Algebraic Solution of Linear Inequalities in One Variable

The governing concept is preserving an inequality while multiplying by a positive number. Start with x/4 ≤ 3. Multiply both sides by 4: 4(x/4) ≤ 4(3), which simplifies to x ≤ 12. Therefore option A is correct. The sign remains unchanged because 4 is positive. The boundary is included because the original relation is ≤; substituting x = 12 gives 12/4 = 3, so equality is satisfied. Values smaller than 12 also work, such as x = 8, which gives 2 ≤ 3. Option B incorrectly reverses the sign, while options C and D confuse multiplying by 4 with dividing by 4 or use the wrong direction. Hence the complete solution set is all real numbers less than or equal to 12.

The governing concept is that multiplication by a negative number reverses the order in an inequality. Begin with x/(−3) < 2. Multiply both sides by −3. Since −3 is negative, the symbol < changes to >, so x > 2(−3) = −6. Therefore option B is correct. A check with x = −5 confirms the result: (−5)/(−3) = 5/3, and 5/3 < 2 is true; also −5 > −6. Option A is the common mistake of multiplying by −3 without reversing the sign. Options C and D have an incorrect endpoint or direction. The endpoint −6 is not included because the original relation is strict; x = −6 would produce 2 < 2, which is false.

The governing concept is solving a one-variable linear inequality by applying the same operation to both sides. Begin with 12 < 4x. Divide both sides by 4, which is positive: 12/4 < 4x/4. Therefore, 3 < x, conventionally written as x > 3. Hence option A is correct. Because the divisor is positive, the inequality sign keeps its direction; it would reverse only if we multiplied or divided by a negative number. Option B incorrectly reverses the relation. Options C and D use 16, which does not result from dividing 12 by 4 and therefore cannot describe the solution. The endpoint x = 3 is excluded because 12 < 12 is false, while x = 4 gives 12 < 16, which is true.

The governing rule is that multiplying or dividing an inequality by a positive number preserves its direction. Start with x/5 > −1. Multiply both sides by 5, which is positive: 5(x/5) > 5(−1). This simplifies to x > −5, so option A is correct. No reversal is permitted because 5 is not negative. Option B reverses the sign incorrectly. Option C changes the negative boundary to a positive one, and option D combines an incorrect boundary with an incorrect direction. A quick check confirms the result: x = 0 satisfies x > −5, and 0/5 = 0 > −1. The boundary x = −5 is excluded because the original relation is strict; substituting it gives −1 > −1, which is false.

Subtracting the same number, \(6\), from both sides of \(b\ge 10\) gives \(b-6\ge 10-6=4\). The inequality sign does not reverse when the same number is added or subtracted from both sides. Thus, A is correct; option B incorrectly reverses the sign. Exam tip: the sign reverses only when both sides are multiplied or divided by a negative number.

When both sides of \(q>2\) are multiplied by the negative number \(-4\), the inequality sign reverses. Thus, \(-4q<-8\) is obtained. Option A is incorrect because it keeps the original sign. Exam tip: Always reverse the inequality sign when multiplying or dividing an inequality by a negative number.

Subtracting 9 from both sides gives \\(x+9-9\le 15-9\\), so \\(x\le 6\\). Since 9 is being subtracted, the inequality sign does not change. Option B incorrectly reverses the sign, while 24 would result from adding 9 instead of subtracting it. Exam tip: the sign changes only when multiplying or dividing by a negative number.

Adding 8 to both sides of \(x-8>2\) gives \(x>10\), so option A is correct. Option C results from incorrectly subtracting 8 instead of adding it. Exam tip: the inequality sign does not change when the same number is added or subtracted from both sides; it reverses only when multiplying or dividing by a negative number.

The governing concept is checking a proposed value by substitution and, when useful, identifying the complete solution interval. Substitute each candidate into 3x + 2 < 14. For x = 3, the left side is 3(3) + 2 = 9 + 2 = 11, and 11 < 14 is true. Thus option C satisfies the inequality. Algebraically, subtracting 2 gives 3x < 12, and dividing by positive 3 gives x < 4, so 3 belongs to the solution set. For x = 4, the statement becomes 14 < 14, false because the inequality is strict. Values 5 and 6 produce left sides 17 and 20, respectively, both greater than 14, so they fail.

