= Solution
The feasible triangle has vertices
$$
v_1=(18/7,2/7),\qquad v_2=(12/11,14/11),\qquad v_3=(2/5,-4/5).
$$
These come from the three pairs of active boundary lines and satisfy the remaining inequalities. Its denominator $3x_1+x_2+2$ is positive at each vertex, with minimum $12/5$, so is positive throughout the triangle because it is an <affine function>.
For a direct <linear programming optimality certificate>, add twice the second inequality to the third to get $-x_1-3x_2\le2$. Thus $x_1+3x_2+2\ge0$, equivalently $2(x_1-x_2)\le3x_1+x_2+2$. Division by the positive denominator gives an objective at most $1/2$. Both inequalities used in the bound are equalities at $v_3$, where the first inequality is also satisfied. Therefore
$$
\boxed{x^*=(2/5,-4/5),\qquad\max\frac{x_1-x_2}{3x_1+x_2+2}=\frac12.}
$$
Equality requires the two bounding inequalities to be tight, so this optimizer is unique.
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