The solution set of inequalities :
\(x + 3 \le 0\) and \(2x + 5 \le 0\); if \(x \in \mathrm{R}\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \((-\infty, -3]\)
Each inequality on its own gives
\(x + 3 \le 0 \Rightarrow x \le -3\)
\( \displaystyle 2x + 5 \le 0 \Rightarrow x \le -\frac{5}{2} \)
Both must hold, so x must satisfy the stronger condition \(x \le -3\) (every \(x \le -3\) is also \(\le -\frac{5}{2}\)). Since x is a real number, the solution set is the interval \((-\infty, -3]\), with −3 included.
Why the other options are incorrect
Option (1) \((-\infty, -3)\) — Incorrect. It leaves out \(x = -3\), which satisfies both inequalities with equality (\(-3 + 3 = 0\) and \(-6 + 5 = -1\)).
Option (2) \(\{\ldots\ldots -5, -4, -3\}\) — Incorrect. It lists only integers, but x is any real number, so values such as −3.5 are also solutions.
Option (4) \(\{-3, -2, -1, 0, \ldots\ldots\}\) — Incorrect. These values go the wrong way: \(x = 0\) gives \(0 + 3 = 3\), which is not \(\le 0\).