Algebraic Solution of Linear Inequalities in One Variable

Expanding the brackets gives \(8x-6<5x+15\). Subtracting \(5x\) from both sides gives \(3x-6<15\), or \(3x<21\). Dividing by 3 yields \(x<7\), so option A is correct. Since the original inequality is strict, \(x=7\) is not included; therefore, \(x\leq7\) is incorrect. Exam tip: the inequality sign changes only when multiplying or dividing by a negative number, not by a positive number.

Expanding the bracket gives \(6-x-2\leq 2x-5\), or \(4-x\leq 2x-5\). Rearranging the terms yields \(9\leq 3x\). Dividing by the positive number 3 gives \(x\geq 3\), so option B is correct. Option A incorrectly reverses the direction of the inequality. Exam tip: the inequality sign changes only when multiplying or dividing by a negative number.

The least common multiple of the denominators 3 and 6 is 6. Multiplying by 6 does not reverse the inequality because 6 is positive: \(2(2x+1)\leq x+9\), so \(4x+2\leq x+9\). Thus, \(3x\leq7\), giving \(x\leq\frac{7}{3}\). Therefore, option A is correct. Option C is incorrect because it excludes \(x=\frac{7}{3}\), even though equality is allowed in the original inequality. Exam tip: reverse the inequality sign only when multiplying or dividing by a negative number; here, 6 is positive.

Multiplying both sides by 6, which is positive, gives \(2(2x+1)\leq x+9\). Thus, \(4x+2\leq x+9\), so \(3x\leq7\) and \(x\leq\frac{7}{3}\). Option B incorrectly reverses the inequality, while C and D replace the non-strict inequality with an unnecessarily strict one. Exam tip: multiplying or dividing an inequality by a positive number does not change its sign.

We have \(2x+3>15\). Subtracting 3 from both sides gives \(2x>12\). Dividing both sides by the positive number 2 gives \(x>6\). Option C is incorrect because “greater than” is a strict inequality, so 6 is not included. Exam tip: the inequality sign does not reverse when multiplying or dividing by a positive number.

The governing concept is solving a compound inequality by applying the same operation to all three parts. Starting with 3 ≤ (x + 2)/5 ≤ 6, multiply every member by 5. Since 5 is positive, neither inequality sign changes, giving 15 ≤ x + 2 ≤ 30. Subtract 2 throughout the compound inequality to obtain 13 ≤ x ≤ 28. Therefore option A is correct, and both endpoints are included because the original signs are non-strict (≤). Option B stops after multiplication and fails to subtract 2; options C and D result from incorrect arithmetic or shifting the expression in the wrong direction. Substitution also checks the endpoints: x = 13 gives the left equality 3, while x = 28 gives the right equality 6, confirming the closed interval [13, 28].

An open circle means that 3 is not included, while shading to the right represents values greater than 3. Hence, \(x>3\) is correct. In \(x\geq3\), the point 3 would be included. Exam tip: use strict signs with open circles.

Simplifying the left-hand side gives \(3(5-x)+2=17-3x\). Thus, \(17-3x<4x-11\), which leads to \(28<7x\), and hence \(x>4\). Therefore, option C is correct. Option D is incorrect because the original inequality is strict, so \(x=4\) is not included. In exams, remember to reverse the inequality sign whenever you multiply or divide by a negative number.

The least common denominator is 10. Multiplying both sides by 10 does not reverse the inequality sign because 10 is positive: \(2(x-4)-(x+1)\geq 20\). Thus, \(2x-8-x-1\geq 20\), so \(x-9\geq 20\), which gives \(x\geq 29\). Therefore, option B is correct. Exam tip: the inequality sign reverses only when both sides are multiplied or divided by a negative number.

Multiplying by (4) gives (-8\leq 3x+1<28). Subtracting (1) and dividing by (3) gives (-3\leq x<9).

The passing condition is \(4x\ge 60\). Dividing both sides by 4 gives \(x\ge 15\), so the minimum number of correct answers is 15. With 14 correct answers, the student would score only 56 marks, which is below 60. In exams, remember that ‘at least’ indicates \(\ge\).

Adding (5) gives (3x>15) and then (x>5). In exams remember that dividing by a positive number does not reverse the sign.

