What is the least integer value of $x$ such that  $\lvert2x+ 7\rvert\le 16$?
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.