Differential equations, solving for a chemical rate of change. Two chemicals $\mathrm{A}$ and $\mathrm{B}$ are put together in a solution where they react to form a compound $\mathrm{X}$. The rate of increase of the mass, $x\,\rm kg$, of $\mathrm{X}$ is proportional to the product of the masses of unreacted $\mathrm{A}$ and $\mathrm{B}$ present at time, $t\,\rm min$. It takes $1\,\rm kg$ of $\mathrm{A}$ and $3\,\rm kg$ of $\mathrm{B}$ to form $4\,\rm kg$ of $\mathrm{X}$. Initially, $2\,\rm kg$ of $\mathrm{A}$ and $3\,\rm kg$ of $\mathrm{B}$ are put together in solution, and $1\,\rm kg$ of $\mathrm{X}$ forms in one minute. 


What I deciphered from this is only:


$A + B = Xt$, only true for $t = 1$ (doesn't help at all).  
$dx/dt = 1$, when $t = 1$.


How can I set up a differential equation expressing $dx/dt$ as a function of $x$?
$dA/dt= (-1/4)(dx/dt)$
$dB/dt=(-3/4)(dx/dt)$
This gives us with $A=2-(1/4)x$ and $B=3-(3/4)x$
We can interpret that $dx/dt =kab$, for some number $k$.
Now we sub in the values of A and B into the equation dx/dt.
This leaves us with $dx/dt= [3k(8 − x)(4 − x)]/16$
$dA/dt= (-1/4)(dx/dt)$
$dB/dt=(-3/4)(dx/dt)$
This gives us with $A=2-(1/4)x$ and $B=3-(3/4)x$
We can interpret that $dx/dt =kab$, for some number $k$.
Now we sub in the values of A and B into the equation dx/dt.
This leaves us with $dx/dt= [3k(8 − x)(4 − x)]/16$