In a chemistry lab, two substances A and B react to form a product…

In a chemistry lab, two substances A and B react to form a product. At time $t$ minutes the concentration of the product is $x$ moles per litre. The rate of the reaction is $\dfrac{dx}{dt} = 0.1\,(5 - x)(3 - x)$, where the constant $0.1$ has the unit litres per mole per minute. The starting concentrations of A and B are $5\ \text{mol/L}$ and $3\ \text{mol/L}$, and $x = 0$ at $t = 0$ minutes. A student separates the variables, so that $0.1\,t = \displaystyle\int \dfrac{dx}{(5 - x)(3 - x)}$, with the integral taken to be $0$ at $x = 0$. To integrate the right side she writes $\dfrac{1}{(5 - x)(3 - x)} = \dfrac{A}{5 - x} + \dfrac{B}{3 - x}$ and finds $A$ and $B$ by the cover-up rule. She then finds the time at which the product reaches $2\ \text{mol/L}$, and the concentration after $5$ minutes. Find $A$ and $B$. Give the time in minutes and the concentration in moles per litre, each to four decimal places.