Algebraic Solution of Linear Inequalities in One Variable

Multiplying by (6) gives (12-2x<3x-6), hence (x>\frac{18}{5}). Move variable terms in the correct direction.

We get (-7\le -3x<9), and dividing by (-3) reverses the order, so (-3<x\le \frac{7}{3}). Division by a negative changes the direction of a compound inequality.

Multiplying first by (2) gives (-8<x+2\le 2), so (-10<x\le 0). A positive multiplier does not change the inequality sign.

Multiplying by (6) gives (4x+3-2(x-2)\ge 3(x+5)), so (x\le -8). After clearing denominators, simplify brackets and signs carefully.

From (-6<5-2x\le 9), we get (-11<-2x\le 4), so (-2\le x<\frac{11}{2}). In compound inequalities, division by a negative reverses the direction.

The direct answer is option A: \(x\ge6\). Start with \(7x-9\ge3x+15\). Subtract \(3x\) from both sides: \(4x-9\ge15\). Add 9: \(4x\ge24\). Divide by positive 4, so the sign does not change: \(x\ge6\). Option A is correct and includes 6 because the original sign is \(\ge\). Option B, \(x\le6\), reverses the direction without a valid reason. Option C, \(x>6\), wrongly excludes 6; checking \(x=6\) gives equality on both sides. Option D, \(x\ge-6\), uses the wrong sign for the answer and includes many values that fail the inequality. The safe method is to collect variable terms on one side and constants on the other. Since the final divisor is positive, no reversal is needed.

(-5x<15), and division by a negative reverses the sign, so (x>-3). Always reverse the sign with a negative coefficient.

Multiplying by (6) gives (2x+8-12\le x-1), so (x\le 3). After clearing fractions, combine all terms carefully.

Expanding the bracket gives \(8x-12>5x+9\). Adding \(5x\) to both sides gives \(3x-12>9\), so \(3x>21\). Since 3 is positive, dividing by 3 does not reverse the inequality sign, giving \(x>7\). The option \(x\geq7\) is incorrect because the original inequality is strict. Exam tip: An inequality sign changes direction only when multiplying or dividing by a negative number.

Multiplying by (4) gives (2(3x-5)\ge x+7+12), so (x\ge \frac{29}{5}). Do not forget to multiply the entire right side.

Simplifying gives (1-2x\le 3x-4), so (5\le 5x) and (x\ge 1). Expand brackets with negative multiplication carefully.

Multiplying by (6) gives (10-8x>x+2), so (x<\frac{8}{9}). Move variables to one side before dividing.

Direct answer: option D, \(x<50\). Multiply the whole inequality by the positive number 10, so its sign does not change: \(10(3x/5+2)<10(x/2+7)\). This gives \(6x+20<5x+70\). Subtract \(5x\): \(x+20<70\). Subtract 20: \(x<50\). Option D is exact. Option A has the inequality direction reversed. Option B uses \(\le\) instead of the required strict \(<\), so it incorrectly includes 50. Option C has an incorrect numerical result; substituting 25 does not mark the boundary. Because 10 is positive, no reversal occurs. Memory cue: reverse an inequality only when multiplying or dividing by a negative number.

Adding (1) to all parts gives (-3\le 2x<10), so (-\frac{3}{2}\le x<5). Keep open and closed endpoints correct.

Multiplying by (5) gives (30-(3x-2)\ge 5), so (x\le 9). Put the full numerator in brackets after subtraction.

On expanding the bracket, \(-2(3-x)=-6+2x\), since \(-2\times(-x)=+2x\). Thus, \(-6+2x+4>x+10\), or \(2x-2>x+10\). Rearranging gives \(x>12\). Therefore, \(x>12\) is the correct option. \(x\ge12\) is incorrect because at \(x=12\) both sides are equal, whereas the inequality is strict. Exam tip: when removing brackets, multiply every term inside by the negative coefficient carefully.

(-0.6x\le -3.6), and division by a negative gives (x\ge 6). The sign reversal rule applies to decimals too.

Multiplying by (10) gives (2x-16+2x+1<30), so (x<\frac{45}{4}). Check signs before combining like terms.

Multiplying by (9) gives (7x+3\ge 6x-3), so (x\ge -6). Multiply the complete numerator while clearing denominators.

(5x<27) gives (x<\frac{27}{5}), so the greatest integer is (5). Check integers near the boundary.

(-4x<-16) gives (x>4), so the least integer is (5). Do not include the boundary in a strict inequality.

The integers are (-3,-2,-1,0,1,2,3), so there are (7) solutions. Include closed endpoints and exclude open endpoints.

(3x\le 18) gives (x\le 6), so natural solutions are from (1) to (6). Count natural numbers from (1) in exams.

The statement translates to \(2x+7>23\). Subtracting \(7\) from both sides gives \(2x>16\). Dividing by \(2\) gives \(x>8\). The option \(x\geq 8\) is incorrect because at \(x=8\), the result is \(23\), not greater than \(23\). Exam tip: the phrase “greater than” always uses \(>\), not \(\geq\).

“At most 250 rupees” means \(15x+40\le250\). Subtracting 40 from both sides gives \(15x\le210\). Since 15 is positive, dividing by 15 does not reverse the inequality sign, so \(x\le14\). The option \(x<14\) is incorrect because \(x=14\) gives a total cost of exactly 250 rupees, which is allowed. Exam tip: translate “at most” as \(\le\).