moment and I assume you like to help a brother out.

I know it leads to b - (c/x) but that doesn't get rid of x...

what do you mean?

You can rewrite it as (b - c/x). Is that all you want?

Are you literally retraded?

No because I want to get rid of x.

I'm just trying to derive how the equation

a(V - fC) / V

can simplify to

a / (1 + af)

and I want to take it one step at a time. I find this is the best way to learn a subject, rather than reading section-by-section through a book.

You were naive once too! Ignorance does not equal retardation.

maybe those symbols stand for other things that are getting substituted in. what do a, V, f, and C represent?

C = aB
B = V - N
N = C(R1 / (R1 + R2))
C = a(V - fC)

And the problem is to derive C / V

This.

Just taking those symbols at face value the above expression does NOT simplify into the bottom one.

>C = a(V - fC)
Are you sure?

Yes.

youtube.com/watch?v=ToEFkUxFURg

18:16

f is assumed to be R1 / (R1 + R2).

The algebra for this I understand.

Well for the specific derivation in the video I will write it out for you:

[math]V_e = V_i - V_f[/math]
[math]V_e + V_f = V_i[/math]
[math]aV_e + aV_f = aV_i[/math]
[math]V_o + aV_f = aV_i[/math]
[math]V_o + afV_o = aV_i[/math]
[math]\frac{V_o + afV_o}{V_i} = a[/math]
[math]\frac{V_o(1 + af)}{V_i} = a[/math]
[math]\frac{V_o}{V_i} = \frac{a}{(1 + af)}[/math]

Wow that is so clever. I understood the algebra and situation and how it got up to this problem in the video, but seeing you do it beautiful. I'm not sure I would have ever come up with that. Is it just through creative practice that you learn to do this?

Sorry, a better question is: can you walk me through your mind and what you were thinking which lead you think each step was necessary?

To be honest it's actually quite a difficult problem, since so many things are recursively defined, probably took me about 15 minutes to come up with the correct approach, after which the problem was quite simple.

My method was to simply work backwards. And yes, you get better at stuff like this with practice.

So we actually start with the last line.
[math]\frac{V_o}{V_i} = \frac{a}{(1+af)}[/math] Next we really don't want (1 + af) in the denominator, so we multiply both sides by that, giving us:
[math]\frac{V_o(1+af)}{V_i} = a[/math] Then we can distribute the [math]V_o[/math] giving us:
[math]\frac{V_o + afV_o}{V_i} = a[/math]
By the same logic I want to eliminate the denominator on the other side, so I multiply both sides by [math]V_i[/math] giving us:
[math]V_o + afV_o = aV_i[/math]
Now our task is to try and eliminate the a and f by using the earlier definitions in the video, so we see that [math]fV_o = Vf[math] meaning we can rewrite this as:
[math]V_o + aV_f = aV_i[/math].
I now want to eliminate the a, and if we had some way of rewriting [math]V_o[/math] in terms of a we could do that, and luckily [math]V_o = aVe[/math] so this gives us:
[math]aV_e + aV_f = aV_i[/math].
Seeing that we now have an a in front of each term we can divide by a, giving us:
[math]V_e + V_f = V_i[/math].
At that point it's quite easy to see that this is just a rewritten form of:
[math]V_e = V_i - V_f[/math].
And we're done!
Simply do this process backwards, and you've got your derivation. The only real "intuitive" steps is to realize working backwards is the correct strategy, and seeing good ways (as well as the need to) eliminate the a and f terms.

Hopefully the LaTeX is fixed now.

In math do you try to avoid denominators at all cost?

Also, what about the pioneers of engineering that said: "OK, we need to take V_e = V_i - V_f, and work it into V_o/V_i somehow." By the reversal of your reversed walkthrough? That is where the real difficulty comes into play, right? I wish to speak to them for a moment.

Generally having things in the denominator is annoying to work with, but the reason I eliminated them was because I knew I wanted to arrive at one of the starting equations, and none of them were written in fraction form. That's the same reason I distribute the [math]V_o[/math], I know none of our starting equations has a "1" in it, so I know I need to distribute it.

That's a lot more difficult yes, since it's not obvious you could arrive at such a simple form. Odds are you'd start with the expression [math]\frac{V_o}{V_i}[/math] and try to simplify it as much as possible through trial and error. They did almost certainly not start with [math]V_e = V_i - V_f[/math].

>brother
found your problem

Sorry, I was trying to relate to you in some way to elicit a response.

Thankfully, helped me in every way I was looking for. (Thank you!)