representation on number line

The open endpoint excludes (-3), and the closed endpoint includes (5). While reading intervals, notice parentheses and square brackets.

In \(-5<x\le 1\), the value \(-5\) is excluded, while \(1\) is included. Thus, the possible integers are \(-4,-3,-2,-1,0,1\), giving a total of 6 values. A common mistake is to count \(-5\) as well, but it is not allowed because the inequality is strict at that endpoint. Exam tip: an open inequality sign excludes the boundary value, whereas \(\le\) or \(\ge\) includes it.

Including both endpoints, the integers are \(-4,-3,-2,-1,0,1,2,3,4\). Hence, their number is \(4-(-4)+1=9\). Option B would result from omitting one endpoint. Exam tip: the number of integers in the inclusive interval \([a,b]\) is \(b-a+1\).

Subtracting (2) from the whole interval subtracts (2) from the endpoints. The open (2) becomes open (0), and (5) remains closed.

Since \(0<x<2\), and the square function is increasing throughout this positive interval, we get \(0^2<x^2<2^2\), or \(0<x^2<4\). Therefore, the correct interval is \((0,4)\). Option B is incorrect because \(x\) cannot equal either 0 or 2, so neither endpoint is included. Exam tip: after transforming an interval, always check whether its endpoints are open or closed.

The closed dot includes ( -4 ) and the open dot excludes ( 3 ). In exams, identify open and closed endpoints first.

The open dot excludes ( 2 ) and the right arrow shows greater values. Use the arrow direction to decide the inequality sign.

Both sides have ( \le ), so both endpoints are closed. In compound inequalities, check both signs separately.

The open dot at \(-2\) means that \(-2\) is not included, so \(-2 < x\). The closed dot at \(6\) means that \(6\) is included, so \(x \le 6\). Therefore, the combined inequality is \(-2 < x \le 6\). In option B, the inclusion status of both endpoints is reversed. Exam tip: use \(<\) or \(>\) for an open dot and \(\le\) or \(\ge\) for a closed dot.

Solving gives ( x<5 ), so ( 5 ) is open and the shading is leftward. First find the boundary point, then check direction.

Dividing by negative ( -2 ) reverses the sign and gives ( x\le -4 ). Do not forget to reverse the sign when multiplying or dividing by a negative.

The first part excludes ( -3 ), and the second part includes ( 4 ). Separate shaded regions usually indicate an OR statement.

Solving \(\frac{x-1}{3}\ge 2\) gives \(x-1\ge 6\), hence \(x\ge 7\). Therefore, \(7\) is included in the solution, so it is shown by a closed dot, with shading to the right for all values greater than \(7\). Option B has an open dot at \(7\), which would represent \(x>7\). Exam tip: use a closed dot for \(\ge\) or \(\le\), and an open dot for \(>\) or \(<\).

Left shading shows smaller values, and a closed dot shows equality. Read the direction and dot type together.

The round bracket excludes ( -6 ), and the square bracket includes ( 1 ). Connect interval brackets directly with dot types.

From \(4-3x<19\), subtracting 4 from both sides gives \(-3x<15\). Dividing by \(-3\) reverses the inequality, so \(x>-5\). Since the inequality is strict, \(-5\) is shown by an open circle, and values greater than \(-5\) are shaded to the right. The close distractor \(x\geq -5\) is incorrect because it includes \(-5\). Exam tip: Always reverse an inequality sign when multiplying or dividing by a negative number.

Both signs are ( < ), so both boundary values are excluded. A strict inequality always gives an open dot.

A filled point at -1 gives \(x\le -1\), while an open point at 4 gives \(x>4\). The two separate rays are joined by “or,” not “and.” Exam tip: filled endpoints use \(\le\) or \(\ge\); open endpoints use \(<\) or \(>\).

For \(x\le -1\text{ or }x>2\), the number line contains one ray extending left from \(-1\) and another ray extending right from \(2\). The part \((-1,2]\) is excluded, so the solution set is disconnected. A single linear inequality normally gives one half-line on the number line. Options A, B, and C each represent only one ray. Exam tip: When you see “or”, check whether the solution is a union of separate parts.

From \(5x+2\le 3x-6\), subtract \(3x\) from both sides to get \(2x+2\le -6\). Subtracting 2 gives \(2x\le -8\), hence \(x\le -4\). Since \(\le\) includes \(-4\), the number line has a closed point at \(-4\) and is shaded to the left for smaller values. Option B incorrectly excludes \(-4\). Exam tip: use a closed point for \(\le\) or \(\ge\), and an open point for \(<\) or \(>\).

\( x\notin (-2,4) \) means \(x\) lies outside the open interval \((-2,4)\). That includes values less than \(-2\) and values greater than \(4\), and also the endpoints \(-2\) and \(4\) because the interval \((-2,4)\) is open. Hence the set is \( (-\infty,-2]\cup[4,\infty) \).

The closed dot includes ( \frac{3}{2} ), and left shading shows smaller values. Read fractional boundaries like ordinary boundaries.

( 6 ) is included and all smaller values are included. ( -\infty ) indicates direction, not a real endpoint.

No number can be both ( \le -1 ) and ( >3 ) at the same time. For AND, take the intersection of the conditions.

The direct answer is option B: -5/2≤x≤7/2. A closed endpoint means that endpoint is included, and inclusion is written with ≤. Since both endpoints are closed, both inequalities must be non-strict. The shaded interval therefore contains every number from -5/2 through 7/2, including the two endpoints. Option A uses < at both ends, so it excludes both endpoints and would describe open dots. Option B includes both ends and is correct. Option C excludes -5/2 but includes 7/2, so it would have a left open and right closed endpoint. Option D includes -5/2 but excludes 7/2, the reverse pattern. The order is also correct because -5/2 is negative and lies to the left of 7/2. Memory cue: closed dot means ≤; open dot means <.