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My algebra is rusty, I'm trying to rearrange an equation, the purpose of which is to calculate the price that would result in a desired margin. Here are my workings:

p = price
c = cost
m = margin
P = profit = p - c

$$m = 100\frac{P}{p}$$ $$m = 100\frac{p-c}{p}$$ $$\frac{m}{100} = \frac{p-c}{p}$$ $$p\frac{m}{100} = p-c$$

$$p = \frac{pm}{100} + c$$

I'm stuck at this point. I can't figure out how to factor out p from the RHS. Can someone show me how to continue?

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  • $\begingroup$ $p = c/(1-m/100)$ $\endgroup$ – Carlos Campos Jun 18 '18 at 12:53
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    $\begingroup$ hint: $\frac{p-c}{p} = 1-\frac{c}{p}$ $\endgroup$ – David Diaz Jun 18 '18 at 13:10
  • $\begingroup$ @DavidDiaz I tried that and I get to $$p = \frac{100c}{m-100}$$ which doesn't seem right $\endgroup$ – Kev Jun 18 '18 at 13:22
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    $\begingroup$ Probably since you should have $$ p = \frac{100 c}{100 - m} $$ $\endgroup$ – Stan Tendijck Jun 18 '18 at 13:29
  • $\begingroup$ @StanTendijck Thank you! I can see where I went wrong. $\endgroup$ – Kev Jun 18 '18 at 13:35
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Here are the full workings to convert the margin equation to the price equation, for anyone interested. Thanks to @DavidDiaz and @StanTendijck for pointing me in the right direction.

$$m = 100\frac{p-c}{p}$$ $$m = 100(1 -\frac{c}{p})$$ $$m = 100 - \frac{100c}{p}$$ $$m-100 = -\frac{100c}{p}$$ $$\frac{1}{p} = -(\frac{m-100}{100c})$$ $$\frac{1}{p} = \frac{100-m}{100c}$$ $$p = \frac{100c}{100-m}$$

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