This is my first question here, so please forgive me if the format etc. are not quite right. I've been attacking an algebra question, and my workings are below. There's a mistake somewhere (I don't know where) because my workings don't reach the correct answer! I'd really appreciate someone letting me know where I'm slipping up:

I have two functions, $G$ as a function of $N$ and $N$ as a function of $l:$

$G(N) = \frac{N}{40}\cdot 100$

$N(l) = \frac{300}{l} + 5$

I need to compose a function that represents $G$ as a function of $l.$ I do it like this:

$$G(l) = \frac{\frac{300}{l}+5}{40}\cdot 100 = \frac{\frac{100 \cdot 300}{l} + 100 \cdot 5}{40}=\frac{\frac{30,000}{l} + 500}{40}= \frac{750}{l} + 12.5$$

And that's it. ...but it's wrong.

Thanks so much for your help with this!

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    $\begingroup$ See meta.math.stackexchange.com/questions/5020/… for formatting. $\endgroup$ – lhf Jul 10 '14 at 12:19
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    $\begingroup$ Why do you say that the answer is wrong? What result did you expect? $\endgroup$ – mfl Jul 10 '14 at 12:23
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    $\begingroup$ The arithmetic adds up... Is there an answer key somewhere? $\endgroup$ – abiessu Jul 10 '14 at 12:28
  • $\begingroup$ Correct as far as eye can see. $\endgroup$ – user21820 Jul 10 '14 at 12:30

You did nothing wrong. The calculation is correct.

Maybe it is easier to see with

$$ G(N) = \frac{5}{2} N = \frac{5}{2} \cdot 300 \cdot \frac{1}{l} + \frac{5}{2} \cdot 5 = 750 \frac{1}{l} + 12.5 $$

  • $\begingroup$ Apparently that actually was right. Sorry everyone. Thanks for your help, and for latexing my question. I'll learn how to use that. How can I mark my question as "done"? $\endgroup$ – Nicholas James Bailey Jul 10 '14 at 16:25

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