Differential Equation Mixture Problem

Question

A large tank starts with $900L$ of water. At time $t = 0$, a solution with a concentration of $1g/L$ is pumped in at a rate of $10L/min$. The solution is then pumped out at $8L/min$. Let x(t) denote the amount of solution in the tank at time $t$. How do I set up an equation for an initial value problem? The general formula is $\frac{dy}{dx}=$rate in - rate out. So that's $\frac{dy}{dx}=(1g/L)(10L/min)-($concentration in tank$)(8L/min)$. My goal is to get the equation an equation in the form $x(t)=...$, how do I do that?