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I apologise maths is not my strong point, there are 2 ways to isolate n

How do I know which is the correct way:

Correct equation:

${n={cv}}$


way 1 divide by n

${c=\frac{n}{v}}$

${\frac{c}{n}=\frac{n}{vn}}$

${\frac{c}{n}={v}}$

${\frac{c}{nc}=\frac{v}{c}}$

${{n}=\frac{v}{c}}$

Way 2 multiply by v:

${c=\frac{n}{v}}$

${cv=\frac{nv}{v}}$

${n={cv}}$

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    $\begingroup$ In the first way, you start with $n=cv$ and end up with $n=$ something else. This should be a sign that something is wrong. $\endgroup$
    – Sandejo
    Mar 8, 2021 at 21:37
  • $\begingroup$ damn sorry it wasn't meant to be that @Sandejo $\endgroup$
    – Nickotine
    Mar 8, 2021 at 21:39

2 Answers 2

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Additional info so OP understands the language.

When you write:

$\require{cancel}$

$${\frac{c}{n}=\frac{n}{vn}}$$

and you do this:

$${\frac{c}{n}=\frac{\cancel{n}}{v\cancel{n}}}\Rightarrow{\frac{c}{n}=\frac{1}{v}}$$

Then you didn't divide by $n$. You cancelled out $n$ because it was in the numerator and the denominator as well. The equation does not change when you do this.

Bonus tip: Say you have equation $a=bc$

When you divide the only thing you mustn't forget is to state that the variable you divide by cannot be zero. So if you divide by $b$:

$$b\neq0$$

And if you divide by $c$

$$c\neq0$$

etc.

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    $\begingroup$ Thank you so much, so this is what happens: ${c=\frac{n}{v}}\Rightarrow{cv=\frac{nv}{v}\Rightarrow{{(n)(1)=cv}}}$ $\endgroup$
    – Nickotine
    Mar 8, 2021 at 23:48
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    $\begingroup$ This is correct, yes. $\endgroup$ Mar 9, 2021 at 5:32
  • $\begingroup$ you’ve been quite kind thank you for the info in the chat, most people don’t give a dumbass like me the time of day, I want to learn but there are some things I don’t notice and I need the help of someone else, would you be willing to chat sometimes about queries I have? @LordCommander $\endgroup$
    – Nickotine
    Mar 9, 2021 at 20:06
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    $\begingroup$ Don't call yourself a dumbass. We all came from there! Sure i would! Here's an advice: don't be discouraged when you get stuck on a problem, math is supposed to be hard. Try to pander over a problem for a longer period of time before posting. While you struggle, your brain actually learns the material. Best of luck :) $\endgroup$ Mar 9, 2021 at 20:26
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$\begingroup$

$\dfrac c{cv}=\dfrac 1v$, not $v$ (which you have in the first way)


Addendum in response to question edit:

$\dfrac nv=c\implies\dfrac nc=v\implies \dfrac {nc}c=vc\implies n=vc$

($\dfrac nc=\dfrac c{cv}$ is not correct, nor is $\dfrac{nc}c=\dfrac vc$, nor $n=\dfrac vc$.)

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    $\begingroup$ I just have 1 last question: ${n=\frac{1}{vc}\Rightarrow{\frac{1}{n}}=\frac{\dfrac{1}{1}}{vc}}\Rightarrow{\frac{1}{n}=vc}$ | I’m used to the concept of what you do on 1 side of the equation you do to the other side. However it seems when you invert that the rules are different, can you please clarify? @LordCommander $\endgroup$
    – Nickotine
    Mar 13, 2021 at 7:35
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    $\begingroup$ @Nickotine what you got here is correct, for example if $n=\frac{1}{5}$, $\frac{1}{n}=\frac{1}{\;\;\frac{1}{5}\;\;}=5$ $\endgroup$ Mar 13, 2021 at 7:48
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    $\begingroup$ Notice how i put a 'big' division sign there, to differentiate between $1/(1/5)$ and $(1/1)/5$. Our case is the former (obviously) because we divide by $n=1/5$ $\endgroup$ Mar 13, 2021 at 7:52
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    $\begingroup$ I’m still getting used to mathJax.... only been using it for maybe 3 or 4 days $\endgroup$
    – Nickotine
    Mar 13, 2021 at 19:10
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    $\begingroup$ say you have $z=z\cdot a$ then you can just divide by $z$ and it will fall out to give you $1=a$_____________($\cdot$ means multiplication) $\endgroup$ Mar 13, 2021 at 20:06

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