representation on number line

The inequality \\(x>1\\) requires \\(x\\) to be strictly greater than 1, so \\(x=1\\) is not allowed. The value \\(x=6\\) is allowed because the condition is \\(x\\le 6\\), and 2 and 5 also lie within the permitted range. Exam tip: An open inequality sign, \\(>\\) or \\(<\\), excludes the boundary, whereas \\(\\le\\) or \\(\\ge\\) includes it.

The inequality \(x\geq -2\) means that \(x\) can be equal to \(-2\) or any greater value. Thus, \(-2\) is included, so a square bracket is used, and the values extend to the right up to infinity. Therefore, the correct interval is \([-2,\infty)\). Exam tip: Use a bracket when the endpoint is included, as with \(\geq\) or \(\leq\); use a parenthesis when it is excluded.

The inequality \(x>1\) includes all real numbers greater than 1 but excludes 1 itself. Therefore, an open parenthesis is used at 1; infinity always takes an open parenthesis, giving \((1,\infty)\). Option B includes 1 and represents \(x\geq1\), so it is incorrect. Exam tip: use open parentheses for \(>\) or \(<\), and closed parentheses for \(\geq\) or \(\leq\).

The inequality x ≥ 0 includes 0 and every positive real number. Therefore, the solution set is [0, ∞); the square bracket shows that 0 is included.

The interval (−∞, 2] contains all real numbers less than or equal to 2. Its square bracket means that 2 is included, so the equivalent inequality is x ≤ 2.

The inequality \(x>-5\) includes all numbers greater than \(-5\), but not \(-5\) itself. Therefore, the interval is \((-5,\infty)\), with an open endpoint at \(-5\).

The inequality \(x\leq 8\) includes 8, so the endpoint 8 must have a closed bracket, \(]\). It also includes every real number less than 8, extending left toward \(-\infty\). Therefore, the interval is \(( -\infty, 8]\). Option B is the closest distractor, but it excludes 8. Exam tip: use a closed bracket when an endpoint is included by \(\leq\) or \(\geq\).

The word “or” means that either inequality may be satisfied, so the solution set is the union of two separate intervals. For x < −1, the endpoint −1 is excluded because the inequality is strict, giving (−∞, −1). Infinity is always written with a parenthesis because it is not a real endpoint. For x ≥ 4, the endpoint 4 is included, giving [4, ∞). Combining them produces (−∞, −1) ∪ [4, ∞), option A. Option B incorrectly includes −1 and excludes 4. Options C and D describe values between the two boundary points rather than values outside them. Thus A correctly represents both the logical connector and endpoint inclusion.

First solve the inequality algebraically. From 5-2x<9, subtracting 5 gives -2x<4. Dividing by -2 reverses the inequality sign, so x>-2. On a number line, a strict inequality excludes its boundary, so -2 is shown with an open circle. Values greater than -2 lie to the right, and that region must be shaded. Therefore option A has both required features: an open circle at -2 and rightward shading. Option B has the wrong direction, while options C and D use closed circles even though equality is not allowed.

Because x belongs to the integers, list only whole-number values, not every real number in the interval. The inequality −3 < x excludes −3, while x ≤ 2 includes 2. The integers strictly greater than −3 and at most 2 are −2, −1, 0, 1, and 2. Hence the solution set is {−2, −1, 0, 1, 2}, option A. Option B incorrectly includes the excluded endpoint −3. Option C omits the included endpoint 2, and option D includes −3 while omitting 2. A useful check is to test both boundary values separately and then list each consecutive integer between them. Therefore A is the complete and correct integer solution set.

Subtracting 5 from all three parts gives \(2-5\le x+5-5<9-5\), so \(-3\le x<4\). Thus, -3 is included and is shown with a square bracket, while 4 is excluded and is shown with a round bracket. Therefore, the correct interval is \([-3,4)\). The options \((-3,4]\), \((-3,4)\), and \([-3,4]\) use incorrect endpoint inclusion. Exam tip: \(\le\) or \(\ge\) gives a closed endpoint, whereas \(<\) or \(>\) gives an open endpoint.

The governing concept is the interpretation of endpoints and direction on a number line. A closed or filled point means that the endpoint itself is included in the solution set, so the relation must contain equality. Shading to the right of -2 represents numbers greater than -2. Combining both observations gives x ≥ -2. For example, -2, 0, and 5 satisfy this inequality, whereas -3 does not. Option B has the correct direction but uses an open endpoint and therefore excludes -2. Options C and D describe the left side of the number line, so they do not match the diagram. Therefore, option A is unambiguously correct.

Subtracting 4 from both sides gives \(x+4>9\Rightarrow x>5\). Thus, 5 is not included, so an open point is placed at 5. All numbers greater than 5 lie to the right, so the shading is directed to the right. Option B is incorrect because a closed point represents inclusion, as in \(x\geq5\), not the strict inequality \(x>5\). Exam tip: use an open point for \(>\) or \(<\), and a closed point for \(\geq\) or \(\leq\).

The interval includes \(-1\) because of \(\leq\), but excludes \(4\) because of \(<\). Thus the integer solutions are \(-1,0,1,2,3\), giving a total of \(5\) solutions. Exam tip: include an endpoint marked by \(\leq\) and exclude one marked by \(<\).

