How to solve this system of linear inequalities? I have the following system of linear inequalities
$$\begin{aligned} \theta_1 + \theta_2 &\geq 0\\ 2 \theta_1 + k \theta_2 &\geq 0\\ \theta_1 + 3\theta_2 & \leq 0\end{aligned}$$
So far, I have tried to take the $3^{\rm rd}$ equation and rewrite it as follows
$$\theta_1 + 2\theta_2 \leq -\theta_2$$
Where can I go from here? And what is the easier solution to this exercise?
 A: $\theta_1 + \theta_2 \ge 0 \cdots(i)$ 
and $\theta_1 + 3\theta_2 \le 0 \implies -\theta_1 - 3\theta_2 \ge 0 \cdots(ii)$
From $(i)$ and $(ii)$,
$\theta_1 + \theta_2 -\theta_1 - 3\theta_2 \ge 0 $

$$-\theta_2 \ge 0 \implies \theta_2 \le 0$$

Now, 
$\theta_1 + \theta_2 \ge 0 \implies \theta_1 \ge -\theta_2 \ge 0$

$$\theta_1\ge0$$

Do similarly with $2\theta_1 + k\theta_2 \ge 0$
A: Try to isolate one of the variables, and then substitute it in another inequality. For instance, the first inequality becomes:
$$
\theta_1 \geq -\theta_2.
$$
Can you take it from here?
A: Calling 
$$
\vec v_1 = (1,1)\\
\vec v_2 = (2,k)\\
\vec v_3 = -(1,3)
$$
now if $\vec v_2 = \lambda_1 \vec v_1 + \lambda_2\vec  v_3$ with $(\lambda_1,\lambda_2) \ge 0$ then the inequality system has non trivial solution otherwise the solution is $\theta_1=\theta_2 = 0$. In the present case we have
$$
\lambda_1 = \frac{6-k}{2} \ge 0\\
\lambda_2 = \frac{2-k}{2} \ge 0
$$
so the condition for non trivial solution is $k \le 2$
Now assuming that the solution is non trivial and calling $\Theta = (\theta_1,\theta_2)$, this solution should verify
$$
\vec v_1\cdot \Theta \ge 0\\
\vec v_2\cdot \Theta \ge 0\\
\vec v_3\cdot \Theta \ge 0
$$
which is verified for
$$
\Theta(\mu_1,\mu_3) = \mu_1\vec v_1+\mu_3\vec v_3 \ge 0
$$
with $\mu_i \ge 0, i=1,3$
