From Boyle’s Law: $$ \begin{equation}\begin{aligned} PV=k_1\\ \end{aligned}\end{equation} $$
From the Pressure Law: $$ \begin{equation}\begin{aligned} P=k_2T\\ \end{aligned}\end{equation} $$
From Charles’ Law: $$ \begin{equation}\begin{aligned} V=k_3T\\ \end{aligned}\end{equation} $$
Multiplying these equations: $$ \begin{equation}\begin{aligned} PV\times P\times V&=k_1\times k_2T\times k_3T\\ P^2V^2&=k_1k_2k_3T^2\\ \frac{P^2V^2}{T^2}&=k_1k_2k_3\\ \end{aligned}\end{equation} $$
Square-rooting both sides: $$ \begin{equation}\begin{aligned} \sqrt{\frac{P^2V^2}{T^2}}&=\sqrt{k_1k_2k_3}\\ \frac{PV}{T}&=\sqrt{k_1k_2k_3}\\ \end{aligned}\end{equation} $$
The square-root of the product of three constants is itself a constant: $$ \begin{equation}\begin{aligned} \frac{PV}{T}=k\\ \end{aligned}\end{equation} $$
Generally: $$ \begin{equation}\begin{aligned} \frac{P_1V_1}{T_1}&=\frac{P_2V_2}{T_2}\\ \end{aligned}\end{equation} $$
Example
A gas at $100\ kPa$ at $20\degree C$ fills a flexible container with an initial volume of $2\ m^3$. If the temperature is raised to $90\degree C$ and the pressure increases to $320\ kPa$, what is the new volume?
Mission details
A gas with pressure $50\ kPa$ at $40\degree C$ fills a flexible container with an initial volume of $1.6\ m^3$. If the volume increases to $2.4\ m^3$ and the pressure changes to $270\ kPa$, what is the new temperature?