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I have fractions like these: $$\frac{12000x}{41}\qquad (1)\\\frac{12000+x}{41}\qquad (2)\\$$ What is the process to convert these to a mixed number (example: $5\frac12$)?

Yivan000
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  • Welcome to MSE. Please read this text about how to ask a good question. – José Carlos Santos Apr 06 '22 at 09:21
  • Note $41$ is prime. You do this in the usual way, dividing the numerator by the denominator to get a quotient and remainder. – Henry Apr 06 '22 at 09:24
  • There is not enough information in question to understand, what author is looking for. $\frac{f(x)}{41}=\lfloor \frac{f(x)}{41}\rfloor + \left(\frac{f(x)}{41}-\lfloor \frac{f(x)}{41}\rfloor\right)$. $\frac{12000x}{41}=\frac{12000}{41}x=292\frac{28}{41}; x$. $\frac{12000+x}{41}=292\frac{28}{41}+\frac{x}{41}$. If $x$ is integer and $0\leq x<13$ then $\frac{12000+x}{41}=292\frac{28+x}{41}$ – Ivan Kaznacheyeu Apr 06 '22 at 09:48
  • For $r \in \Bbb{R},~$ let $\lfloor r\rfloor$ (i.e. the floor function) denote the largest integer $\leq r$. Then, $$\left\lfloor \frac{12000}{41}\right\rfloor = 292 \implies \frac{12000}{41} = 292 + \frac{28}{41}.$$ Therefore, $$\frac{12000x}{41} = 292x + \frac{28x}{41},$$ while $$\frac{12000 + x}{41} = 292 + \frac{28}{41} + \frac{x}{41}.$$ – user2661923 Apr 06 '22 at 10:14
  • Re my previous comment, and as (also) indicated by the comment of @IvanKaznacheyeu, since $x$ is (presumably) a variable with no fixed value, you can not be more precise than the expressions in my previous comment. – user2661923 Apr 06 '22 at 10:18

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