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Get $p$. How can one solve these type of questions? What is the easiest and quickest method?

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What's wrong with $2p^2=(2p-1)(p+3)$? Of course under the conditions $p \neq 0$ and $p+3 \neq 0$. From a geometrical point of view, you are looking for the intersection points of two hyperbolas, and it is natural that you should solve a second order equation. –  Siminore Oct 6 '12 at 12:36
You've gotten answers, but generally, you want to simplify in an efficient way that produces correct results. For ratio problems like this, the typical approach is cross multiply, simplify, gather like terms and then solve resulting problem. –  Amzoti Oct 6 '12 at 12:51

3 Answers 3

The easiest and quickest method is by "componendo / divideno". [See Hall and Knight.]

From the original equation, we do $\dfrac {p - (p + 3)} {p + 3} = \dfrac {2p - 1 - (2p)} {2p}$

Therefore, $\dfrac {- 3} {p + 3} = \dfrac {- 1} {2p}$

Then, $6p = p + 3$, and hence the result.

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The given equation is


Equation is solving if $p\neq 0$ and $p+3\neq 0\implies p\neq -3$ $$\begin{align*} p\cdot 2p&=(2p-1)(p+3)\\ 2p^2&=(2p-1)(p+3)\\ 2p^2&=2p^2-p+6p-3\\ 2p^2&=2p^2+5p-3\\ 2p^2-2p^2&=5p-3\\ 3&=5p\\ p&=\frac{3}{5} \end{align*}$$

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Thank you sir! And how would one solve a similar equation for example: $ \dfrac{p-1}{q} = \dfrac{q-1}{2p-1} = \dfrac{3}{5} $ I am guessing $ (p-1)(2p-1) = q(q-1) - \dfrac{3}{5} $ Is this correct? –  kamal Oct 6 '12 at 12:52
@kamal This is actually two equations. $\frac{p-1}{q}=\frac{3}{5}$ and $\frac{q-1}{2p-1}=\frac{3}{5}$. –  Patrick Li Oct 6 '12 at 13:00
So would you have to solve this: $(p-1)5 - 3q = (q-1)5 - (2p-1)3 $ Or is this again incorrect? –  JohnPhteven Oct 6 '12 at 13:21
@kamal As Patrick has pointed out, you have two equations in p and q. To solve them, use the first: $\frac{p-1}{q}=\frac{3}{5} \implies p=\frac{3}{5}q+1$ We may substitute this into our second equation, to give: $\frac{q-1}{2p-1}=\frac{3}{5} \implies \frac{q-1}{\frac{6}{5}q+2-1}=\frac{q-1}{\frac{6}{5}q+1}=\frac{3}{5}$ Thus $q-1=\frac{18}{25}q+\frac{3}{5}$ And finally $\frac{7}{25}q=\frac{8}{5} \implies q=\frac{40}{7}$ Substituting into our first equation, $p=\frac{3}{5}\frac{40}{7}+1=\frac{31}{7}$ –  Daniel Littlewood Oct 6 '12 at 13:25

You can deal with it as following:

$$\frac{p}{p+3}=\frac{2p-1}{2p}\Rightarrow 1-\frac{p}{p+3}=1-\frac{2p-1}{2p}\Rightarrow \frac{3}{p+3}=\frac{1}{2p}\Rightarrow 6p=p+3\Rightarrow p=\frac{3}{5}$$

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