representation on number line

(|2x+5|>9) gives (2x+5<-9) or (2x+5>9), so (x<-7) or (x>2). In exams, modulus with (>) gives open outer parts.

The common part is from (-2) to (3), and both endpoints are included. In exams, take only the common shaded region for (\cap).

The open dot excludes (-9), and the closed dot includes (-1). In exams, use a round bracket for an open endpoint.

For \(5x-8\ge 2x+10\), add \(2x\) to both sides to get \(3x-8\ge 10\). Adding \(8\) gives \(3x\ge 18\), so \(x\ge 6\). Hence, mark a closed dot at \(6\), since it is included, and shade to the right. Option A has the correct direction but an open dot, which excludes \(6\). Exam tip: use a closed dot for \(\ge\) or \(\le\), and an open dot for \(>\) or \(<\).

The first part is open before (-4), and the second starts from (3) inclusive. In exams, read (\cup) as or.

Multiplying all parts by (3) gives (-6\le x+5<12), so (-11\le x<7). In exams, apply the same operation to every part of a compound inequality.

Integers greater than (-\frac{5}{2}) start at (-2), and the last integer up to (\frac{13}{3}) is (4). In exams, choose valid integers at fractional boundaries.

The given interval includes (-6) and excludes (2), so the complement excludes (-6) and includes (2). In exams, check endpoints carefully while taking complements.

The direct answer is B: \((-5,11)\). The rule \(|u|<8\) is equivalent to \(-8<u<8\). Put u=x-3: \(-8<x-3<8\). Add 3 to all three parts, giving \(-5<x<11\). The endpoints are open because equality is not allowed: x=-5 gives absolute value 8, and x=11 also gives absolute value 8. Option A, \([-5,11]\), incorrectly includes both endpoints. Option B correctly shows the open interval between them. Option C shows the two outside rays, which would describe values with distance greater than 8. Option D shows outside rays including endpoints, which would describe distance at least 8. The correct idea is that a strict absolute-value inequality gives the inside region. Memory cue: less than means inside and open; greater than means outside.

(-9\le4x+1\le9) gives (-10\le4x\le8), so (-\frac{5}{2}\le x\le2). In exams, modulus with (\le) gives a closed interval.

The common part is from (-2) to (3), and both endpoints are excluded. In exams, take only the overlapping region for (\cap).

(8-6x-3>-7) gives (5-6x>-7), so (x<2). In exams, reverse the sign when dividing by a negative coefficient.

The first condition gives (x<5), and the second gives (x\ge-4), so the common solution is ([-4,5)). In exams, and means the common shaded part.

The first part includes (-3), and the second starts after (-3), so no gap remains. In exams, merge connected intervals.

The closed dot includes (5), and the open dot excludes (14). In exams, decide (\le) or (<) from the endpoint.

The removed part does not remove (2), but it removes (7). In exams, pay attention to the endpoints removed in set difference.

(x^2\ge64) gives (|x|\ge8), so two closed outer rays are obtained. In exams, include boundary points when the sign is (\ge).

The natural numbers are (4,5,6,7,8,9,10), so there are (7) points. In exams, do not count the strict upper boundary.

No number can be less than or equal to (7) and greater than (7) at the same time. In exams, contradictory conditions have an empty intersection.

For (x\ne9), the whole number line is taken except (9). In exams, show the removed point with an open circle.

The first inequality gives (-2\le x\le4), and the second gives (x>0), so the common part is ((0,4]). In exams, apply the extra condition as an intersection.

The first interval includes (-5), and the second starts after (-5), so there is no gap. In exams, merge touching intervals.

(|x+2|\le5) gives (-7\le x\le3), so there are (11) integers. In exams, count endpoints in a closed interval.

In (x\le-2), (-2) is included, and in (4<x<10), both endpoints are excluded. In exams, write or using union.

(11-3x\ge-10) gives (-3x\ge-21), so (x\le7). In exams, reverse the inequality while dividing by a negative.