representation on number line

The square bracket at −2 means that endpoint is included, so x can equal −2. The round bracket at 5 means that endpoint is excluded. Thus the interval is represented by −2 ≤ x < 5, making option A correct.

For option A, adding 6 gives 3x ≤ 6, and dividing by 3 gives x ≤ 2. This is exactly the interval (−∞, 2], which includes 2.

This interval contains all real numbers less than −6 and includes −6 itself, as shown by the square bracket. Therefore, it is written as x ≤ −6.

In the open interval \((2,7)\), neither endpoint 2 nor endpoint 7 is included, so \(2<x<7\) is correct. Option A includes both endpoints, option C includes 2, and option D includes 7. Exam tip: round brackets indicate that the corresponding endpoint is excluded.

The interval for \(x\leq -4\) is \(\left(-\infty,-4\right]\) because \(-4\) is included. The interval for \(x>1\) is \(\left(1,\infty\right)\) because \(1\) is excluded. The word “or” means the union of these two intervals, giving \(\left(-\infty,-4\right]\cup\left(1,\infty\right)\). Option B reverses the inclusion status of both finite endpoints. Exam tip: use square brackets for \(\leq\) or \(\geq\), and parentheses for strict inequalities.

In the interval \([0,6)\), 0 is included because it has a square bracket, whereas 6 is excluded because it has a round bracket. Both 5 and \(\frac{1}{2}\) lie between 0 and 6, so they are included. Exam tip: a square bracket means included, while a round bracket means excluded.

(x\le -3) includes (-3), while (x>1) does not include (1). In exams, remember closed dot for (\le) and open dot for (>).

A round bracket means open endpoint and a square bracket means closed endpoint. In exams, use this rule while converting interval notation to inequality.

From (-3x<-6), dividing by a negative reverses the sign. In exams, reverse the inequality when dividing by a negative.

A single linear inequality in one variable, such as \(ax+b<0\) with \(a\ne0\), gives a ray of the form \(x<c\), \(x\le c\), \(x>c\), or \(x\ge c\). Option C, \(-1<x\le4\), is a bounded interval with two finite endpoints. It requires both conditions \(x>-1\) and \(x\le4\) simultaneously. Options A, B, and D are each single-boundary rays. Exam tip: A solution set with two finite endpoints usually requires more than one inequality.

Both endpoints are open, so (3) and (8) are not included. In exams, write open circles with round brackets.

(|x-2|\le 5) gives (-5\le x-2\le 5), hence (-3\le x\le 7). In exams, absolute value with (\le) gives a closed middle interval.

(|x+1|>4) gives two outer regions, and endpoints are not included. In exams, absolute value with (>) is shaded outward.

Multiplying by (2) gives (x-1\le 6), so (x\le 7). In exams, multiplying by a positive number does not reverse the sign.

Solving gives (5-x\ge 6), then (-x\ge 1), so (x\le -1). In exams, remember to reverse the sign when removing (-x).

In intersection, the common part of both intervals is ([-4,0]). In exams, look for the overlapping region for (\cap).

The open dot excludes (-2), and the closed dot includes (6). In exams, read endpoint symbols carefully.

(4-2x\le 10) gives (-2x\le 6), so (x\ge -3). In exams, reverse the sign when dividing by a negative.

The first part is before (-1), and the second part is closed from (4) to (9). In exams, read separate intervals separately.

The direct answer is option A: the interval is [-1,3). Start with -2≤3x+1<10. Subtract 1 from all three parts: -3≤3x<9. Divide all parts by positive 3, so the inequality signs do not change: -1≤x<3. The left endpoint is included because equality is allowed at -1, so it is shown with a closed point or square bracket. The right endpoint is excluded because the sign is <, so it is shown with an open point or round bracket. Option A matches exactly. Option B reverses both endpoint inclusions and is wrong. Option C uses the original bounds -2 and 10 without solving for x. Option D describes outside the interval and also uses incorrect endpoint directions. Exam cue: for a positive divisor, keep the signs; use square brackets for ≤ and round brackets for <.

(2x<7) gives (x<\frac{7}{2}), so the greatest integer is (3). In exams, choose the integer just before the boundary.

(6\le 2x\le 14) gives (3\le x\le 7). In exams, if both signs are inclusive, the final interval stays closed.

The complement of ([2,6)) excludes (2) but includes (6). In exams, check endpoints carefully when taking complements.

(x^2\ge 16) means (|x|\ge 4), giving two closed outer regions. In exams, include (-4) and (4) for (\ge).

The direct answer is option B, x≠2. The union (-∞,2)∪(2,∞) contains every real number less than 2 and every real number greater than 2. The number 2 itself is excluded because neither interval includes its endpoint. That is exactly the meaning of x not equal to 2. Option A, x=2, represents only the single point 2, which is the opposite of this interval. Option C, x≤2, includes all numbers below 2 and also includes 2, but it excludes numbers above 2. Option D, x≥2, includes 2 and numbers above it, but excludes numbers below 2. The open-looking gap at 2 is the important clue. Memory cue: two rays extending in both directions with only one point missing mean ‘all real numbers except that point’.