representation on number line

Correct answer: C. \\(x=1\\)

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.

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

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.

Correct answer: A. \((1,\infty)\)

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 \(\leq\).

Correct answer: A. [0, ∞)

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.

Correct answer: A. x ≤ 2

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.

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

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 \(-5\).

Correct answer: A. \(( -\infty, 8]\)

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 \(\geq\).

Correct answer: A. (−∞, −1) ∪ [4, ∞)

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.

Correct answer: A. open circle at (x=-2) with right shading

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.

Correct answer: A. {−2, −1, 0, 1, 2}

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.

Correct answer: A. [-3,4)

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.

Correct answer: A. x ≥ -2

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.

Correct answer: A. Open point at 5 and shading to the right

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 \(\leq\).

Correct answer: A. 5

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 \(<\).

Correct answer: B. Closed circle at \(-2\) and shading to the right

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.

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

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.

Correct answer: B. \((-\infty,3]\)

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.

Correct answer: B. (-4\le x<3)

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.

Correct answer: B. ([-3,5))

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

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

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 ∪.

Correct answer: B. Closed circle at 5 with shading to the right

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 \(\le\).

Correct answer: A. \(( -\infty,7)\)

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 \(\geq\).

Correct answer: B. ((-\infty,-5)\cup(1,\infty))

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

Correct answer: B. (x\le -7) or (x\ge 3)

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

Correct answer: A. ([-5,13))

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.