Algebraic Solution of Linear Inequalities in One Variable

Subtracting 4 from both sides of x + 4 < 9 gives x < 5, so option A is correct. Option B incorrectly reverses the inequality sign; subtracting 4 does not change its direction. Exam tip: the direction of an inequality changes only when both sides are multiplied or divided by a negative number.

Adding 3 to both sides of \(x-3\le 7\) gives \(x\le 10\). The inequality sign does not change when the same number is added to both sides. Option C reverses the direction incorrectly, while option D wrongly excludes \(x=10\). Exam tip: the inequality sign reverses only when both sides are multiplied or divided by a negative number.

Dividing both sides of 2x > 10 by the positive number 2 gives x > 5. The inequality sign does not reverse when dividing by a positive number; it reverses only when multiplying or dividing by a negative number. Exam tip: divide by the coefficient of the variable and check its sign.

The governing rule is that an inequality sign reverses whenever both sides are divided or multiplied by a negative number. Start with −2x<8. Divide both sides by −2: the left side becomes x and the right side becomes −4. Because −2 is negative, the symbol < changes to >, giving x>−4. A substitution check confirms the result: x=0 satisfies the original inequality because 0<8, and 0 is greater than −4. In contrast, x=−5 does not satisfy the original inequality because 10<8 is false, and −5 is not greater than −4. Option A forgets the sign reversal. Options C and D also use incorrect arithmetic or direction. Therefore option B is the unique correct solution set.

Substituting \(x=4\) gives \(3\times4=12\), so \(12\le12\) is true. In contrast, \(x=5\) gives \(15\le12\), which is false. In an exam, test each option by substituting its value directly into the inequality.

Solving \(2x+1>9\) gives \(2x>8\), so \(x>4\). Therefore, \(x=5\) satisfies the inequality. Note that \(x=4\) gives \(2x+1=9\), so it is not valid for the strict inequality \(>\). In an exam, carefully distinguish between \(>\) and \(\geq\).

Subtracting 1 from both sides of \(x+1\ge 6\) gives \(x\ge 5\). The inequality sign remains unchanged when the same number is added or subtracted from both sides, so option 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.

The governing concept is solving a linear inequality while reversing its sign when dividing by a negative number. Begin with 5−x>2. Subtract 5 from both sides to obtain −x>−3. To isolate x, divide both sides by −1. Since −1 is negative, the inequality direction reverses, producing x<3. A check with x=0 gives 5−0=5>2, and 0<3, so the result includes a valid value. A check with x=4 gives 5−4=1, and 1>2 is false; 4 is also not less than 3. Option B has the wrong direction. Options C and D result from mishandling the constant terms or the sign. Therefore option A is the complete solution.

The governing concept is the algebraic solution of a linear inequality in one variable. Apply inverse operations while preserving the inequality relation. From 4x + 2 ≤ 14, subtract 2 from both sides to obtain 4x ≤ 12. Dividing by the positive number 4 gives x ≤ 3; the sign does not reverse because the divisor is positive. The endpoint is included because the original symbol is ≤. Substitution confirms the result: x = 3 gives 4(3) + 2 = 14, which satisfies 14 ≤ 14, while x = 4 gives 18 ≤ 14, which is false. Option B reverses the direction unnecessarily, and option C uses the wrong boundary.

The governing concept is sign reversal when an inequality is divided by a negative number. Start with 3 − 2x ≤ 9. Subtract 3 from both sides to get −2x ≤ 6. To isolate x, divide both sides by −2. Since −2 is negative, the inequality symbol reverses, giving x ≥ −3. Option A is therefore correct. A substitution check supports this: x = 0 gives 3 ≤ 9, which is true, and 0 is indeed at least −3. However, x = −4 gives 11 ≤ 9, which is false, and −4 does not satisfy x ≥ −3. Option B is the common error of forgetting sign reversal; options C and D use an incorrect boundary or direction.

The inequality \(-2<x\) excludes \(-2\), while \(x\le 2\) includes \(2\). Therefore, the integer values are \(-1,0,1,2\). Exam tip: a strict sign \(<\) or \(>\) excludes the endpoint, whereas \(\le\) or \(\ge\) includes it. Option A is incorrect because it includes \(-2\).

Subtracting 7 from both sides of \(x+7>7\) gives \(x>0\), so option A is correct. Option B reverses the inequality sign incorrectly. Exam tip: Adding or subtracting the same number on both sides does not reverse the inequality sign.

Multiplying both sides of \(\frac{x}{3} \le 2\) by the positive number 3 gives \(x \le 6\). The inequality sign remains unchanged when multiplying by a positive number, so option A is correct. Option C results from incorrectly dividing 2 by 3. Exam tip: reverse the inequality sign only when multiplying or dividing by a negative number.

