# Solving simple equation

I can't seem to solve/simplify step by step to get from equation 3) to 4) as they do in this paper. As per the paper:

3) $$p = \frac{p' + (r-R)p}{1+r}$$

Because both sides of equation 3) involve the current price, $$p$$, the equation can be solved as follows:

4) $$p = \frac{p'}{1+R}$$

Could somebody please step by step show how I'd solve this step please?

• This looks like a homework question. We are not here to do your homework. – FUBAR Jul 11 at 13:34
• Is your first equation $$p=p'+\frac{(r-R)p}{1+r}$$? – Dr. Sonnhard Graubner Jul 11 at 13:36
• Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Jul 11 at 13:38
• Welcome to StackExchange. A couple of things: 1. Please use MathJax when formatting questions. It makes reading them much easier. Here is a link: math.meta.stackexchange.com/questions/5020/… 2. As FUBAR mentioned, this is not a homework question answering site. Please try to show work when you ask questions so that we know you've put effort into trying to solve them. – scoopfaze Jul 11 at 13:38
• Ah thank you for the info! Yeah, I wasn't sure about how to format (only used programming subs). This isn't homework, I'm just trying to understand a paper: rhsmith.umd.edu/files/Documents/Centers/CFP/ICIConf2013/… Apologies, I'm really not sure of where to start. I thought I'd try and * 1/p on both sides to remove the p on the right side of the equation. Then I thought maybe both 'r's cancel out so they disappear. But I wasn't sure how R moved to the bottom of the equation... How just p was on the left... etc. Apologies, I can't explain! – bennyjim Jul 11 at 13:52

## 1 Answer

We start with the original equation: \begin{align*} p = \frac{p' + (r-R)p}{1+r} \end{align*} We want to obtain $$p$$ as a function of the other values. So, we first want to collect all terms with $$p$$ on the same side. We do this by splitting the fraction and bringing one term over. \begin{align*} p &= \frac{p'}{1+r} + \frac{(r-R)p}{1+r} \\ p - \frac{(r-R)p}{1+r} &= \frac{p'}{1+r} \end{align*} Now, we can factor out the $$p$$ on the left side, and put everything on the same denominator. \begin{align*} p \left( 1 - \frac{(r-R)}{1+r} \right) &= \frac{p'}{1+r} \\ p \left( \frac{1+r}{1+r} - \frac{r-R}{1+r} \right) &= \frac{p'}{1+r} \\ p \left( \frac{1+R}{1+r} \right) &= \frac{p'}{1+r} \end{align*} (Note of course that $$(1+r)-(r-R) = (1+r) + (-r +R) = 1+R$$.). Finally, we multiply both sides of the equation by $$(1+r)/(1+R)$$ to get: \begin{align*} p \left( \frac{1+R}{1+r} \right) \cdot \frac{1+r}{1+R} &= \frac{p'}{1+r} \cdot \frac{1+r}{1+R} \\ p = \frac{p'}{1+R} \end{align*} And this is the desired result.

• thank you very much! I understand each step except for one bit... the first step after ‘Now, we can factor out the p on the left side, and put everything on the same denominator’, has p followed by a bracket and then a one and minus - I.e. ‘p(1-....’ . I’m not sure where the sudden ‘1’ is coming from? – bennyjim Jul 11 at 18:55
• @bennyjim Try multiplying out. In general, we have that $pa+pb = p(a+b)$; this is a specific case where $a=1$. I.e., $p+pb = p(1+b)$. – Sambo Jul 11 at 19:41
• @bennyjim P.S. I edited your question earlier; does it match your original intent? Also note that you can click the check mark next to an answer to accept it if it fully answers your question. – Sambo Jul 11 at 19:44
• Thank you!!! All makes sense. Question answered. – bennyjim Jul 11 at 22:20