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

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

Answer: (a) \(x\leqslant-2\) or \(x\geqslant3\). (b) \(\frac12\lt x\lt 3\).

(a) Factorise the quadratic:

\(x^2-x-6=(x-3)(x+2).\)

The critical values are

\(x=-2\quad\text{and}\quad x=3.\)

Since the coefficient of \(x^2\) is positive, the parabola opens upwards. Therefore the expression is non-negative outside the two roots, including the roots themselves.

Hence

\(x\leqslant-2\quad\text{or}\quad x\geqslant3.\)

(b) For

\(|3x-4|\lt x+2,\)

the right-hand side must be positive, and the inequality is equivalent to

\(-(x+2)\lt3x-4\lt x+2.\)

From the left inequality,

\(-x-2\lt3x-4.\)

Adding \(x+4\) to both sides gives

\(2\lt4x,\)

so

\(x\gt\dfrac12.\)

From the right inequality,

\(3x-4\lt x+2.\)

Therefore

\(2x\lt6,\)

so

\(x\lt3.\)

Combining the two inequalities gives

\(\dfrac12\lt x\lt3.\)