Answer: \(x\lt -2\) or \(x\gt 2\).

For

\(y=5+|3x-2|,\)

the vertex occurs when \(3x-2=0\), so

\(x=\frac23.\)

At this point,

\(y=5.\)

So the vertex is \(\left(\frac23,5\right)\). Also, when \(x=0\),

\(y=5+|-2|=7,\)

so the \(y\)-intercept is \((0,7)\). The graph is a V-shape.

The straight line \(y=11-x\) has \(y\)-intercept \((0,11)\) and gradient \(-1\).

To solve the inequality algebraically, first find the intersection points.

If \(x\lt \frac23\), then

\(5+|3x-2|=5-(3x-2)=7-3x.\)

So

\(11-x=7-3x,\)

giving

\(x=-2.\)

If \(x\geq\frac23\), then

\(5+|3x-2|=5+(3x-2)=3x+3.\)

So

\(11-x=3x+3,\)

giving

\(x=2.\)

The inequality

\(11-x\lt 5+|3x-2|\)

means that the straight line is below the modulus graph. This happens outside the two intersection points, so

\(x\lt -2\quad\text{or}\quad x\gt 2.\)