$$\dfrac{p}{p+3}=\dfrac{2p-1}{2p}$$
Get $p$. How can one solve these type of questions? What is the easiest and quickest method?
$$\dfrac{p}{p+3}=\dfrac{2p-1}{2p}$$
Get $p$. How can one solve these type of questions? What is the easiest and quickest method?
The given equation is
$$\frac{p}{p+3}=\frac{2p-1}{2p}$$
Equation is solving if $p\neq 0$ and $p+3\neq 0\implies p\neq -3$ $$\begin{align*} p\cdot 2p&=(2p-1)(p+3)\\ 2p^2&=(2p-1)(p+3)\\ 2p^2&=2p^2-p+6p-3\\ 2p^2&=2p^2+5p-3\\ 2p^2-2p^2&=5p-3\\ 3&=5p\\ p&=\frac{3}{5} \end{align*}$$
You can deal with it as following:
$$\frac{p}{p+3}=\frac{2p-1}{2p}\Rightarrow 1-\frac{p}{p+3}=1-\frac{2p-1}{2p}\Rightarrow \frac{3}{p+3}=\frac{1}{2p}\Rightarrow 6p=p+3\Rightarrow p=\frac{3}{5}$$
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.