Algebraic Solution of Linear Inequalities in One Variable

The governing concept is performing equivalent operations on both sides of a one-variable linear inequality. Begin with 2x + 7 ≤ 3. Subtract 7 from both sides, giving 2x ≤ 3 − 7, or 2x ≤ −4. Now divide both sides by the positive number 2. Since the divisor is positive, the inequality sign does not change, and the result is x ≤ −2. The equality is included because the original symbol is ≤; therefore x = −2 must be part of the solution. Substitution verifies the boundary: 2(−2) + 7 = 3, so equality holds. A value such as x = −3 also works, while x = 0 does not. Options B and D reverse the direction incorrectly, and option C wrongly excludes −2. Hence option A is correct.

The direct answer is A: \(x\leq-3\). Start with \(4-3x\geq13\). Subtract 4 from both sides: \(-3x\geq9\). Now divide by \(-3\). When an inequality is divided by a negative number, its direction reverses, so \(x\leq-3\). Option A is exactly this result. Option B, \(x\geq-3\), forgets the reversal. Option C, \(x<-3\), incorrectly excludes equality; \(x=-3\) gives \(4-3(-3)=13\), so equality is allowed. Option D, \(x>-3\), has both the wrong direction and wrong strictness. A quick check confirms the result: choose \(x=-4\); then \(4+12=16\geq13\), so values below -3 work. Choose \(x=0\); then 4 is not at least 13, so it does not work. Memory cue: dividing an inequality by a negative flips the sign.

The direct answer is option A: x>6. An inequality compares two quantities, and its sign tells which side is larger. Start with x+1<2x-5. Subtract x from both sides: 1<x-5. Add 5 to both sides: 6<x, which is the same as x>6. Addition and subtraction do not reverse the sign. A is correct because it gives exactly all numbers greater than 6; for example, x=7 works. B, x<6, is the opposite direction and fails for x=7. C, x≥6, incorrectly includes x=6, but substituting 6 gives 7<7, which is false. D, x≤6, also has the wrong direction and includes values that do not satisfy the inequality. The endpoint 6 is excluded because the original sign is strict (<), not ≤. Memory cue: after isolating x, test one simple value such as 7 to confirm the direction.

Subtracting \(2x\) from both sides gives \(x-4>1\). Adding 4 to both sides yields \(x>5\), so option A is correct. Remember that adding or subtracting a positive number does not reverse the inequality sign; reversal occurs only when multiplying or dividing by a negative number.

Given \(8 \ge x+3\). Subtract \(3\) from both sides: \(8-3 \ge x+3-3\Rightarrow 5 \ge x\). This is the same as \(x \le 5\).

Subtracting (1) gives (10<2x), so (5<x), meaning (x>5). Read the sign correctly when writing (x) on the left.

Expand and simplify: \(2(x+3)=2x+6\), so \(2x+6\le x+10\). Subtract \(x\) from both sides to get \(x+6\le 10\). Subtract 6 to obtain \(x\le 4\). Thus the solution is \(x\le 4\). On the nearest distractor: option C \(x<4\) excludes \(x=4\) which is a valid solution because the inequality allows equality. Exam tip: adding or subtracting terms does not change the inequality direction; only multiply/divide by a negative number reverses the inequality sign.

Expanding the bracket gives \(3x-3>x+5\). Subtracting \(x\) from both sides and adding 3 gives \(2x>8\). Dividing by 2 yields \(x>4\), so option A is correct. Option C is incorrect because the strict inequality does not include the boundary value \(x=4\). Exam tip: For \(>\) or \(<\), the boundary point is not included in the solution set.