Substituting \(x=5\) gives \(5(5)-1=24\), and \(24\ge24\) is true. Therefore, \(x=5\) satisfies the inequality. For the other values, the left-hand side is less than 24. Exam tip: \(\ge\) means “greater than or equal to,” so the boundary value is included.

Subtracting 4 from both sides of \(x+4\ge 13\) gives \(x\ge 9\). The inequality sign does not change when the same number is added or subtracted from both sides, and 9 is included because the sign is \(\ge\). Option B incorrectly reverses the sign. Exam tip: reverse the inequality sign only when multiplying or dividing by a negative number.

The governing concept is solving a linear inequality by applying equivalent operations to both sides. Starting with 2x + 5 ≤ 17, subtract 5 from both sides: 2x ≤ 12. Then divide both sides by the positive number 2: x ≤ 6. Because division by a positive number does not reverse the inequality sign, the result remains “less than or equal to.” Therefore option A is correct. Option B incorrectly reverses the direction, while options C and D result from treating 5 as though it were not subtracted before division or from using an incorrect operation. A quick check with x = 6 gives 17, which satisfies the inclusive inequality.

The governing rule is that dividing or multiplying an inequality by a negative number reverses its direction. Begin with 15 − 3x ≥ 6. Subtract 15 from both sides to obtain −3x ≥ −9. Now divide both sides by −3. Since the divisor is negative, the sign changes from ≥ to ≤, giving x ≤ 3. Thus option A is correct. A useful check is x = 3: the left side becomes 15 − 9 = 6, so equality is allowed. A value such as x = 2 gives 9 ≥ 6 and also works. Option B forgets the reversal caused by negative division; options C and D use an incorrect magnitude or sign.

The inequality \(-4<x\le 1\) includes every integer greater than \(-4\) and less than or equal to \(1\). Thus, the possible values are \(-3,-2,-1,0,1\). Option A is incorrect because \(-4\) is excluded by the strict inequality \(x>-4\). Exam tip: a \(<\) or \(>\) sign excludes the endpoint, while \(\le\) or \(\ge\) includes it.

Subtracting 11 from both sides of \\(x+11>11\\) gives \\(x>0\\). The inequality sign does not change when the same number is subtracted from both sides. Therefore, option A is correct; option B reverses the sign incorrectly. Exam tip: The inequality sign reverses only when both sides are multiplied or divided by a negative number.

The governing concept is reversal of an inequality when multiplying or dividing by a negative number. Starting from x/−5 > 3, multiply both sides by −5 to remove the denominator. Because −5 is negative, the sign reverses: x < 3(−5) = −15. Hence option B is correct. A check confirms the result: if x = −16, then (−16)/(−5) = 3.2, which is greater than 3, so it satisfies the original inequality. If x = −15, the quotient is exactly 3 and is excluded because the original sign is strict. Option A makes the common error of keeping the sign unchanged; options C and D use the wrong sign or magnitude.

The governing concept is isolating the variable in a linear inequality while preserving its direction. From 18 < 6x, divide both sides by 6, which is positive. This gives 3 < x. Rewriting the same relationship with x first gives x > 3, so option A is correct. Since the divisor is positive, the inequality sign does not reverse. The endpoint x = 3 is not included because the original relation is strict: 18 is not less than 18. A value such as x = 4 gives 18 < 24 and works, while x = 2 gives 18 < 12 and fails. Options B reverses the sign without justification, and C and D incorrectly add rather than divide by the coefficient 6.