Subtracting 7 from both sides gives \\(2x\leq8\\). Dividing both sides by the positive number 2 does not reverse the inequality sign, so \\(x\leq4\\). Option B reverses the sign incorrectly, while option C excludes the equality case. Exam tip: the inequality sign changes direction only when multiplying or dividing by a negative number.

Subtracting (5) gives (-2x\geq6) and dividing by (-2) gives (x\leq-3). Do not forget to reverse the sign for a negative coefficient.

Subtracting (2) gives (\frac{x}{3}>4) and multiplying by (3) gives (x>12). Multiplying by a positive number keeps the sign unchanged.

The answer is option A, x≤11. The inequality says that half of x−1 is at most 5. Since the denominator 2 is positive, multiplying both sides by 2 does not change the inequality sign: x−1≤10. Adding 1 to both sides gives x≤11. This includes 11 and every smaller number. Option A is correct because it states exactly this solution set. Option B, x≥11, reverses the direction without any valid reason; the sign changes only when multiplying or dividing by a negative number. Option C, x≤9, comes from an arithmetic mistake, because adding 1 to 10 gives 11, not 9. Option D, x≥9, has both the wrong boundary and the wrong direction. A quick check confirms the result: x=11 gives (11−1)/2=5, which satisfies the “less than or equal to” sign, while x=12 gives 5.5 and does not satisfy it. Remember: multiplying by a positive number keeps the inequality sign unchanged.

Multiplying by (5) gives (2x+3\geq15) and then (2x\geq12) gives (x\geq6). A positive denominator does not reverse the sign.

The answer is option A, x>3. Begin with (7−3x)/2<−1. Multiply both sides by 2. Because 2 is positive, the inequality sign stays the same: 7−3x<−2. Subtract 7 from both sides: −3x<−9. Now divide by −3. Division by a negative number reverses the sign, so x>3. Option A is correct. Option B, x<3, forgets to reverse the sign when dividing by −3. Option C, x>−3, has the wrong boundary because −9 divided by −3 is 3, not −3. Option D, x<−3, has both the wrong direction and the wrong boundary. Check x=4: (7−12)/2=−5/2, which is less than −1, so x=4 works. Check x=3: (7−9)/2=−1, but the inequality is strict, so equality is not accepted. Thus the solution is every number greater than 3, not 3 itself. The key exam rule is that division or multiplication by a negative number reverses an inequality sign.

Bringing like terms together gives (3x\leq15) so (x\leq5). Keep variable terms on one side carefully.

Subtracting \(2x\) from both sides gives \(4x+1>13\). Subtracting \(1\) and then dividing by the positive number \(4\) gives \(x>3\), so option A is correct. Option C is incorrect because it includes \(x=3\), whereas the strict sign \(>\) excludes equality. Exam tip: The inequality sign reverses only when multiplying or dividing by a negative number; that does not occur here.

Expanding gives \(3x-6\geq2x+5\) and hence \(x\geq11\). Watch the signs while opening brackets.

The governing concept is solving a linear inequality using the distributive law and collecting like terms. Expand the left side: 2(3x + 1) = 6x + 2. The inequality becomes 6x + 2 < 5x + 9. Subtract 5x from both sides to get x + 2 < 9, and then subtract 2 to obtain x < 7. Therefore option A is correct. No sign reversal occurs because the variable is not divided by a negative number; effectively, the final coefficient of x is the positive number 1. Options B and D reverse the inequality direction, while options C and D also incorrectly change the sign of the boundary value. A quick check confirms the result: x = 0 satisfies 2 < 9, whereas x = 8 gives 50 < 49, which is false.

Expanding the brackets gives \(5x+10-3\leq 2x+16\), or \(5x+7\leq 2x+16\). Subtracting \(2x\) from both sides yields \(3x\leq 9\). Dividing by the positive number \(3\) gives \(x\leq 3\), so option A is correct. Option B incorrectly reverses the inequality sign; the sign changes only when multiplying or dividing by a negative number. Exam tip: always check whether the number used to divide an inequality is positive or negative.

The left side becomes (5+2x) so (5+2x>3x-4) gives (x<9). Signs change while opening a negative bracket.

Subtracting (3) from all parts gives (-1<x\leq5). In a compound inequality apply the same operation to every part.