representation on number line

The direct answer is option C: x≥3, with a closed point at 3 and shading to the right. Solve 7-2x≤1. Subtract 7: -2x≤-6. Divide by -2, and reverse the sign because the divisor is negative: x≥3. Equality is allowed in the original inequality, so x=3 belongs to the solution set; therefore the point is closed. Numbers greater than 3 also work, so shade right. Option A incorrectly shades left and uses the wrong direction. Option B excludes 3 even though equality is allowed. Option D has the correct rightward direction but incorrectly uses a strict sign and an open point. Always check a value: x=4 gives 7-8=-1≤1, confirming the right side.

In a union, both separate intervals are shaded. Check the open or closed endpoints of each interval separately.

Solving the inequality gives 2x - 7 > -1, so 2x > 6 and x > 3. Therefore, the boundary point is 3. Since the inequality sign is >, 3 is not included in the solution set, so an open dot is used. All numbers greater than 3 lie to the right of 3, so the shading extends to the right. Exam tip: use an open dot for > or < and a closed dot for ≥ or ≤.

Solving gives ( 7<-7x\le 21 ), then signs reverse to ( -1>x\ge -3 ). In order, write ( -3\le x<-1 ).

The direct answer is option C, the empty set ∅. On a number line, the notation (3,3) means the strict interval 3<x<3. No real number can be greater than 3 and less than 3 at the same time. Therefore there are no points to shade and the solution set is empty. Option A, {3}, would be correct for the single equation x=3 or a closed singleton, not for an open interval (3,3). Option B, ℝ, would mean every real number is included, which is clearly not the case. Option C correctly represents no possible value. Option D, (-∞,3), includes all numbers less than 3 and is much larger than the stated interval. A useful distinction is that [3,3] would contain the single number 3, while (3,3) contains nothing. Memory cue: identical open endpoints make an empty interval; identical closed endpoints make a one-point set.

Solving gives ( 8\le 4x\le 16 ), so ( 2\le x\le 4 ). Since both signs include equality, both endpoints are closed.

Intersect means take the common part of both intervals. \( ( -\infty,-2) \) contains all numbers less than \(-2\), but not \(-2\). \( [-5,4] \) contains all numbers between \(-5\) and \(4\), including both ends. The overlap therefore goes from \(-5\) (included) up to \(-2\) (excluded), giving \( [-5,-2) \).

Solving gives ( 3-2x\le -5 ), then ( -2x\le -8 ) gives ( x\ge 4 ). The final negative division reverses the direction.

We need values that satisfy both inequalities simultaneously. Since \(x\ge -6\), \(-6\) is included; since \(x<1\), \(1\) is excluded. Hence, the common interval is \([-6,1)\). Option \((-6,1]\) reverses the endpoint inclusion. Exam tip: use a square bracket for \(\ge\) or \(\le\), and a round bracket for \(<\) or \(>\).

The square bracket includes \( -\frac{1}{3} \), and the round bracket excludes \( \frac{2}{3} \). Read both endpoints separately in a mixed interval.

( -2 ) is closed, so use ( \le ), and ( 5 ) is open, so use ( < ). Decide the left and right endpoint signs separately.

Solving gives ( 4<-4x<16 ), and division by a negative gives ( -1>x>-4 ). In order, write ( -4<x<-1 ).

The left sign is ( < ), so it is open, and the right sign is ( \le ), so it is closed. The same rule applies to fractional endpoints.

The common part is ( -1<x<0 ) because ( -1 ) is open in the second interval and ( 0 ) is open in the first. In intersection, keep the common limits precisely.

Solving gives ( -\frac{x}{3}>3 ), so ( -x>9 ) and ( x<-9 ). Multiplication by ( -1 ) reverses the direction.

The interval [2,5] includes both 2 and 5, so the number line has closed endpoints at these values. Since 3 is removed from the interval, it must be marked with an open hole. Equivalently, the set is [2,3) ∪ (3,5]. Exam tip: a square bracket indicates inclusion, while a parenthesis or open hole indicates exclusion.

Both conditions are true together only at ( x=1 ). When two rays meet only at the boundary, the result is one closed dot.

Multiplying by ( 4 ) gives ( -20<x-2\le 0 ), so ( -18<x\le 2 ). Multiplication by a positive number does not reverse signs.

The two intervals together cover the entire number line including ( -7 ). In a union, all covered regions are included.

The first inequality gives ( x<4 ), and the second gives ( x\ge -2 ). If both must be true, shade ( -2\le x<4 ).

An open circle means that \(-2\) and \(3\) are not included in the solution set. The left ray represents \(x<-2\), while the right ray represents \(x>3\). Since either of these separate regions may be chosen, the connector is \(\lor\). Option C would have closed endpoints because it uses \(\le\) and \(\ge\). Exam tip: separate shaded regions usually indicate \(\lor\), whereas one continuous middle region usually indicates \(\land\).

Solving gives ( 5-3x\ge -4 ), then ( -3x\ge -9 ) gives ( x\le 3 ). Reverse the sign when dividing by a negative.

Taking the complement removes ( -2 ) and includes ( 6 ). In complements, open and closed endpoint behavior changes.

Solving gives ( 6\le 3x<15 ), so ( 2\le x<5 ). Apply the same operation to every part of a compound inequality.

The open dot excludes ( -6 ), and the closed dot includes ( 2 ). In exams, first identify whether endpoints are open or closed.