Working out a reverse formula

My math skills are getting rusty. I am trying to work out what the formula should be for calculating price, $P$, based on a formula I used to calculate margin, $\mu$, with a parameter, cost, $C$.

$$\mu = 39.32\text{, }\ P = 177.99 \text{, }\ C = 108.00.$$

$$\mu = \frac{100(P-C)}{P}$$

I know $P$ should be $177.99$ when $C$ is $108.00$ and $\mu$ is $32.39$. But what is the formula to calculate it? $$P = F(\mu,C)$$

After going through several math tutorials I was able to simplify the formula a little. But I am still at a loss as to how to get price to one side of the equation by itself.

$$P=\frac{P\cdot\mu}{100} +C$$

Hint:

A nice formatting helps me understanding, perhaps for you too: $$\mbox{price} = \frac{\mbox{margin}}{100} \mbox{price} + \mbox{cost} \iff \\ \mbox{price} - \frac{\mbox{margin}}{100} \mbox{price} = \mbox{cost} \iff \\ 1 \cdot \mbox{price} + \left(- \frac{\mbox{margin}}{100}\right) \mbox{price} = \mbox{cost}$$ Now use the distributive law $$A \cdot C + B \cdot C = (A + B) \cdot C$$ for the left hand side, to consolidate terms with price. You just have to match $A$, $B$ and $C$ with the term $$1 \cdot \mbox{price} + \left(- \frac{\mbox{margin}}{100}\right) \mbox{price}$$ and apply the law. Start with $C$ because it shows up twice and is the reason why we do this, because we want to have $C$ show up only once, like in the right hand side of the distributive law.

An example: $5 \cdot 7 + 3 \cdot 7 = (5+3)\cdot 7$. thus $A =5$, $B=3$, $C=7$.

Solution:

$$1 \cdot \mbox{price} + \left(- \frac{\mbox{margin}}{100}\right) \mbox{price} = \mbox{cost} \iff \\ \left( 1 + \left(- \frac{\mbox{margin}}{100}\right)\right) \mbox{price} = \mbox{cost} \iff \\ \mbox{price} = \frac{\mbox{cost}}{\left( 1 + \left(- \frac{\mbox{margin}}{100}\right)\right)} \iff \\ \mbox{price} = \frac{\mbox{cost}}{\frac{100 -\mbox{margin}}{100}} = \frac{100}{100 -\mbox{margin}} \mbox{cost}$$

Note: This answer uses the variable names from the OP's version of the question.

• Thanks, but it is the consolidation of terms I am having problems with. None of the tutorials I have found explains exactly what to do when the variable is in both the enumerator and denominator on one side of the equation. Note also I am trying to solve for price, not cost. – NightOwl888 Mar 28 '15 at 20:03
• I don't even really understand how that is supposed to be helpful. I need an answer, not a deep understanding of the problem. All I need is to get the price over to one side of the equation by itself. – NightOwl888 Mar 28 '15 at 20:09
• Yes, but a little pain of insight is not that bad. You will be happy once it makes click. – mvw Mar 28 '15 at 20:11
• Sorry, I don't get paid for insight. I get paid for results. I have been stuck on this for 5 hours and I need an answer so I can deliver the end product. I appreciate you trying to be helpful with teaching, but it has been far too many years since I have studied distributive law to make it work. If I saw the answer, I could probably reverse engineer distributive law, but without the formula I am seeking I fail to see how to get there. – NightOwl888 Mar 28 '15 at 20:24
• This aims to be not a solution providing, but insight providing site. Anyway. – mvw Mar 28 '15 at 20:27

Hint: Notice that

$$\frac{\mathrm{price} - \mathrm{cost}}{\mathrm{price}} = \frac{\mathrm{price}}{\mathrm{price}} - \frac{ \mathrm{cost}}{\mathrm{price}} = 1 - \frac{ \mathrm{cost}}{\mathrm{price}}$$

There should only be 1 "price" left after this.

Note that $$\mu = \frac{100(P-C)}P = \frac{100 P - 100 C}P = 100 \frac PP - 100 \frac CP = 100 - 100 \frac CP$$ So rearranging gives $$100\frac CP = 100 - \mu$$ Divide by $100 C$ to get $$\frac1P = \frac{100 - \mu}{100 C}$$ And finally take reciprocals on both sides: $$P = \frac{100 C}{100 - \mu}$$