The governing concept is that multiplying an inequality by a negative number reverses its direction. Begin with x/(−4) > 1. Multiplying both sides by −4 gives x < −4, because 1 × (−4) = −4 and the sign changes from > to <. Thus option B is correct. A numerical check makes the result clear: x = −5 gives (−5)/(−4) = 1.25, and 1.25 > 1 is true; also, −5 < −4. In contrast, x = −3 gives 0.75 > 1, which is false. Option A is obtained by forgetting the reversal. Options C and D have an incorrect endpoint or sign and do not describe the solution set of the original inequality.

The governing property is that multiplying both sides of an inequality by a positive number preserves the inequality direction. Starting with p < 3, multiply each side by 6: 6p < 6 × 3, so 6p < 18. Therefore option A is correct. The direction would reverse only if both sides were multiplied or divided by a negative number. Option B incorrectly reverses the sign even though 6 is positive. Option C multiplies only the right side and leaves p unchanged, so it is not an equivalent transformation. Option D changes both the coefficient and direction without justification. The operation must be applied equally to both sides.

Subtracting 6 from both sides of \(x+6<14\) gives \(x<8\). Option B reverses the inequality sign incorrectly, while the options containing 20 result from adding 6 instead of subtracting it. Exam tip: Adding or subtracting a number does not reverse the inequality sign.

Adding 5 to both sides of \(x-5\ge 1\) gives \(x\ge 1+5=6\). The inequality sign does not change when the same number is added or subtracted from both sides. Therefore, option A is correct; option B reverses the sign incorrectly. Exam tip: the inequality sign reverses only when multiplying or dividing by a negative number.

Dividing both sides of \(3x<21\) by the positive number 3 gives \(x<7\). The inequality sign remains unchanged when multiplying or dividing by a positive number. Remember that the sign reverses only when multiplying or dividing by a negative number.

Divide both sides by −5. Because −5 is negative, reverse the inequality sign: x ≤ 15 ÷ (−5) = −3. Therefore, option B is correct.

The governing concept is solving and checking a strict linear inequality. Starting with 4x − 2 > 10, add 2 to both sides to obtain 4x > 12. Divide by the positive number 4, giving x > 3. Among the choices, only x = 4 is greater than 3, so option C is correct. Direct substitution confirms it: 4(4) − 2 = 14, and 14 > 10 is true. At x = 3, the expression equals 10, but equality is not allowed because the symbol is strictly >. At x = 2 and x = 1, the expressions are 6 and 2, respectively, and both fail the condition. The boundary value 3 is excluded.

Subtracting 2 from both sides of \(x+2\ge 9\) gives \(x\ge 7\), so option A is correct. Option B incorrectly reverses the inequality sign. Remember that adding or subtracting a number does not change the direction of an inequality; it changes only when multiplying or dividing by a negative number.

The governing rule is that an inequality reverses direction when both sides are multiplied or divided by a negative number. Start with 8 − x < 3. Subtract 8 from both sides to obtain −x < −5. To isolate x, divide both sides by −1; because the divisor is negative, the sign reverses, giving x > 5. Hence option A is correct. A quick check confirms this: if x = 6, then 8 − 6 = 2 and 2 < 3. Option B results from forgetting the sign reversal. Options C and D incorrectly use 11, which can arise from adding instead of subtracting or mishandling the constant. The final solution contains every number greater than 5.

The governing concept is solving a two-step linear inequality while preserving its direction under positive division. Begin with 6x + 4 ≤ 22. Subtract 4 from both sides: 6x ≤ 18. Divide both sides by 6, which is positive, so the sign remains unchanged: x ≤ 3. Therefore option A is correct. Substitution at the boundary gives 6(3) + 4 = 22, and equality is allowed because the original sign is ≤. Values less than 3 also work, while values greater than 3 do not. Options B and D reverse the direction without reason, and option C uses an incorrect result of the division.

The governing concept is reversal of an inequality when dividing by a negative coefficient. From 10 − 2x ≥ 4, subtract 10 from both sides to get −2x ≥ −6. Divide by −2; because the divisor is negative, the inequality reverses and gives x ≤ 3. Thus option A is correct. The endpoint is included because the original symbol is ≥. Checking x = 3 gives 10 − 6 = 4, so equality holds. Checking x = 2 gives 6 ≥ 4, which is true, while x = 4 gives 2 ≥ 4, which is false. Option B results from failing to reverse the sign, and options C and D use the wrong boundary or direction.

Subtracting 9 from both sides of x + 9 > 9 gives x > 0. Therefore, x > 0 is the correct solution. Option B reverses the meaning of the inequality, while the options involving 18 result from incorrectly combining the constants. Exam tip: adding or subtracting the same number on both sides does not reverse the inequality sign.