Subtracting 4 from both sides of \(3x+4\ge -2\) gives \(3x\ge -6\). Dividing by 3 gives \(x\ge -2\). Since \(\ge\) includes \(-2\), the point \(-2\) is shown with a closed circle, and values greater than \(-2\) are shaded to the right. An open circle is used for strict inequalities such as \(>\) or \(<\). Exam tip: First write the inequality in terms of \(x\), then use the sign to choose the circle and direction.

The direct answer is option A, [-2,∞). Solve step by step: 4-3x≤10. Subtract 4 from both sides to get -3x≤6. Now divide by -3. When an inequality is divided by a negative number, its sign reverses, so x≥-2. The solution includes -2 and all larger numbers, which is written [-2,∞). Option A is correct because the square bracket includes -2 and the interval extends rightward. Option B, (-∞,-2], represents x≤-2, the opposite direction. Option C, (-2,∞), points right but excludes -2, which is wrong because equality is allowed. Option D, (-∞,-2), points left and excludes -2, so it is doubly incorrect. On the number line, use a filled dot at -2 and shade right. The key mistake is forgetting to reverse the sign when dividing by -3.

Adding 1 to both sides of \(5x-1\le14\) gives \(5x\le15\). Dividing by 5 gives \(x\le3\). Hence, all real numbers less than or equal to 3 are included, represented by \((-\infty,3]\). Option A excludes 3, so it represents \(x<3\). Exam tip: Use a square bracket at an endpoint when \(\le\) or \(\ge\) includes that endpoint.

The direct answer is option B: \(-4\le x<3\). In interval notation, a square bracket means the endpoint is included, while a round parenthesis means it is excluded. The interval \([-4,3)\) therefore includes \(-4\) but does not include 3. Translating this into inequalities gives \(-4\le x<3\). Option A, \(-4<x<3\), excludes \(-4\), so it does not match the square bracket. Option B includes \(-4\) with \(\le\) and excludes 3 with \(<\), exactly matching the notation. Option C, \(-4<x\le3\), reverses both endpoint conditions. Option D, \(-4\le x\le3\), wrongly includes 3. To check the meaning, test the endpoints: \(x=-4\) belongs, but \(x=3\) does not. Memory cue: read the left bracket first and the right bracket second; square means included, round means excluded.

Dividing all parts by (2) gives (-3\le x<5). Hence the left endpoint is closed and the right endpoint is open.

The direct answer is option A, (-∞,-2]∪(4,∞). The word ‘or’ means that either separate condition may be true, so the solution sets are joined with union, ∪. First, x≤-2 includes -2 because equality is allowed; its interval is (-∞,-2]. Second, x>4 excludes 4 because the inequality is strict; its interval is (4,∞). Option A correctly combines both parts. Option B excludes -2 and includes 4, reversing both endpoint rules. Option C, [-2,4], contains the middle section and does not represent either outer condition. Option D, (-2,4), also shows only the middle section and excludes both endpoints, so it is incorrect. On the number line, place a filled dot at -2 and shade left; place an open circle at 4 and shade right. Memory cue: ‘or’ usually creates separate regions joined by ∪.

In \(\frac{x-2}{3}\ge 1\), 3 is positive, so multiplying both sides by 3 does not reverse the inequality: \(x-2\ge 3\). Thus, \(x\ge 5\). Since 5 is included in the solution, the number line has a closed circle at 5 and shading to the right for values greater than 5. Option A would represent \(x>5\) because it uses an open circle. Exam tip: Use a closed circle for \(\ge\) or \(\le\).

In \(\frac{2x+1}{5}<3\), 5 is positive, so multiplying both sides by 5 does not reverse the inequality: \(2x+1<15\). Thus, \(2x<14\), giving \(x<7\). Since the inequality is strict, 7 is not included; the correct interval is \(( -\infty,7)\). The interval \(( -\infty,7]\) is incorrect because it includes 7. Exam tip: use a round bracket for \(<\) or \(>\), and a square bracket for \(\leq\) or \(\geq\).

Outward shading shows two separate regions and open circles exclude the endpoints. Hence ((-\infty,-5)\cup(1,\infty)) is correct.

Closed circles include both endpoints and outward shading shows two outside regions. So inclusive inequalities with or are used.

The direct answer is option A: \([-5,13)\). Start with \(-2\le\frac{x-1}{3}<4\). Since 3 is positive, multiply every part by 3 without changing either inequality sign: \(-6\le x-1<12\). Add 1 to all three parts: \(-5\le x<13\). The left inequality is non-strict, so \(-5\) is included and receives a square bracket. The right inequality is strict, so 13 is excluded and receives a round parenthesis. Therefore the interval is \([-5,13)\). Option A matches exactly. Option B, \((-5,13]\), reverses both endpoint inclusions. Option C, \([-7,11)\), results from incorrect arithmetic and does not follow from multiplying and adding. Option D, \((-7,11]\), has both wrong values and wrong endpoint types. A quick check: at \(x=-5\), the fraction is \(-2\), so the left boundary is included; at \(x=13\), it is 4, but the right inequality is strict, so 13 is excluded. Remember to multiply the entire chained inequality and preserve the signs when the multiplier is positive.