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

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

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    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.2012-10-06

3 Answers 3

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The given equation is

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

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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    @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}$2012-10-06
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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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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.