Algebraic Solution of Linear Inequalities in One Variable

(3x\ge 7) gives (x\ge \frac{7}{3}), so the least integer is (3). Take the first integer above the boundary.

Subtracting 5 from both sides of \(5-x>1\) gives \(-x>-4\). Multiplying both sides by \(-1\) reverses the inequality sign, so \(x<4\). The greatest integer less than 4 is 3; therefore, 3 is correct. Exam tip: Always reverse the inequality sign when multiplying or dividing by a negative number.

Substitute \(x=3\):
- \(x<2\) becomes \(3<2\) (false).
- \(x=3\) is an equality, not the given inequality type.
- \(x>2\) becomes \(3>2\) (true).
- \(x\le 0\) becomes \(3\le 0\) (false).
So \(x>2\) is the true inequality.

The direct answer is option A: x≥0. Begin with x+7≥7. Subtract 7 from both sides: x≥0. Subtraction does not change the inequality direction. A is correct and includes 0, because 0+7=7 and 7≥7 is true. B, x≤0, reverses the direction without a reason and wrongly excludes positive solutions such as x=1. C, x≥14, comes from an incorrect operation; the 7 is removed by subtracting 7, not by adding 7 to obtain 14. It also wrongly excludes x=1. D, x≤14, has the wrong direction and includes negative values such as -1, for which -1+7=6≥7 is false. Equality matters: the original sign is ≥, so the boundary value 0 is included. On a number line, shade 0 and everything to its right. Memory cue: subtract the constant, preserve the sign, and include the endpoint for ≥ or ≤.

Adding (4) to both sides gives (x<3). Even with negative numbers, addition does not change the sign.

Subtracting 6 from both sides of \(x+6>-2\) gives \(x>-2-6=-8\). Therefore, the correct solution is \(x>-8\). The inequality sign does not change when the same number is subtracted from both sides; it reverses only when multiplying or dividing by a negative number. Exam tip: move the constant term to the other side and change its sign.

In (-3x>12), we divide by the negative number (-3), so the sign reverses. This rule is very important in inequalities.

The governing concept is solving a one-variable linear inequality while preserving its direction. Begin with 8 + 2x ≤ 20. Subtract 8 from both sides to isolate the variable term: 2x ≤ 12. Divide both sides by the positive number 2, obtaining x ≤ 6. Since the divisor is positive, the inequality sign remains unchanged; therefore option A is correct. Option B reverses the sign incorrectly. Options C and D use 14, which comes from subtracting in the wrong direction or failing to simplify the variable coefficient. Checking the boundary value x = 6 gives 8 + 12 = 20, and a smaller value such as x = 0 gives 8 ≤ 20, confirming the solution region.

Let the number be (x). Adding (5) gives (x+5), and greater than means (>). Convert words into algebra first.

The statement gives (2x<18), and dividing by (2) gives (x<9). Dividing by a positive number does not change the sign.

After spending, the amount is (x-30), and at least (70) means (\ge 70). In words, at least is written using (\ge).

The cost of four pens is \(4x\) rupees. “Not more than 100” means \(4x\le 100\). Dividing both sides by the positive number 4 gives \(x\le 25\), so option A is correct. In exams, remember that “not more than” indicates the \(\le\) sign, not the strict sign \(<\).

The statement gives (x-8\ge 0), and adding (8) gives (x\ge 8). Use (\ge) for greater than or equal to.

Subtracting (2) from (7x+2\le 23) gives (7x\le 21), and dividing by (7) gives (x\le 3). Division by a positive number does not change the sign.

Subtracting (5) gives (-4x<12), and dividing by (-4) gives (x>-3). The inequality sign reverses when dividing by a negative number.

Multiplying by (4) gives (x+3<8), and subtracting (3) gives (x<5). The sign stays the same when clearing a positive denominator.

The direct answer is option A, x\ge3. Begin with 2x+1\le5x-8. Subtract 2x from both sides: 1\le3x-8. Add 8 to both sides: 9\le3x. Divide by positive 3: 3\le x, which is the same as x\ge3. Option A is correct. Option B, x\le3, reverses the direction without a valid reason; no negative number was used. Option C, x\ge-3, has the wrong boundary because the correct division gives 3, not -3. Option D, x\le-3, has both the wrong sign direction and wrong boundary. Check x=3: left side is 7 and right side is 7, so equality is allowed. Check x=4: 9\le12 is true, while x=2 gives 5\le2, false. Memory cue: only multiplication or division by a negative number reverses an inequality.

