The given inequality is:

$3 |x| \leq |x - 1| + 5$

To solve this, we need to consider the behavior of the absolute value function. Absolute value functions have different cases based on the sign of the expression inside them, so we will break the problem down into intervals and solve for each case.

Case 1: $x \geq 1$

• In this case, both $|x| = x$ and $|x - 1| = x - 1$.

Substituting these into the inequality: $3x \leq (x - 1) + 5$ Simplifying: $3x \leq x + 4$ Subtract $x$ from both sides: $2x \leq 4$ Divide by 2: $x \leq 2$

In this case, $x \geq 1$ and ( x