Algebraic Solution of Linear Inequalities in One Variable

The direct answer is A: \(x\le13\). Begin with \(x-4\le9\). To remove -4 from the left side, add 4 to both sides: \(x-4+4\le9+4\). This simplifies to \(x\le13\). Addition preserves the direction of an inequality, so the symbol does not change. Option A is correct and includes x=13 because the original sign is \(\le\). Option B, \(x\ge13\), reverses the relation and would include values larger than 13, which do not work; for x=14, the original statement becomes 10\le9, false. Option C, \(x<13\), wrongly excludes x=13, which actually gives 9\le9. Option D, \(x>13\), is also the wrong direction and excludes all valid smaller values. Important rule: adding or subtracting the same number does not reverse the sign; multiplying or dividing by a negative number does. Exam cue: keep the inequality sign unchanged when shifting terms by addition or subtraction.

The governing concept is that subtracting the same quantity from both sides preserves an inequality, and division by a positive number preserves its direction. Starting with 5x + 2 ≥ 17, subtract 2 from both sides to get 5x ≥ 15. Divide by 5, which is positive, and obtain x ≥ 3. The equality sign remains because x = 3 gives 5(3) + 2 = 17, so the original statement is true at the boundary. Therefore option A is correct. Option C incorrectly changes the inclusive relation to a strict one and excludes x = 3. Options B and D reverse the direction, placing the solution on the wrong side of the boundary. In interval notation, the solution is [3, ∞).

Adding 7 to both sides gives \(4x<16\). Dividing by 4 yields \(x<4\), and the inequality sign remains unchanged because 4 is positive. Therefore, option A is correct. The distractor \(x\le4\) is incorrect because the original inequality is strict and does not include equality. In an exam, always check whether division by a negative number requires reversing the inequality sign.

Subtracting (7) gives (-2x\le -6), then division by (-2) reverses the sign to (x\ge 3). Be careful with a negative coefficient.

For the inequality \(2x+3<x+8\), subtracting \(x\) from both sides gives \(x+3<8\). Subtracting \(3\) from both sides then gives \(x<5\), so option A is correct. Option B incorrectly reverses the inequality sign; no multiplication or division by a negative number has occurred. Exam tip: Reverse the inequality sign only when multiplying or dividing by a negative number.

Subtracting \\(2x\\) from both sides gives \\(4x-1\ge 11\\). Adding 1 to both sides gives \\(4x\ge 12\\), and dividing by 4 gives \\(x\ge 3\\). Since 4 is positive, the inequality sign does not change. Remember that \\(x=3\\) is also included, so \\(x>3\\) is incorrect.

Dividing both sides by the positive number 3 gives \\(x+2>5\\). Subtracting 2 from both sides gives \\(x>3\\), so option A is correct. Option B incorrectly reverses the inequality sign; division by a positive number does not reverse it. Exam tip: Reverse the inequality sign only when multiplying or dividing by a negative number.

The governing concept is equivalence in solving a linear inequality: the same operation may be applied to both sides, and division by a positive number leaves the inequality sign unchanged. Divide 2(x − 5) ≤ 8 by 2 to obtain x − 5 ≤ 4. Adding 5 to both sides gives x ≤ 9. Therefore option A is correct. The endpoint 9 must be included because the original relation is ≤; direct substitution gives 2(9 − 5) = 8, which is true. Options B and D place the solution on the wrong side of 9, while option C incorrectly excludes the valid boundary value. The complete solution is every real number less than or equal to 9.

The direct answer is A: \(x\ge9\). Start with \((x-1)/4\ge2\). Multiply both sides by 4. Since 4 is positive, the inequality direction remains unchanged: \(x-1\ge8\). Add 1 to both sides: \(x\ge9\). Thus A is correct, and x=9 is included because the original sign is \(\ge\); substitution gives \((9-1)/4=2\). Option B, \(x\le9\), reverses the direction and is false for x=10, which satisfies the original inequality. Option C, \(x>9\), wrongly excludes the valid boundary x=9. Option D, \(x<9\), has both the wrong direction and excludes valid larger values. The denominator 4 is positive, so no sign reversal occurs. If the multiplier or divisor had been negative, the sign would have to reverse. Memory cue: positive multiplication keeps the sign; negative multiplication flips it.

Multiplying both sides by 5 gives \(2x+1\le 15\). Thus, \(2x\le 14\), so \(x\le 7\). Since 5 is positive, the inequality sign remains unchanged. Option B reverses the sign incorrectly, while option C wrongly excludes \(x=7\). In an exam, check whether the multiplier is positive or negative before deciding whether to reverse the inequality sign.

Multiplying both sides by (-4) reverses the sign and gives (x>-8). Reverse the sign when multiplying by a negative number.

The direct answer is A: \(x<3\). Solve step by step: \(9-3x>0\). Subtract 9 from both sides to get \(-3x>-9\). Now divide by -3. Because the divisor is negative, the inequality sign must reverse: \(x<3\). Therefore A is correct. A direct check confirms it: if x=2, then \(9-3(2)=3>0\). Option B, \(x>3\), has the reversed direction and fails for x=4, giving \(9-12=-3>0\), which is false. Option C, \(x\le3\), incorrectly includes x=3; at x=3 the expression is 0, and 0>0 is false. Option D, \(x\ge3\), also includes the invalid boundary and values above it. The strict sign > means equality is not allowed. The key rule is that multiplying or dividing by a negative number reverses the inequality. Memory cue: “negative crossing flips the sign.”