The statement gives (x+3\le 16), and subtracting (3) gives (x\le 13). Not more than means (\le).

Subtracting (9) from (9-2x\le 1) gives (-2x\le -8), and dividing by (-2) gives (x\ge 4). The sign reverses when dividing by a negative number.

The governing concept is preservation of an inequality while performing equivalent algebraic operations. Start with 4x − 9 ≤ 2x + 7. Subtract 2x from both sides to obtain 2x − 9 ≤ 7. Add 9 to both sides, giving 2x ≤ 16. Dividing by the positive number 2 does not reverse the inequality sign, so x ≤ 8. Therefore option A is correct. The endpoint 8 is included because the original symbol is ≤; substituting x = 8 gives 32 − 9 = 23 and 16 + 7 = 23. Options B and D reverse the direction incorrectly, while option C excludes the valid endpoint and uses a strict inequality.

The governing idea is to perform the same operation on both sides while preserving the order. Starting from 15 ≤ 4x - 1, add 1 to both sides, obtaining 16 ≤ 4x. Dividing by the positive number 4 does not reverse the inequality, so 4 ≤ x. Written with x on the left, this is x ≥ 4; hence option A is correct. The equality sign remains because the original inequality includes equality. Option B reverses the direction, while options C and D either omit the boundary or point in the wrong direction. Checking x = 4 gives 15 ≤ 15, confirming that 4 is included.

The governing concept is solving an inequality and then testing the proposed values against the resulting condition. Starting with 4x - 6 > 10, add 6 to both sides to obtain 4x > 16. Dividing by the positive number 4 does not change the inequality direction, so x > 4. Among the choices, x = 5 is the only value greater than 4. The value x = 4 is not accepted because the inequality is strict; equality would not satisfy >. The values 3 and 2 are below the required boundary. Direct substitution confirms the result: 4(5) - 6 = 14, and 14 > 10. Therefore option A is the unique correct answer, while the other options fail the original inequality.

The governing concept is solving a linear inequality and observing that division by a negative number reverses its sign. From 12 - 3x ≤ 0, subtract 12 from both sides to get -3x ≤ -12. Dividing by -3 changes ≤ into ≥, giving x ≥ 4. Of the listed values, only x = 4 satisfies this condition; 3, 2 and 1 are all less than 4. Equality is allowed because the original sign is ≤, so the endpoint x = 4 belongs to the solution set. Substitution provides a direct check: 12 - 3(4) = 12 - 12 = 0, and 0 ≤ 0 is true. Therefore option A is correct, whereas each other option makes the original inequality false.

The governing concept is solving a linear inequality while preserving its direction. Start with x/6 − 1 ≥ 2. Add 1 to both sides, giving x/6 ≥ 3. Multiply both sides by 6. Because 6 is positive, the inequality sign remains unchanged: x ≥ 18. Thus every real number greater than or equal to 18 satisfies the original inequality, and 18 itself is included because the symbol is ≥. Option C is correct. Option A and option D reverse the direction incorrectly, while option B excludes 18 and uses a strict inequality. A sign reversal would be required only if we multiplied or divided by a negative number, which does not happen here. On a number line, the answer is shown by a closed point at 18 with shading to the right.

The governing concept is solving a linear inequality and expressing its answer in set-builder notation. Begin with 7x + 2 ≥ 3x − 18. Subtract 3x from both sides to obtain 4x + 2 ≥ −18. Then subtract 2, giving 4x ≥ −20. Dividing by the positive number 4 preserves the inequality sign, so x ≥ −5. In set notation, the solution is {x : x ≥ −5}, which includes −5 because equality is permitted by the original ≥ sign. Therefore option D is correct. Option A reverses the direction, option B incorrectly excludes the endpoint, and option C uses the wrong strict direction. A useful check is x = −5: both sides become −33, so equality holds and −5 must be included.