The concentrations of two chemicals $A$ and $B$ as functions of time are denoted by $x$ and $y$ respectively. Each alone decays at a rate proportional to its concentration. Put together, they also interact to form a third substance, at a rate proportional to the product of their concentrations. All this is expressed in the equations: $$\frac{dx}{dt} = -2 x - x y, \qquad \frac{dy}{dt} = -3 y - x y.$$

Note that $(x,y) = (0,0)$ is the only equilibrium state (both concentrations must be positive or zero).

(a) Find an equation for $y(x)$:
$\frac{dy}{dx} =$

(b) Solve your equation to obtain a relationship between $x$ and $y$. Letting $x(0) = 4$ and $y(0) = 8$, give the relationship as an equation involving $x$ and $y$:

(Your answer will be an equation involving $x$ and $y$.)

Note that one implication of your relationship should be that if $x$ and $y$ are positive and very small (so that $x$ is very much smaller than $\ln(x)$), $y^{2}/x^{3}$ is roughly constant.

(c) Suppose that $x = y$. From your equation, what are $x$ and $y$ in this case?
$x = y =$