jmc

Clearly, the inequality has some solutions for which $2x+7$ is negative. For example, if $x=-4$, then $2x+7=-1$, so $|2x+7|=1$, which is less than 16. As we make $x$ even smaller, $2x+7$ gets even more less than zero, so $|2x+7|$ gets larger. But how small can we make $x$? To figure this out, we note that if $2x+7$ is negative, then $|2x+7|=-(2x+7)$. Then, our inequality becomes $-(2x+7)\le 16$. Multiplying both sides by $-1$ (and flipping the direction of the inequality symbol) gives $2x+7\ge -16$. Subtracting 7 and then dividing by 2 gives $x\ge -11.5$. So, the smallest possible integer value for $x$ is $\boxed{-11}$. Checking, we see that when $x=-11$, we have $|2x+7|=15$, which is less than 16.