Algebraic Solution of Linear Inequalities in One Variable

Subtracting 3 from both sides of \(2x+3>11\) gives \(2x>8\), and dividing by 2 gives \(x>4\). Since \(x\) is an integer, 4 is not included; the smallest integer greater than 4 is 5. Exam tip: For a strict inequality \(>\), the boundary value is excluded.

Subtracting 1 from both sides gives \(9<3x\). Dividing by 3, which is positive, does not reverse the inequality sign, so we obtain \(3<x\), equivalent to \(x>3\). Therefore, option A is correct. Exam tip: the inequality sign changes direction only when multiplying or dividing by a negative number, not by a positive number.

Subtracting 5 from both sides of \(x+5<12\) gives \(x<7\). Together with \(0\leq x\), the possible integer values of \(x\) are \(\{0,1,2,3,4,5,6\}\). The value 7 is excluded because \(7<7\) is false. Exam tip: In a strict inequality such as \(<\), the boundary value is not included.

Subtracting \\(x\\) from both sides gives \\(2x-2>8\\). Adding 2 to both sides gives \\(2x>10\\), and dividing by 2 yields \\(x>5\\). Since the original inequality is strict, \\(x=5\\) is not included; therefore, \\(x\\geq5\\) is incorrect. Exam tip: Dividing an inequality by a positive number does not reverse its sign.

For option A, subtracting 3 from both sides of \(x+3\leq 0\) gives \(x\leq -3\). Option B gives \(x\leq 3\), option C gives \(x\geq -3\), and option D gives \(x\geq 3\). In an exam, carefully track the sign when moving a constant to the other side.

Dividing both sides by the positive number \(5\) does not reverse the inequality sign: \(x\leq \frac{-20}{5}=-4\). Therefore, \(x\leq -4\) is correct. In option B, the sign has been reversed unnecessarily. Exam tip: the inequality sign reverses only when multiplying or dividing by a negative number.

Dividing both sides by \\(-5\\) reverses the inequality sign because \\(-5\\) is negative: \\(x \leq -3\\)。 Option B is incorrect because it does not reverse the sign. Exam tip: Always reverse an inequality sign when multiplying or dividing by a negative number.

The important concept is that multiplying or dividing an inequality by a negative number reverses its sign. From 4 − x/2 > 1, subtract 4 from both sides to obtain −x/2 > −3. Multiplying both sides by 2 gives −x > −6. Now multiply by −1; because the multiplier is negative, the inequality reverses: x < 6. Hence option A is correct. The boundary value x = 6 is excluded because the original sign is strict. Option B forgets the reversal caused by the negative coefficient, while C and D incorrectly include the boundary and/or use the wrong direction.

Starting with \\(4x+6>22\\), subtracting 6 from both sides gives \\(4x>16\\). Dividing by the positive number 4 gives \\(x>4\\), so option A is correct. Option C is incorrect because including \\(x=4\\) would make the left side equal to 22, not greater than 22. Exam tip: dividing or multiplying an inequality by a negative number reverses its sign; a positive number does not.

Multiplication or division by a negative number reverses order, so the inequality sign must be reversed. For example, multiplying \(2<5\) by \(-1\) gives \(-2>-5\). Exam tip: immediately flip the sign when using a negative factor.

Divide both sides by −6. Because the divisor is negative, reverse the inequality sign: x < 18 ÷ (−6) = −3. Therefore, option A is correct.

Solving \(3x-2<13\) gives \(3x<15\), hence \(x<5\). Therefore, \(x=5\) is not a solution because substitution gives \(3(5)-2=13\), and \(13<13\) is false. In contrast, \(x=4\) satisfies the inequality since \(10<13\). Exam tip: In strict inequalities such as '<' or '>', the boundary value is not included.

The whole numbers are W = {0, 1, 2, 3, …}. Those that satisfy x ≤ 4 are 0, 1, 2, 3, and 4, so the solution set is option A.

