representation on number line

The direct answer is option B: \([0,8]\). The common part means the numbers that belong to both intervals at the same time. The first interval, \((-4,8]\), contains numbers greater than -4 and up to and including 8. The second interval, \([0,10)\), contains numbers from 0 including 0 to numbers less than 10. The overlap begins at 0 because the second interval includes 0 and 0 is also inside the first interval. The overlap ends at 8 because the first interval includes 8 and 8 is also inside the second interval. Thus the intersection is \([0,8]\). Option A, \((-4,10)\), is too large because it includes numbers such as -3 and 9 that are not in both intervals. Option B includes exactly the common part. Option C, \((0,8)\), wrongly excludes both endpoints, although 0 and 8 are included. Option D, \([-4,8]\), wrongly includes values from -4 to 0 that are not in the second interval; -4 itself is also excluded from the first interval. Memory cue: for intersection, take the larger left endpoint and the smaller right endpoint, carefully checking brackets.

The first interval ends before (2), and the second starts after (2). Therefore there is no common point.

In \(1\le x<7\), \(x\) can equal 1 because the symbol \(\le\) includes 1. However, \(x=7\) is not allowed because \(<7\) excludes 7. Therefore, the integer points are \(1,2,3,4,5,6\). Option B wrongly omits 1 and includes 7. Exam tip: Include an endpoint with \(\le\) or \(\ge\), but exclude an endpoint with \(<\) or \(>\).

Solve the inequality: \(7-2x>1\Rightarrow -2x>-6\). Dividing by \(-2\) reverses the inequality sign, giving \(x<3\). Hence, the correct interval is \(( -\infty,3 )\), and 3 is not included. \(( -\infty,3 ]\) is incorrect because it includes \(x=3\). Exam tip: Always reverse an inequality sign when multiplying or dividing by a negative number.

The governing concept is solving a linear inequality and reversing its sign when dividing by a negative number. Start with 6 - 4x ≥ -10. Subtract 6 from both sides to obtain -4x ≥ -16. Dividing by -4 reverses the inequality from ≥ to ≤, giving x ≤ 4. The equality sign means 4 is included, so the endpoint is represented by a closed bracket or a filled point. Every number smaller than 4 is included, which gives the interval (-∞, 4]. Therefore option B is correct. Option A excludes the allowed endpoint 4. Options C and D shade values to the right and reflect either a sign-reversal mistake or incorrect direction. The result can also be checked by substituting a value such as x=0.

Adding 1 to all parts of \(2<3x-1\le 11\) gives \(3<3x\le 12\). Dividing all parts by 3 gives \(1<x\le 4\). Thus, 1 is not included whereas 4 is included, so the correct interval is \((1,4]\). Option B incorrectly includes 1. Exam tip: use parentheses \((\ )\) for an excluded endpoint and square brackets \([\ ]\) for an included endpoint.

An intersection contains only the numbers common to both intervals. \((-\infty,0]\) contains all numbers up to and including 0, while \([-3,4)\) contains numbers from -3 up to 4. Their common part is therefore \([-3,0]\). Both -3 and 0 are included in both intervals, so square brackets are used at both ends. Exam tip: For an intersection, keep only the overlap and check endpoint inclusion separately.

In \((1,5]\), 1 is excluded and 5 is included; in \([5,9)\), 5 is included and 9 is excluded. Since the intervals meet at 5, their union contains every number between 1 and 9, but excludes both endpoints. Hence, the simplified interval is \((1,9)\). \([1,9)\) is incorrect because it includes 1. Exam tip: For a union, check carefully whether adjoining intervals share or include the joining point.

The direct answer is option A: \([-6,10]\). Begin with \(-1\le\frac{x+2}{4}\le3\). Because 4 is positive, multiplying all three parts by 4 does not reverse either inequality: \(-4\le x+2\le12\). Subtract 2 throughout: \(-6\le x\le10\). Both original inequalities were non-strict, so both endpoints remain included. Therefore the interval is \([-6,10]\). Option A is correct and uses square brackets at both ends. Option B, \((-6,10)\), wrongly excludes both endpoints. Option C, \([-2,12]\), results from forgetting that 2 must be subtracted after clearing the denominator. Option D has the same incorrect endpoints and also excludes them. The positive multiplier 4 is important; a negative multiplier would reverse inequality signs. Exam cue: solve a compound inequality in the same order on all three parts, then choose brackets from inclusion.

The interval [-2, 6] includes every real number from −2 through 6, including both endpoints. The set-difference symbol \\ means that the element 3 must be removed from this interval. On a number line, this is shown by shading the whole segment from −2 to 6, placing closed endpoints at −2 and 6, and marking 3 with an open circle or a small gap. Thus option A gives the correct description. Option B incorrectly keeps 3, option C shows only one point rather than the entire interval, and option D includes numbers outside the stated bounds.

