Q.Solve $|2x + 1| \ge 5$ for real $x$, and write the solution set using interval notation.

[!TLDR] Use $|x|\ge a \iff x\le -a$ or $x\ge a$ applied to the expression $2x+1$. [!ANSWER] …

The inequality $|2x+1|\ge 5$ has the form $|E|\ge a$ with $E=2x+1$ and $a=5$. By the slack modulus rule, this splits into two separate cases: $2x+1 \le -5$ or $2x+1 \ge 5$. Solving the first: $2x \le -6 \Rightarrow x \le -3$. Solving the second: $2x \ge 4 \Rightarrow x \ge 2$. Since the original condition was an "or" (not "and\ …

Split into the two cases $2x+1\le -5$ and $2x+1\ge 5$ (an "or", not "and", since the modulus inequality is a "g …

Students often treat a $|E|\ge a$ inequality like a compound "and" inequality and try to write a single interval between the two boundary values, which is wrong — a "greater t …