The governing concept is clearing a positive denominator in a linear inequality. Starting with (x + 2)/3 < 5, multiply both sides by 3. Because 3 is positive, the inequality direction remains unchanged: x + 2 < 15. Subtracting 2 from both sides gives x < 13. The endpoint 13 is excluded because the original relation is strict, <, rather than ≤. Substitution makes this clear: x = 12 gives 14/3 < 5, which is true, whereas x = 13 gives 15/3 = 5, which is not strictly less than 5. Options B and D reverse the direction without justification, and option C incorrectly includes the endpoint. The denominator is a fixed positive number, so no sign reversal occurs. Therefore option A is the complete solution.

The governing concept is solving a fractional linear inequality by multiplying by a positive denominator. From (3x − 2)/2 ≥ 5, multiply both sides by 2. Since 2 is positive, the inequality sign remains ≥, producing 3x − 2 ≥ 10. Add 2 to both sides to obtain 3x ≥ 12, and divide by positive 3 to get x ≥ 4. Equality must be retained because the original symbol includes equality. Checking the boundary, x = 4 gives (12 − 2)/2 = 5, so it satisfies the inequality exactly. Values larger than 4 also work, while values smaller than 4 do not. Option B reverses the direction, option C incorrectly excludes x = 4, and option D gives both the wrong direction and endpoint. Hence option A is correct.

The direct answer is A: \(\{0,1,2,3\}\). In school mathematics, whole numbers are the non-negative integers: \(0,1,2,3,\ldots\). The condition \(x\le3\) keeps every whole number from 0 through 3 and excludes 4 and all larger numbers. Thus the finite answer set is \(\{0,1,2,3\}\). Option A is correct. Option B, \(\{1,2,3\}\), wrongly omits 0; 0 is a whole number and satisfies \(0\le3\). Option C, \(\{0,1,2\}\), wrongly omits 3, which is allowed because the sign is \(\le\), not \(<\). Option D, \(\{3,4,5\}\), includes numbers greater than 3 and therefore violates the condition, although it includes 3. The phrase whole number is important: it starts at 0, unlike positive counting numbers, which usually start at 1. Memory cue: whole numbers are 0 and the counting numbers; with \(\le3\), stop at and include 3.

The direct answer is option A: x>5. The aim is to leave x alone. Begin with x-7>-2. Add 7 to both sides: x>-2+7, so x>5. Adding the same number to both sides preserves the order because the distance between the two sides does not change. A is correct and includes values such as 6, which satisfies 6-7>-2 because -1>-2. B, x<5, reverses the required direction; a value such as 4 gives -3>-2, which is false. C, x≥5, wrongly includes the boundary x=5; substitution gives -2>-2, false because the sign is >, not ≥. D, x≤5, has the wrong direction and also includes many invalid values. No sign reversal occurs here because we only add a positive number, not multiply or divide by a negative number. Exam cue: when adding or subtracting, keep the inequality sign unchanged; reverse it only when multiplying or dividing by a negative number.

The direct answer is A: \(x\le-7\). Start with \(x+6\le-1\). Subtract 6 from both sides: \(x+6-6\le-1-6\). The left side becomes \(x\), and the right side becomes \(-7\), so \(x\le-7\). Option A is correct. Option B reverses the inequality without a valid reason; no multiplication or division by a negative number occurred. Option C changes the inclusive sign to a strict sign and would incorrectly exclude \(x=-7\). Option D has the wrong direction and also excludes the valid boundary. Check the boundary: if \(x=-7\), then \(x+6=-1\), so equality is allowed. A quick test also confirms the direction: \(x=-8\) works, while \(x=-6\) gives 0, which does not satisfy \(0\le-1\). Remember that subtracting the same number from both sides preserves the inequality sign.

The governing concept is reversal of an inequality when dividing by a negative number. In −4x < 16, divide both sides by −4. Because the divisor is negative, the symbol < changes to >, giving x > −4. This can also be checked by testing the boundary: x = −4 gives −4(−4) = 16, but the inequality is strict, so the boundary itself is excluded. A value such as x = 0 gives 0 < 16, which is true, confirming that values greater than −4 belong to the solution set. Values less than −4 fail; for example, x = −5 gives 20 < 16, which is false. Option B forgets to reverse the sign, while options C and D mishandle the sign of the constant. Therefore option A is correct.