The governing concept is solving a fractional linear inequality by multiplying both sides by the denominator. In x/7 > 2, multiply both sides by 7 to obtain x > 14. Since 7 is positive, the inequality direction remains unchanged. Therefore option A is correct. A check with x = 15 gives 15/7, which is greater than 2, while x = 14 gives exactly 2 and is excluded because the original sign is strict. Option B reverses the direction without a negative multiplier. Options C and D incorrectly divide by 7 again or confuse multiplication with division. The denominator is a positive constant, so multiplying by it is valid and produces the unique solution x > 14.

The governing rule is that adding or subtracting the same number from both sides of an inequality preserves the inequality sign. Starting with x + 3 ≥ 10, subtract 3 from both sides: x + 3 − 3 ≥ 10 − 3. Hence x ≥ 7. The endpoint 7 is included because the original relation uses ≥; indeed, x = 7 gives 7 + 3 = 10, which satisfies the condition. Values greater than 7 also work, while values below 7 do not. Option A is therefore correct. Option B reverses the direction incorrectly, option C wrongly excludes the valid endpoint 7, and option D gives the opposite solution set.

The governing concept is maintaining an inequality while carrying out equivalent operations on both sides. Start with 3x + 5 > 14. Subtract 5 from both sides to obtain 3x > 9. Then divide by the positive number 3; the direction remains unchanged, giving x > 3. Therefore option A is correct. Since the original symbol is strictly greater than, x = 3 is not included: substituting it gives 3(3) + 5 = 14, and 14 > 14 is false. Option B reverses the direction without justification. Option C incorrectly includes the endpoint, and option D both reverses the direction and includes the endpoint. Any value such as x = 4 satisfies the original inequality.

The governing rule is that an inequality remains equivalent when we add or subtract the same quantity from both sides. Starting with 2x − 7 < 5, add 7 to both sides: 2x < 12. Now divide both sides by 2. Since 2 is positive, the inequality symbol keeps its direction, giving x < 6. Therefore option A is correct. Option B reverses the solution without justification, option C incorrectly changes the sign of the answer, and option D includes x = 6 even though substituting x = 6 gives 12 − 7 = 5, which does not satisfy the strict inequality.

Solve the inequality by collecting variable terms on one side and constants on the other. Starting with 5x − 2 > 3x + 8, subtract 3x from both sides to obtain 2x − 2 > 8. Add 2 to both sides, giving 2x > 10. Dividing by the positive number 2 preserves the direction, so x > 5. Therefore option A is correct. The endpoint is excluded because the original sign is strict: when x = 5, both sides equal 23, so 23 > 23 is false. Options B and D reverse or mis-handle the direction, while C incorrectly includes the equality case.

The governing procedure for a one-variable linear inequality is to isolate the variable while preserving the inequality rule. Starting with 3x+2>11, subtract 2 from both sides to obtain 3x>9. Dividing by the positive number 3 does not reverse the sign, so x>3. The endpoint x=3 is excluded because it would give 3(3)+2=11, whereas the original statement requires a value strictly greater than 11. Option B reverses the direction without justification, and options C and D incorrectly include or reverse the boundary. Hence option A gives the correct solution set.

The phrase “at least 13” means greater than or equal to 13, so the inequality is 2x + 5 ≥ 13. Subtracting 5 from both sides gives 2x ≥ 8. Dividing by 2, which is positive, preserves the direction of the inequality and yields x ≥ 4. The equality case must be included: when x = 4, the left side is 2(4) + 5 = 13, which satisfies the phrase “at least.” Thus option A is correct. Option B incorrectly excludes x = 4 by using a strict sign. Options C and D reverse the direction and describe values that generally make 2x + 5 less than 13. The endpoint and direction are both essential.

The governing concept is maintaining equivalence while isolating the variable in a linear inequality. Since 5 is added to x, subtract 5 from both sides: x + 5 − 5 > 12 − 5. This simplifies to x > 7. The operation is subtraction of the same number, so the direction of the inequality does not change. Therefore option A is correct. Option B gives the opposite direction, while option C incorrectly includes x = 7; substituting 7 into the original statement gives 12 > 12, which is false. Option D has both the wrong direction and the wrong endpoint condition. Any number greater than 7 satisfies the inequality.