representation on number line

Correct answer: A. ((-\infty,-4]\cup(5,\infty))

(\le -4) gives a closed endpoint, and (>5) gives an open endpoint. In exams, always use round brackets on the infinite side.

Correct answer: A. (x<2), open dot at (2) shaded left

The direct answer is option A: x<2, with an open point at 2 and shading to the left. Solve 7-3x>1. Subtract 7 from both sides: -3x>-6. Divide by -3; division by a negative number reverses the inequality, giving x<2. Since 2 itself is not allowed, the point at 2 is open. Values smaller than 2, such as 0, satisfy the inequality, so the shading goes left. Option A shows exactly this. Option B has the wrong direction because it says x>2. Option C includes 2, although strict > does not permit equality. Option D has both the wrong direction and an included endpoint. The key rule is that multiplying or dividing an inequality by a negative number reverses its sign.

Correct answer: A. ((2,3])

The common part is from (2) to (3), but (2) is excluded and (3) is included. In exams, check endpoint inclusion separately in intersections.

Correct answer: B. (x\le -7)

A closed ray includes the endpoint, and a left ray means less than or equal. In exams, check both ray direction and endpoint.

Correct answer: A. (x>-4), open dot at (-4) shaded right

Multiplying by (-3) reverses the inequality to (2x+5>-3), so (x>-4). In exams, reverse the sign when removing a negative denominator.

Correct answer: A. ((-\infty,0)\cup[2,5))

In (x<0), (0) is open; in (2\le x<5), (2) is closed and (5) is open. In exams, convert or into union.

Correct answer: B. ({1,2,3,4,5,6})

Usually in Class (11), (\mathbb{N}={1,2,3,\ldots}), so points from (1) to (6) are marked. In exams, if the domain is natural numbers, mark only natural points.

Correct answer: A. ([4,10))

The direct answer is option A: [4,10). Begin with 0≤(x-4)/3<2. Multiply every part by positive 3; the signs remain unchanged: 0≤x-4<6. Add 4 to all parts: 4≤x<10. Thus 4 is included and receives a closed point or square bracket, while 10 is excluded and receives an open point or round bracket. Option A is exactly [4,10). Option B incorrectly excludes 4 and includes 10, reversing both endpoint rules. Option C gives the bounds from the unsolved middle expression rather than the x-values. Option D represents values outside the interval and is unrelated to the solution set. A useful check is to test x=4: it gives 0, so it must be included; x=10 gives 2, but the upper sign is strict, so it must be excluded.

Correct answer: A. ((-\infty,5))

The two intervals meet at (-2), and (5) is not included. In exams, merge touching intervals when the common endpoint is included.

Correct answer: A. ((-\infty,\infty))

(x<1) and (x\ge 1) together give all real numbers. In exams, the union of complementary rays is the whole number line.

Correct answer: A. (\emptyset)

No (x) can be greater than (3) and less than or equal to (3) at the same time. In exams, the intersection of contradictory conditions is empty.

Correct answer: A. (x>-2), open dot at (-2) shaded right

(2-2x<x+8) gives (-3x<6), so (x>-2). In exams, change the sign while dividing by a negative coefficient.

Correct answer: A. ([-5,1))

The common part is from (-5) to (1), including (-5) and excluding (1). In exams, both conditions must be satisfied in an intersection.

Correct answer: A. (x<-1) or (x>4)

Open dots exclude the endpoints, and outer shading gives two rays. In exams, read outer shading as an or inequality.

Correct answer: A. ([-4,3))

The intervals overlap and together extend from (-4) to (3). In exams, the union of overlapping intervals forms one larger interval.

Correct answer: A. ((-\infty,-1)\cup[2,4])

Removing ([-1,2)) removes (-1) but keeps (2). In exams, check endpoints carefully in set difference.

Correct answer: C. (7)

The points are (-3,-2,-1,0,1,2,3), so there are (7) points. In exams, include both endpoints while counting integers.

Correct answer: A. ((-5,-2])

(4\le -2x<10) gives (-5<x\le -2) after negative division. In exams, endpoints also reverse when dividing a compound inequality by a negative.

Correct answer: A. ((-2,10))

The intervals join at (6) with no gap. In exams, intervals merge when the common endpoint is included.

Correct answer: A. (2\le x<5)

The common part is from (2) to (5), including (2) and excluding (5). In exams, read intersection as a simultaneous condition.

Correct answer: B. (x\le -13), closed dot at (-13) shaded left

Multiplying by (15) gives (3x+6\ge 5x-20), so (x\le 13), not (x\ge 13). In exams, recheck the final sign and arithmetic.

Correct answer: A. (x<11), open dot at (11) shaded left

Multiplying by (4) gives (3x-1<2x+10), so (x<11). In exams, after clearing denominators, a strict sign gives an open endpoint.

Correct answer: A. ([4,10))

Multiplying all parts by (2) gives (14\le 3x+2<32), so (4\le x<10). In exams, apply the same operation to all parts of a compound inequality.

Correct answer: A. ({-4,-3,-2,-1,0,1,2})

Integers greater than (-\frac{9}{2}) start at (-4), and (2) is the last integer up to (\frac{5}{2}). In exams, choose the nearest valid integer at fractional boundaries.

Correct answer: A. ((-3,1])

The common part of the two intervals is from (-3) to (1), excluding (-3) and including (1). In exams, check endpoint inclusion separately in intersections.