Solve the compound inequality 1/6 < (2x - 13)/12 <= 2/3.

Step-by-step solution

1. 1

   Treat it as two inequalities joined at the hip. The unknown appears only in the middle, so every operation must be applied to all three parts at once. That keeps the chain equivalent without ever splitting it up.

2. 2

   Multiply through by 12 to clear all denominators. $12$ is the least common multiple of $6$, $12$ and $3$, and it is positive, so the inequality directions are unchanged: $12\cdot\frac16 < 12\cdot\frac{2x-13}{12} \le 12\cdot\frac23 \Longrightarrow 2 < 2x-13 \le 8.$

3. 3

   Add 13 to every part. $2+13 < 2x \le 8+13 \Longrightarrow 15 < 2x \le 21.$ Adding a constant never affects the direction of an inequality.

4. 4

   Divide every part by 2. Since $2>0$ the signs stay as they are: $\frac{15}{2} < x \le \frac{21}{2}, \qquad\text{i.e.}\qquad 7.5 < x \le 10.5.$

5. 5

   Keep the endpoint types straight. The left end came from a strict $<$, so $7.5$ is excluded; the right end came from $\le$, so $10.5$ is included. In interval notation that is $\left(\tfrac{15}{2},\tfrac{21}{2}\right]$.

6. 6

   Check the boundaries numerically. At $x=7.5$: $\tfrac{2(7.5)-13}{12} = \tfrac{2}{12} = \tfrac16$ - equal, so correctly excluded. At $x=10.5$: $\tfrac{2(10.5)-13}{12} = \tfrac{8}{12} = \tfrac23$ - equal, so correctly included. At $x=7.4$ the middle value is $0.15 < \tfrac16$, outside; at $x=10.6$ it is $0.6833 > \tfrac23$, also outside.