Given the system of ODEs, $$x'=x(1-x-y)$$ $$y'=y(x-1),$$ $Q=\{(x,y):x\ge 0, y\ge 0\}$, and $S=(x,y)\in Q:x+y\le k$, $k>1$, I need to show that $S$ is invariant under this system of ODEs.

**Attempted solution:** $x'=x(1-(x+y))\ge x(1-k) \le 0$. Suppose $y'>0$, then $x<xk$ implies that $x-x^2-xy\ge x-xk \iff y\le (k-x)$. Hence, $y' = y(x-1) \le (k-x)(x-1)$. $y'$ changes sign at $y'=0$, thus we consider $(k-x)(x-1)=0$, whose roots are $x=k$ and $x=1$. But $x\ne k$ since $x'$ is decreasing, so $y'$ changes sign at $x=1$.

And so I don't know what to do next. I realize that we need to show that $x$ is decreasing faster than $y$ is increasing. We can solve the equation for $x(t)$ to obtain $x(t)= x_0 e^{1-k}$, and then solve for $y(t)$ and compare. Would that be practical?