Algebraic Solution of Linear Inequalities in One Variable

Dividing every part of \(12<4x\leq 28\) by the positive number \(4\) gives \(3<x\leq 7\). Since the divisor is positive, the inequality signs do not reverse. Option B incorrectly reverses the endpoint inclusions. Exam tip: reverse an inequality sign only when multiplying or dividing by a negative number.

Expanding the brackets gives \(4x+8-3x+15\geq 20\), so \(x+23\geq 20\). Subtracting 23 from both sides yields \(x\geq -3\), making option A correct. Option B reverses the inequality incorrectly, while options C and D result from an incorrect subtraction. Exam tip: when removing a minus sign before a bracket, change the sign of every term inside that bracket.

From the first inequality, \(x-2<4\) gives \(x<6\). From the second, \(x+5\geq 1\) gives \(x\geq -4\). Therefore, the common solution is \(-4\leq x<6\), written in interval notation as \([-4,6)\). Remember that the square bracket includes \(-4\), while the parenthesis excludes \(6\).

‘Not less than’ means ‘greater than or equal to’, so the inequality is \(2x-5\geq x+3\). Subtracting \(x\) from both sides gives \(x-5\geq 3\), and adding 5 gives \(x\geq 8\). Option B is incorrect because it uses the strict inequality \(>\), which excludes equality. Exam tip: phrases such as ‘at least’ and ‘not less than’ correspond to \(\geq\).

Simplifying the inequality gives \(4x-8>3x+7\), so \(x>15\). The values 16, 20, and 25 are greater than 15 and are therefore solutions, whereas \(x=14\) does not satisfy the condition. Exam tip: with a strict sign \(>\), the boundary value 15 itself is also excluded.

Expanding the brackets gives \(3x+6-5=3x+1\) on the left and \(2x-2+8=2x+6\) on the right. Thus, \(3x+1\geq 2x+6\), which simplifies to \(x\geq 5\). Option B reverses the inequality without justification, while 9 results from incorrect simplification of the constants. Exam tip: reverse the inequality sign only when multiplying or dividing by a negative number.

Option B gives \(3x+1>25\), so \(3x>24\) and hence \(x>8\). Dividing by the positive number 3 keeps the inequality sign unchanged. The other options give, respectively, \(x\geq 8\), \(x<8\), and \(x\leq 8\).

Subtracting \(1.5\) from both sides gives \(0.2x\leq 1.6\). Dividing by the positive number \(0.2\) yields \(x\leq 8\). The inequality sign does not reverse because the divisor is positive, so \(x\geq 8\) is incorrect. Exam tip: reverse the inequality sign only when multiplying or dividing by a negative number.

Adding 3 to both sides of \(2a-3<7\) gives \(2a<10\). Dividing both sides by the positive number 2 yields \(a<5\), so option A is correct. Option B incorrectly reverses the inequality; the sign reverses only when multiplying or dividing by a negative number. Exam tip: always reverse the inequality sign when dividing by a negative quantity.

Rearranging gives \(6-4x\geq 2x-12\Rightarrow 18\geq 6x\). Dividing both sides by the positive number 6 yields \(3\geq x\), or \(x\leq 3\). 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 divisor is positive or negative before deciding the direction of the inequality.

Given \(3x-7>11\), add 7 to both sides to get \(3x>18\). Dividing by the positive number 3 does not reverse the inequality, so \(x>6\). The option \(x\geq6\) is incorrect because equality is not included; substituting \(x=6\) gives \(11>11\), which is false. Exam tip: an inequality sign changes direction only when multiplying or dividing by a negative number.

(-2x\le 8) and dividing by a negative reverses the sign so (x\ge -4). Watch the sign reversal in exams.

Given \(\frac{x}{3}+4<7\), subtracting 4 from both sides gives \(\frac{x}{3}<3\). Multiplying both sides by the positive number 3 does not change the inequality sign, so \(x<9\). The option \(x\leq 9\) is incorrect because the original inequality is strict. Exam tip: Reverse an inequality sign only when multiplying or dividing by a negative number.

From (2x-5\ge 12), (2x\ge 17), so (x\ge \frac{17}{2}). Multiplying by a positive denominator does not change the sign.

On expanding the bracket, \(7-3x-3>2x-6\), so \(4-3x>2x-6\). Adding \(3x\) and \(6\) to both sides gives \(10>5x\). Since 5 is positive, dividing by 5 does not reverse the inequality sign, giving \(x<2\). The option \(x\leq2\) is incorrect because the original inequality is strict and does not include equality. Exam tip: Reverse an inequality sign only when multiplying or dividing by a negative number.

Subtracting \(2x\) from both sides gives \(5<x-4\). Adding \(4\) to both sides gives \(9<x\), or \(x>9\). The option \(x\geq 9\) is not correct because the original inequality is strict; at \(x=9\), both sides are equal. Exam tip: Reverse an inequality sign only when multiplying or dividing by a negative number.

The direct answer is option D: \(x\ge\frac{15}{2}\). Expand the right side: \(2(2x+7)=4x+14\). Thus \(6x-1\ge4x+14\). Subtract \(4x\): \(2x-1\ge14\). Add 1: \(2x\ge15\). Divide by positive 2, so the inequality sign stays the same: \(x\ge15/2\). Option A reverses the direction and has the wrong boundary. Option B uses a strict greater-than sign, although equality is allowed. Option C has the wrong numerator, 13 instead of 15. Option D includes equality and has the correct value. Sign reversal is needed only when multiplying or dividing by a negative number.

The correct answer is option A: x ≥ −13. Because 5 and 3 are positive, multiply the inequality by 15 without changing its direction: 3(x − 2) ≤ 5(x + 4). Expand: 3x − 6 ≤ 5x + 20. Subtract 3x: −6 ≤ 2x. Subtract 20: −26 ≤ 2x. Divide by positive 2: −13 ≤ x, which is the same as x ≥ −13. Option A matches this closed solution because equality is included. Option B, x ≤ −13, reverses the direction and fails for x = 0. Option C, x ≤ 13, has the wrong sign and wrong boundary. Option D, x > −13, wrongly excludes x = −13, even though equality satisfies the original inequality. Check: at x = −13 both sides are −3, so equality is valid. Memory cue: multiplying by a positive number keeps the inequality sign; dividing by a negative number reverses it.

(2(3x+1)>x-5) gives (6x+2>x-5), so (x>-\frac{7}{5}). Multiply both sides correctly while clearing denominators.

(-0.3x<-0.9), and division by a negative gives (x>3). The sign reversal rule also applies with decimals.

(-4x\ge 16), and dividing by a negative gives (x\le -4). Check the inequality sign in the final step.

(-5x<-10), so reversing the sign gives (x>2). Always reverse the sign when dividing by a negative.

(7-8x\le 3x-4) gives (11\le 11x), hence (x\ge 1). Be careful with signs while expanding a negative multiple.

Subtracting (1) from all parts gives (-4<2x\le 8), so (-2<x\le 4). Apply the same operation to all three parts.

Adding (2) to all parts gives (6\le 3x<15), so (2\le x<5). Keep open and closed endpoints correct.