= Solution
Introduce nonnegative <slack variables> $s_1,s_2$ and let $z$ denote profit. We display each <simplex tableau> as coefficients of its row equation, with the coefficient of $z$ in the objective equation implicitly one. Thus a negative coefficient in the last row permits an improving entering variable. The initial <simplex tableau> is
$$
\begin{array}{c|rrrr|r}
\text{basis}&x_1&x_2&s_1&s_2&\text{RHS}\\\hline
s_1&1&2&1&0&6\\
s_2&2&1&0&1&6\\\hline
z&-3&-4&0&0&0
\end{array}.
$$
Use the <simplex method> with $x_2$ entering. The <simplex ratio test> gives ratios $6/2=3$ and $6/1=6$, so $s_1$ leaves. Divide its row by two and eliminate the other $x_2$ entries:
$$
\begin{array}{c|rrrr|r}
\text{basis}&x_1&x_2&s_1&s_2&\text{RHS}\\\hline
x_2&1/2&1&1/2&0&3\\
s_2&3/2&0&-1/2&1&3\\\hline
z&-1&0&2&0&12
\end{array}.
$$
Next $x_1$ enters. The positive-column ratios are $3/(1/2)=6$ and $3/(3/2)=2$, so $s_2$ leaves. After this pivot and a row reordering,
$$
\begin{array}{c|rrrr|r}
\text{basis}&x_1&x_2&s_1&s_2&\text{RHS}\\\hline
x_1&1&0&-1/3&2/3&2\\
x_2&0&1&2/3&-1/3&2\\\hline
z&0&0&5/3&2/3&14
\end{array}.
$$
The last row is $z=14-(5/3)s_1-(2/3)s_2$, so every nonnegative feasible solution has $z\leq14$. Setting the nonbasic variables to zero gives
$$
\boxed{x_1=x_2=2,\qquad z_{\max}=14.}
$$
The equality certificate also shows uniqueness, since both nonbasic objective coefficients are strictly unfavorable.
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