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October 7

silly question

Hello, when I try solving the equation a/(b+c)=d/c for b using pencil and paper, I get b=ac/d-c (like this:
a/(b+c)=d/c
(multiply by b+c) a=d(b+c)/c
(multiply by c) ac=d(b+c)
(divide by d) ac/d=b+c
(substract c) ac/d-c=b.)
(One could also flip both fractions and save a step: a/(b+c)=d/c; (b+c)/a=c/d; b+c=ac/d; b=ac/d-c.)
However, when I input the formula into WolframAlpha, telling it to solve for b, it says that b=c(a-d)/d. Is it doing this for no reason (the way the equation rearranging algorithm works) or is it because the latter form is preferred in refined circles (is it?) and WA maybe even takes extra steps to arrive at it? Thanks in advance. Asmrulz (talk) 08:47, 7 October 2012 (UTC)[reply]

You have the right answer if you, as I hope, put the "-c" outside the fraction. A few algebraic manipulations of your answer is all that is needed to arrive at the answer your program gave

You started with

b={\frac {ac}{d}}-c

convert the "

-c

" into a fraction which has a common denominator with the first fraction

b={\frac {ac}{d}}-{\frac {cd}{d}}

Now you arrive at the program's answer

b={\frac {ac-cd}{d}}={\frac {c(a-d)}{d}}

Sjakkalle (Check!) 09:38, 7 October 2012 (UTC)[reply]

Also, you asked if there was any reason for this. Well, there is no "best" standard, but in many cases a single fraction (one term) is a tidier expression than a single fraction with something subtracted from it (two terms). Sjakkalle (Check!) 10:00, 7 October 2012 (UTC)[reply]

In some cases where there are multiple ways to write something one of them is better. In this case they're about the same and it boils down to personal preference and context (e.g., if a-d has some special meaning it makes sense to use it in the expression). Without any context I'd mildly prefer the version you gave.

Mathematica (what Alpha is based on) has some rules to quantify how complicated an expression is; when simplifying something (which I assume Alpha is doing automatically) it will give the equivalent expression with the minimal complexity. It so happens that with the rules they're using the particular form they gave is considered simpler. -- Meni Rosenfeld (talk) 20:42, 9 October 2012 (UTC)[reply]

simple math wikicode

I just wasted a 45 minutes trying to find the code for a simple division.
I saw a chart in an article that has the wrong division results and I want to show the correct math.
I don't just want to change the results in the chart, I want to show the math with the math code in the article code.
1,152,000,000/365 $\neq$ 1,152,000,000/365.26,
so if I change the results to the later I don't just want to replace the numbers,
I want to show that how it was calculated in the article code. I know it's possible,
I have seen it with the automatic conversion of Celsius to Fahrenheit, and I did find the addition code: {{Addition|1|1|1}}
Somebody please help.
24.79.38.15 (talk) 11:03, 7 October 2012 (UTC)[reply]

So in the mean time I find that this works {{addition|360/365.26}} = {{addition|360/365.26}}
But it has too many sig difs. I can`t believe this is so hard to find.
24.79.38.15 (talk) 11:26, 7 October 2012 (UTC)[reply]

Maybe {{round|360/365.26|3}}, giving 0.986? — Quondum 12:32, 7 October 2012 (UTC)[reply]

Thank you very much, I will use that. And thanks for ``Roaming the foam.`` 24.79.38.15 (talk) 14:33, 7 October 2012 (UTC)[reply]