Adding 11 to both sides gives \(8x\geq 5x+21\). Subtracting \(5x\) from both sides gives \(3x\geq 21\), and dividing by the positive number 3 yields \(x\geq 7\). The inequality sign does not reverse because the divisor is positive, so \(x\leq 7\) is incorrect. Exam tip: reverse the inequality sign only when multiplying or dividing by a negative number.

Adding 1 to both sides gives \(\frac{x}{4}>3\). Multiplying both sides by the positive number 4 yields \(x>12\), so option A is correct. Option C is incorrect because \(x=12\) gives equality, whereas the inequality requires a strict greater-than sign. Exam tip: Multiplication or division by a positive number does not reverse the inequality sign.

Multiplying \(\frac{x+3}{2}>6\) by 2 gives \(x+3>12\), so \(x>9\). The smallest integer greater than 9 is 10. Therefore, the correct answer is 10. Remember that the strict sign ‘>’ excludes 9; with ‘\(\geq\)’, 9 would also be included.

Subtracting \(4x\) and \(5\) from the respective sides of \(6x+5<4x+19\) gives \(2x<14\). Dividing both sides by the positive number \(2\) yields \(x<7\), so option A is correct. The inequality sign does not reverse because the divisor is positive; reversal is required only when multiplying or dividing by a negative number. Exam tip: Verify the result by testing a value such as \(x=0\) in the original inequality.

Subtracting \(5x\) from both sides gives \(4x-4\leq 20\). Adding 4 and dividing by 4 yields \(x\leq 6\). Thus every integer less than or equal to 6 is a solution, and the greatest integer solution is 6. Exam tip: after solving an inequality, identify the largest integer included in the solution set.

Dividing both sides by the positive number 4 gives \(x+1>7\). Subtracting 1 from both sides yields \(x>6\), so option A is correct. Option C results from incorrectly treating 7 as the value of \(x\). Exam tip: the inequality sign reverses only when multiplying or dividing by a negative number.

Expanding the bracket gives \(6x-15\geq x+10\). Subtracting \(x\) from both sides gives \(5x-15\geq 10\), or \(5x\geq 25\), so \(x\geq 5\). Remember that dividing an inequality by a positive number does not reverse its sign.

The governing rule is that multiplying or dividing an inequality by a negative number reverses its inequality sign. Begin with 15−5x<0. Subtract 15 from both sides: −5x<−15. Now divide both sides by −5; because the divisor is negative, the sign changes from < to >, giving x>3. Thus option A is correct. A quick check supports it: taking x=4 gives 15−20=−5<0, while x=2 gives 15−10=5, which is not less than zero. Options B, C, and D either reverse the direction incorrectly or include the boundary value 3, although 15−5(3)=0 and the original inequality is strict.

Adding 1 to both sides gives \(4x>16\), and dividing by 4 gives \(x>4\). Since the inequality is strict, 4 is not included; the smallest integer greater than 4 is 5. Exam tip: In a strict inequality such as \(>\) or \(<\), the boundary value is excluded.

Subtracting 3 from both sides gives \(9 \geq 3x\). Dividing both sides by the positive number 3 gives \(3 \geq x\), which is equivalent to \(x \leq 3\). Option B incorrectly reverses the inequality sign; division by a positive number does not change its direction. Exam tip: the inequality sign reverses only when multiplying or dividing by a negative number.

The governing concept is maintaining an inequality while performing equivalent operations. Start with (3x−2)/4≥4. Since 4 is positive, multiply both sides by 4 without reversing the sign: 3x−2≥16. Add 2 to both sides to get 3x≥18. Divide by positive 3, again keeping the sign unchanged, and obtain x≥6. Thus option A is correct. The equality is retained throughout, so 6 is included. Options C and D incorrectly replace ≥ by a strict sign, while B reverses the direction without any negative multiplication or division. Substitution verifies the boundary: when x=6, the left side is (18−2)/4=4, satisfying the original inequality.

Subtracting 4 from both sides gives \(10<2x\). Dividing both sides by the positive number 2 gives \(5<x\), which is equivalent to \(x>5\). Therefore, option A is correct. Option C is incorrect because the original inequality does not include equality. Exam tip: The inequality sign remains unchanged when both sides are multiplied or divided by a positive number.