What is the least integer value of $x$ such that  $\lvert2x+ 7\rvert\le 16$?
To solve this inequality, I need to get rid of the absolute value sign and find the possible values of $x$ that make the expression inside the absolute value less than or equal to 16. One way to do this is to split the inequality into two cases: one where the expression inside the absolute value is positive or zero, and one where it is negative. If $2x+ 7\ge 0$, then I can just drop the absolute value sign and solve for $x$ as usual: $2x+ 7\le 16$, subtracting 7 from both sides gives $2x\le 9$, and dividing by 2 gives $x\le 4.5$. If $2x+ 7< 0$, then I need to flip the inequality sign and change the sign of the expression inside the absolute value: $-(2x+ 7)\le 16$, multiplying by -1 gives $2x+ 7\ge -16$, subtracting 7 from both sides gives $2x\ge -23$, and dividing by 2 gives $x\ge -11.5$. So the possible values of $x$ are those that satisfy both $x\le 4.5$ and $x\ge -11.5$. The least integer value of $x$ that satisfies this condition is -11, since -11.5 is not an integer. 