Answer: \(x\leqslant-\dfrac52\) or \(x\geqslant-\dfrac14\).

The inequality

\(|3x+3|\geqslant |x-2|\)

means that the graph of \(y=|3x+3|\) must be on or above the graph of \(y=|x-2|\).

The graph of \(y=|3x+3|\) has vertex at \(x=-1\). The graph of \(y=|x-2|\) has vertex at \(x=2\). The solution changes at the points where the two graphs intersect.

Find one intersection using the left branch of \(y=|x-2|\) and the right branch of \(y=|3x+3|\):

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

Thus

\(4x=-1,\)

so

\(x=-\dfrac14.\)

Find the other intersection using the left branch of \(y=|3x+3|\) and the left branch of \(y=|x-2|\):

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

So

\(-3x-3=-x+2,\)

which gives

\(-2x=5,\qquad x=-\dfrac52.\)

From the graph, \(|3x+3|\) is greater than or equal to \(|x-2|\) outside the interval between these two intersection points.

Therefore

\(x\leqslant-\dfrac52\quad\text{or}\quad x\geqslant-\dfrac14.\)