It takes all numbers on both sides but leaves out (2). On the number line, (2) has an open circle.

The governing concept is the distance interpretation of absolute value. The statement |x| < 3 means that the distance of x from zero is less than 3. Therefore x must be greater than -3 and less than 3, so -3 < x < 3. Since the inequality is strict, neither -3 nor 3 belongs to the solution set. On a number line, both endpoints are shown with open circles and only the segment between them is shaded. In interval notation this is (-3, 3), making option A correct. Option B incorrectly includes both endpoints and would represent |x| ≤ 3. Option C represents values outside the interval, where |x| > 3, and option D fails to impose the lower limit.

Adding 6 gives \(3x>6\), so \(x>2\). Since 2 is not included for a strict ‘>’ inequality, use an open circle and shade right. Exam tip: solve first, then check whether the boundary is included.

A filled point means that
-2 is included, while shading to the right represents values greater than
-2. Hence \(x\ge-2\). In \(x>-2\), the endpoint would be open. Exam tip: filled endpoint always includes equality.

(|x|\ge 2) means (x\le -2) or (x\ge 2). Equality is included, so the outer endpoints are closed.

The governing concept is the union of intervals: it contains every number belonging to at least one of the given intervals. The first interval includes all numbers from -8 through -2, including both endpoints. The second interval contains numbers strictly greater than -2 and less than 5, so -2 is excluded there but is already included by the first interval. Thus there is no gap at -2. The left endpoint -8 remains included, while 5 remains excluded because the second interval has an open right endpoint. Therefore the union is [-8,5), which is option A. Options B and D incorrectly exclude -8, while option C incorrectly includes 5.

In \(x>-2\), the value −2 is not included, so an open circle is used at −2. All values of \(x\) are greater than −2, so the shading extends to the right. Option B includes −2 with a filled circle; it represents \(x\geq -2\). Exam tip: use an open circle for ‘>’ or ‘<’, and a filled circle for ‘≥’ or ‘≤’.

Simplifying \(2(x-1)\le x+4\) gives \(2x-2\le x+4\). Subtracting \(x\) from both sides gives \(x\le 6\). Hence, all values less than or equal to 6 are solutions, represented by \(( -\infty,6]\). Option C is the closest distractor, but it excludes 6. Exam tip: Use a closed bracket when the inequality contains \(\le\) or \(\ge\).

Solve the inequality: \(3(x+2)>2x-1\Rightarrow 3x+6>2x-1\Rightarrow x>-7\). Hence, all numbers to the right of \(-7\) are represented by \(( -7,\infty )\). The value \(-7\) is not included because the sign is \(>\), not \(\ge\); therefore, option C is incorrect. Exam tip: A strict inequality \((>,<)\) always has an open endpoint.

Simplifying gives (3-2x\ge 1), then (-2x\ge -2), and (x\le 1). The sign reverses when dividing by a negative.

An open circle means that the endpoint itself is not part of the solution. Shading between two endpoints means that all numbers strictly between them satisfy the condition. Therefore the represented interval is \((2,9)\), not including either 2 or 9. Hence option C is correct. If the circles were filled or closed, the corresponding endpoints would be included.

To read the number line, first inspect the circles and then inspect the shaded region. Both circles are open, so neither boundary belongs to the set. The shading lies only between the two points, so values such as 3, 5, and 8 are included, while 2 and 9 are excluded. This is exactly the meaning of “both endpoints are not included.”

The original interval includes both \(-3\) and \(7\). Removing only \(0\) creates an open point at \(0\), while \(-3\) and \(7\) remain included. Hence, the set is \([-3,0)\cup(0,7]\). In option B, \(0\) is still included, while option C incorrectly excludes \(-3\) and \(7\) as well. Exam tip: When a point is removed from an interval, use an open parenthesis at that point.

The second interval includes (-4), so no gap remains in the middle. The combined interval is ([-10,2]).

The governing concept is intersection of intervals: an intersection contains only the real numbers that belong to both given sets. The first interval, (-2,6), contains numbers greater than -2 and less than 6; it excludes both endpoints. The second interval, [1,8], contains numbers from 1 through 8 and includes both endpoints. The common region begins at 1, because 1 belongs to both intervals, so the left endpoint is closed. It ends at 6, but 6 is excluded by (-2,6), so the right endpoint is open. Therefore the intersection is [1,6), which is option A. Option B reverses the endpoint inclusion, option C incorrectly excludes 1, and option D represents a union-like broad range rather than the common part.

Since ([2,5]) includes (2) and (5), the outside part will not include them. Thus both boundaries are open.