解不等式 $\displaystyle x - 5 < \frac{2x}{5} \le \frac{x}{2} + \frac{1}{5}$。 答案 ...

1. Separate the compound inequality into two inequalities.

$x-5<\frac{2x}{5}\quad \text{and}\quad \frac{2x}{5}\leqslant \frac{x}{2}+\frac{1}{5}$

2. Solve the left inequality to find the upper bound for $x$.

$x-5<\frac{2x}{5}$
Multiply by $5$:
$5x-25<2x$
$3x<25$
$x<\frac{25}{3}$

3. Solve the right inequality to find the lower bound for $x$.

$\frac{2x}{5}\leqslant \frac{x}{2}+\frac{1}{5}$
Multiply by $10$:
$4x\leqslant 5x+2$
$-x\leqslant 2$
Divide by $-1$, reversing the inequality sign:
$x\geqslant -2$

4. Combine the two solution ranges.

Since both inequalities must be true,
$x\geqslant -2 \quad \text{and}\quad x<\frac{25}{3}$
Hence,
$-2\leqslant x<\frac{25}{3}$