To solve the inequality \(|2x-1| \leq |x+3|\), we must consider the different cases based on the critical points \(2x-1 = 0\) and \(x+3 = 0\). These yield \(x = \frac{1}{2}\) and \(x = -3\), respectively. These points divide the real number line into intervals which we will consider separately:

1. **Interval 1: \(x < -3\)**

Here, both expressions inside the absolute values are negative. Thus, \(|2x-1| = -(2x-1) = -2x+1\) and \(|x+3| = -(x+3) = -x-3\). The inequality becomes:
\[
-2x + 1 \leq -x - 3
\]
Solving this inequality gives:
\[
-2x + 1 \leq -x - 3 \quad \Rightarrow \quad -2x + x \leq -3 - 1 \quad \Rightarrow \quad -x \leq -4 \quad \Rightarrow \quad x \geq 4
\]
This is a contradiction since \(x < -3\) in this interval. Therefore, there are no solutions in this interval.

2. **Interval 2: \(-3 \leq x \leq \frac{1}{2}\)**

In this interval, \(x+3\) is non-negative, and \(2x-1\) changes sign at \(x = \frac{1}{2}\). We need to separate further into two parts:

- **Sub-interval 2a: \(-3 \leq x \leq \frac{1}{2}\)**

For \(x < \frac{1}{2}\), \(|2x-1| = -(2x-1) = -2x+1\), \(|x+3| = x+3\). Consider the inequality:
\[
-2x + 1 \leq x + 3
\]
Solving this gives:
\[
-2x + 1 \leq x + 3 \quad \Rightarrow \quad -2x - x \leq 3 - 1 \quad \Rightarrow \quad -3x \leq 2 \quad \Rightarrow \quad x \geq -\frac{2}{3}
\]
Therefore, in this sub-interval, the solution is \(-\frac{2}{3} \leq x \leq \frac{1}{2}\).

- **Sub-interval 2b: \(x = \frac{1}{2}\)**

At this point, \(|2x-1| = 0\) and \(|x+3| = x+3 = \frac{1}{2} + 3 = \frac{7}{2}\). Clearly, \(0 \leq \frac{7}{2}\) holds true.

3. **Interval 3: \(x > \frac{1}{2}\)**

Here, \(2x-1\) is positive and \(x+3\) remains positive, so \(|2x-1| = 2x-1\) and \(|x+3| = x+3\). The inequality becomes:
\[
2x - 1 \leq x + 3
\]
Solving this gives:
\[
2x - 1 \leq x + 3 \quad \Rightarrow \quad 2x - x \leq 3 + 1 \quad \Rightarrow \quad x \leq 4
\]
Thus, in this interval, the solution is \(\frac{1}{2} < x \leq 4\).

Combining all valid intervals, the solution to the inequality \(|2x-1| \leq |x+3|\) is \(-\frac{2}{3} \leq x \leq 4\).