The governing concept is isolating the variable while reversing the sign when dividing by a negative coefficient. Starting with 6 − 2x ≥ 0, subtract 6 from both sides to obtain −2x ≥ −6. Dividing by −2 reverses ≥ to ≤, so x ≤ 3. The equality is included because the original relation is ≥. A boundary check gives 6 − 2(3) = 0, confirming that x = 3 belongs to the solution set. Testing x = 2 gives 2 ≥ 0, which is true, while x = 4 gives −2 ≥ 0, which is false. Option B results from forgetting the sign reversal, and options C and D incorrectly remove or reverse the endpoint and direction. Therefore option A correctly represents all solutions.

Subtracting \(3x\) from both sides gives \(4x+2<18\). Subtracting 2 and then dividing by the positive number 4 gives \(4x<16\), so \(x<4\). The inequality sign does not reverse because the divisor is positive; therefore, option B has the wrong direction. Exam tip: reverse the inequality sign only when multiplying or dividing by a negative number.

The governing concept is collecting variable terms on one side and constants on the other while preserving the inequality. Begin with 9x − 5 ≥ 4x + 20. Subtract 4x from both sides to get 5x − 5 ≥ 20. Add 5 to both sides, giving 5x ≥ 25. Dividing by positive 5 does not change the direction, so x ≥ 5. The equality is retained because the original sign is ≥. Substitution confirms the boundary: when x = 5, both sides equal 40. A larger value, such as x = 6, satisfies the relation, while x = 4 gives 31 ≥ 36, which is false. Option B reverses the direction, and options C and D incorrectly exclude or reverse the boundary. Thus option A is the complete solution.

The governing concept is translating a word condition into a linear inequality and identifying the least value allowed by the solution set. The condition is x + 15 ≥ 50. Subtract 15 from both sides to obtain x ≥ 35. Therefore x can be 35 or any larger amount, and the minimum possible value is exactly 35. The endpoint is included because the symbol is ≥: if Ravi has 35 rupees, then x + 15 = 35 + 15 = 50, satisfying the condition. The value 34 gives 49, which is insufficient. Although 50 and 65 also satisfy the inequality, they are not minimum values because both are greater than 35. Thus option A is correct. The important distinction is between a value that satisfies the condition and the smallest value that satisfies it.

Twice means (2x) and less than means (<), so (2x<14) and (x<7). Convert words into the correct mathematical sign.

Adding (9) to both sides gives (2x\ge 10), so (x\ge 5). Dividing by a positive number does not change the sign.

Multiplying both sides by (6) gives (x-4\le 12), so (x\le 16). Multiplying by a positive denominator keeps the sign unchanged.

The inequality gives (3x<15), so (x<5) and the greatest integer is (4). A strict inequality does not include the boundary value.

The governing concept is solving a linear inequality while preserving its direction under positive operations. First expand the bracket: 2(x + 4) = 2x + 8, so the inequality becomes 2x + 8 > x + 13. Subtract x from both sides to obtain x + 8 > 13. Subtract 8 from both sides, giving x > 5. Therefore option A is correct. Since the operations used were adding or subtracting the same quantities, the inequality sign did not change. The equality value x = 5 is excluded because it gives 2(9) = 18, while 5 + 13 = 18, producing equality rather than the required strict greater-than relation. The remaining options either reverse the direction or include the endpoint.

Subtracting (2x) from both sides gives (2x+7<19), so (x<6). Keeping variable terms on one side makes the solution faster.

Subtracting 8 from both sides of \\(x+8>20\\) gives \\(x>12\\). The inequality sign does not change when the same number is added to or subtracted from both sides. Option C is incorrect because substituting \\(x=12\\) gives \\(20>20\\), which is false. Exam tip: The inequality sign reverses only when both sides are multiplied or divided by a negative number.