The governing concept is collecting variable terms on one side and constants on the other without changing the inequality through addition or subtraction. From 2x + 5 ≤ x + 12, subtract x from both sides to get x + 5 ≤ 12. Next subtract 5 from both sides, giving x ≤ 7. Therefore option A is correct. Equality is included because x = 7 changes the original statement to 14 + 5 ≤ 7 + 12, or 19 ≤ 19, which is true. Options C and D wrongly exclude or reverse the boundary, whereas B gives the opposite direction. All real numbers up to and including 7 satisfy the inequality.

The governing concept is that multiplying or dividing an inequality by a negative number reverses its direction. Begin with 10 − x < 4. Subtract 10 from both sides to obtain −x < −6. To isolate x, divide by −1; because the divisor is negative, the sign reverses, giving x > 6. Thus option A is correct. The reversal is essential: forgetting it would produce x < 6, option B. Options C and D also mishandle the strict sign or direction. A quick check confirms the result: x = 7 gives 3 < 4, true, while x = 5 gives 5 < 4, false. Hence precisely numbers greater than 6 satisfy the inequality.

The governing concept is algebraic equivalence for inequalities: expand brackets, collect like terms, and divide by a positive coefficient without reversing the sign. Expanding gives 4x + 4 ≥ 2x + 10. Subtract 2x from both sides to obtain 2x + 4 ≥ 10, and subtract 4 to get 2x ≥ 6. Dividing by positive 2 yields x ≥ 3, so option A is correct. The equality is retained because x = 3 makes both sides equal to 16. Option C incorrectly excludes this valid boundary, while B and D reverse the direction. The complete solution is every real number at least 3.

The governing concept is that equivalent algebraic operations preserve a linear inequality, except that division by a negative number would reverse its sign. First expand the bracket: 5x − 10 < 3x + 4. Subtract 3x from both sides to get 2x − 10 < 4. Add 10, giving 2x < 14. Since the divisor 2 is positive, division produces x < 7 without reversing the sign. Hence option A is correct. The strict inequality excludes 7: when x = 7, the original statement becomes 5(5) < 25, or 25 < 25, which is false. Option B reverses the direction, and options C and D incorrectly include the endpoint.

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

Dividing by (2) gives (x\ge 5), so the least integer is (5). The equality sign includes the boundary value.

The governing concept is the algebraic solution of a strict linear inequality and its representation in interval notation. Start with 3x + 1 > 10. Subtract 1 from both sides to obtain 3x > 9, and divide by the positive number 3, giving x > 3. Because the inequality is strict, x = 3 is not included in the solution set, so a round or open bracket is required at 3. The solution continues without bound toward positive infinity, which is always written with a parenthesis. Therefore option A, (3, ∞), is correct. Option B incorrectly includes 3, while options C and D describe values less than 3.

The governing concept is converting an algebraic inequality into interval notation while correctly treating its boundary. Starting with 4x − 8 ≤ 0, add 8 to both sides to obtain 4x ≤ 8. Divide by the positive number 4, giving x ≤ 2. Numbers less than 2 and the number 2 itself are included, while numbers greater than 2 are excluded. Therefore the interval is (−∞, 2], which is option A. The square bracket at 2 records inclusion because the relation is ≤. Option B wrongly excludes 2, and C and D describe values to the right of 2 rather than to its left. Infinity always uses a parenthesis, never a square bracket.

The governing concept is solving a linear inequality while preserving the correct order relation. Starting with 5 − x ≥ 2, subtract 5 from both sides to obtain −x ≥ −3. Multiplying or dividing an inequality by −1 reverses its sign, so x ≤ 3. Therefore every value less than or equal to 3 satisfies the inequality, while values greater than 3 do not. Testing the options confirms this: x = 3 gives 2 ≥ 2, x = 2 gives 3 ≥ 2, and x = 1 gives 4 ≥ 2. For x = 4, the left side is 1, and 1 ≥ 2 is false. Hence option A is the only value that does not satisfy the inequality. The other options are distractors because they lie within the solution set.

To isolate (x), subtract (9) from both sides. Choosing the inverse operation saves time in exams.

The governing concept is the order property of inequalities: multiplying or dividing both sides by a negative number reverses the inequality sign. In −5x ≤ 20, division by −5 gives x ≥ −4, not x ≤ −4. The reversal is necessary because the order of numbers changes when both are multiplied by a negative quantity; for example, 2 > 1 becomes −2 < −1. Adding or subtracting the same number from both sides does not reverse the sign, so options C and D state incorrect rules. Option B is also incorrect because it ignores the negative divisor. Thus option A correctly identifies the rule. A quick check confirms the result: x = −4 gives −5(−4) = 20, and larger x values make −5x smaller, so they satisfy −5x ≤ 20.

Dividing both sides by (3) gives (4\le x), that is (x\ge 4). Write the final answer starting with (x) in standard form.

Divide both sides by \(6\) (positive number, so the inequality sign does not change): \(18>6x\Rightarrow 18/6>x\Rightarrow 3>x\). Hence \(x<3\).