Solve the inequalities $x \le \dfrac{x+4}{3} \le 2x-1$. Answer ...

1. Split the compound inequality into two inequalities.

$x \leqslant \frac{x+4}{3} \leqslant 2x-1$
This means
$x \leqslant \frac{x+4}{3} \quad \text{and} \quad \frac{x+4}{3} \leqslant 2x-1.$

2. Solve the left inequality.

$x \leqslant \frac{x+4}{3}$
Multiply by $3$:
$3x \leqslant x+4$
$2x \leqslant 4$
$x \leqslant 2.$

3. Solve the right inequality and combine both conditions.

$\frac{x+4}{3} \leqslant 2x-1$
Multiply by $3$:
$x+4 \leqslant 6x-3$
$7 \leqslant 5x$
$x \geqslant \frac{7}{5}.$
Combining with $x \leqslant 2$ gives
$\frac{7}{5} \leqslant x \leqslant 2.$