To solve the compound inequality $3(2x+5) \leq 18$ and $2(x-7) < -6$, start by simplifying both sides of each inequality. For the first inequality, expanding the left side gives $6x + 15 \leq 18$. Subtracting $15$ yields $6x \leq 3$, which simplifies to $x \leq \frac{1}{2}$. For the second inequality, expanding gives $2x - 14 < -6$. Adding $14$ results in $2x < 8$, which simplifies to $x < 4$. Graphing each solution shows that the values satisfying both conditions must be in the overlap. The graph of $x \leq \frac{1}{2}$ is shaded to the left with a right bracket at $\frac{1}{2}$, and $x < 4$ is shaded to the left with a right parenthesis at $4$. The intersection of these two graphs is simply the region where $x \leq \frac{1}{2}$, because any number less than or equal to $\frac{1}{2}$ is also automatically less than $4$. In interval notation, this final combined solution is written as $(-\infty, \frac{1}{2}]$.