representation on number line

The square bracket includes \(\frac{1}{2}\) and the round bracket excludes \(4\). In exams convert brackets directly into endpoints.

Solving gives (x>4) so (4) is not included. In exams first find the boundary and then decide the direction.

The value (-6) is included and (-1) is excluded so (-6\le x<-1) is correct. In exams read negative endpoints in number line order.

Multiplying by (5) and subtracting (3) gives (7\le x<12). In exams positive multiplication does not reverse the inequality sign.

For or take the union of both separate parts. In exams keep the (\le) endpoint closed and the (>) endpoint open.

Dividing by (-3) reverses both signs and gives (-2\le x<5). In exams fix the endpoint order after negative division.

Both open dots show that both endpoints are excluded. In exams write an open-open segment using round brackets.

The value (\frac{5}{3}) is included and all smaller values are taken. In exams understand ((-\infty,a]) as (x\le a).

The value (-2) is excluded but (3) is included so the integers are from (-1) to (3). In exams choose integer points according to open and closed endpoints.

Solve the inequality step by step while preserving the inequality direction. Starting with 2x − 3 ≥ 7, add 3 to both sides to obtain 2x ≥ 10. Dividing by the positive number 2 gives x ≥ 5; the sign does not reverse because the divisor is positive. Since equality is included by ≥, the value 5 belongs to the solution set and must be shown with a closed dot. Values greater than 5 lie to the right, so that side is shaded. Therefore option A is correct. An open dot would represent a strict inequality, and shading left would represent x ≤ 5 or x < 5.

Since 9 = 3², the inequality x² < 9 becomes x² < 3². For real numbers, this is equivalent to |x| < 3, which means −3 < x < 3. Every number strictly inside this interval has a square less than 9. The endpoints must be excluded: when x = −3 or x = 3, x² = 9, and equality does not satisfy the strict sign <. On a number line, both endpoints are therefore shown with open circles, with the segment between them selected. Values outside the interval have absolute value greater than 3 and square greater than 9. Thus option A, (−3, 3), is correct; option B incorrectly includes both endpoints.

Expand the brackets carefully: 3(x − 2) ≤ 2(x + 5) gives 3x − 6 ≤ 2x + 10. Subtracting 2x from both sides gives x − 6 ≤ 10, and adding 6 gives x ≤ 16. The symbol ≤ includes equality, so x = 16 belongs to the solution set and must be marked by a closed dot. All values smaller than 16 lie to the left of 16, so that side of the number line is shaded. Therefore option A gives the correct algebraic result and graphical representation. Option B reverses the solution direction, while options C and D use open dots, which would incorrectly exclude 16.

To solve the inequality, first multiply both sides by 3. Since 3 is positive, the inequality sign remains unchanged: 2x − 7 > 15. Adding 7 gives 2x > 22, and dividing by the positive number 2 gives x > 11. The strict sign > means that x = 11 is not included, so 11 must be marked with an open circle. Values greater than 11 lie to the right of 11 on the number line, so that ray is shaded to the right. A closed point would represent ≥ or ≤ and is therefore incorrect. Shading left would represent x < 11 or x ≤ 11. Hence option A correctly combines the algebraic solution and its number-line representation.

Multiply both sides of (4x − 9)/5 > 7 by 5. Because 5 is positive, the direction of the inequality remains the same: 4x − 9 > 35. Add 9 to both sides to obtain 4x > 44, and divide by the positive number 4 to get x > 11. The strict inequality excludes x = 11, so the number line must show an open circle at 11. Since all values greater than 11 are located to the right, the ray extending rightward is shaded. A closed circle would incorrectly include 11 and would correspond to ≥. Shading to the left would represent a value less than 11. Therefore the complete and correct representation is given by option A.

The governing concept is solving a compound linear inequality while preserving endpoint inclusion. Start with -3≤2x-1<9. Add 1 to all three parts, giving -2≤2x<10. Since division by the positive number 2 does not reverse either inequality sign, divide throughout by 2 to obtain -1≤x<5. The left endpoint is included, so it receives a square bracket; the right endpoint is excluded because the inequality is strict there, so it receives a parenthesis. Thus the interval is [-1,5), making option A correct. Options B reverses the endpoint inclusion, while C and D use incorrect boundary values.

The governing concept is solving a linear inequality and translating its solution to a number line. Expand both sides: 3(x-2)=3x-6 and 2(x+5)=2x+10. Thus 3x-6≥2x+10. Subtract 2x from both sides and add 6, giving x≥16. Equality is allowed, so the point 16 must be shown with a closed dot. Values greater than 16 satisfy the inequality, so the ray is shaded to the right. Therefore option A is correct. Option B incorrectly excludes 16, while options C and D shade the wrong side and represent values less than 16.

The governing concept is converting an absolute-value inequality into a double inequality. For |A|<5, the equivalent condition is -5<A<5. Substituting A=2x-3 gives -5<2x-3<5. Add 3 throughout to obtain -2<2x<8, and divide by the positive number 2 to get -1<x<4. Both inequalities are strict, so neither endpoint is included; the interval is therefore (-1,4). Hence option D is correct. Options A and C incorrectly use closed endpoints and also have wrong limits. Option B has the endpoints in the wrong positions and does not result from the required algebra.

The governing concept is solving a fractional linear inequality while checking the sign of the denominator. Since 5 is positive, multiplying both sides by 5 preserves the inequality direction: 2x-3<5. Add 3 to obtain 2x<8, then divide by the positive number 2 to get x<4. The strict sign means 4 is not included, so the number line has an open point at 4. All values smaller than 4 satisfy the condition, so shading extends to the left. Therefore option A is correct. Option B wrongly includes 4, while options C and D reverse the solution direction and describe values greater than 4.

The governing concept is intersection, which keeps only values common to both sets. The interval (-∞,-1] contains all numbers less than or equal to -1. The interval [-1,6) contains all numbers from -1 through values less than 6, including -1. The only common value is therefore -1 itself: numbers below -1 fail the second interval, and numbers above -1 fail the first interval. Since -1 is included in both intervals, it is represented by a closed dot. Thus option B is correct. Option A includes many values greater than -1, option C includes values not in the second interval, and option D incorrectly claims that the intersection is empty.

The inequality |x + 2| > 3 states that the distance of x from −2 is greater than 3. For an absolute-value inequality with a greater-than sign, split it into two cases: x + 2 < −3 or x + 2 > 3. Solving the first inequality gives x < −5, while solving the second gives x > 1. Since the original sign is strict, the boundary values −5 and 1 are excluded; they are represented by open circles on the number line. The solution is therefore the union of two outer intervals, extending left from −5 and right from 1. Hence option D is correct. Options A and B describe the central region, which would correspond to a less-than condition, and option C incorrectly includes the endpoints.

The governing rule is that multiplying or dividing an inequality by a positive number preserves its direction. Starting with (x + 7)/4 ≥ -2, multiply both sides by 4, which is positive: x + 7 ≥ -8. Subtracting 7 from both sides gives x ≥ -15. The equality is included because the sign is ≥, so -15 must be represented by a closed dot. Numbers greater than -15 lie to the right, and that region is shaded. Thus option C is the complete and correct number-line representation. Option A reverses the direction and places the shading on the wrong side. Options B and D use open dots, incorrectly excluding the boundary value -15; D also shades the opposite side.