We have the following LES $I$ \begin{align} 3x_1-2x_2+4x_3-x_4&=1\\ -\frac43x_2-\frac{13}3x_3+\frac13x_4&=-\frac{19}3 \end{align} with the solution set $\mathcal{L}_{I}$, which contains all solutions of $I$. Let $\mathcal{L}_2$ be some other solution, which is defined as the following: $$\mathcal{L}_2=\{(x_1,x_2,x_3,x_4)\mid(x_1,x_2,x_3,x_4)=(3,3,1,6)+\lambda_1(3,3,-1,-1)+\lambda_2(2,3,1,4),\;\lambda_1,\lambda_2\in\mathbb{R}\}$$
We want to determine whether $\mathcal{L}_2\subseteq \mathcal{L}_{I}$ or not.
Therefore, we will insert the tuples into $I$. If all equations hold for the inserted values $x_1,\cdots,x_4$, the solution set $\mathcal{L}_2$ is a subset of all solutions.: First equation holds. Moving on to the next one. We can simplify the term $$-\frac{4}3(3+3\lambda_1+3\lambda_2)-\frac{13}{3}(1-\lambda_1+\lambda_2)+\frac13(6-\lambda_1+4\lambda_2)=-\frac{19}3$$ to $$-\frac{19}3-7\lambda_2=-\frac{19}3$$ How do we go on now? Can we choose $\lambda_2=0$ and assume $\mathcal{L}_2\subseteq \mathcal{L}_I$ because all equations hold for $\lambda_2=0$ or is this in fact the sign for $\mathcal{L}_2\nsubseteq \mathcal{L}_I$?
If you choose $\lambda_1=0,\lambda_2=-1$ you get $x \in \mathcal{L}_2$ of the form $(1, 0, 0, 2)$, which does not solve the second equation, therefore $\mathcal{L}_2\nsubseteq \mathcal{L}_I$.