Algebraic Solution of Linear Inequalities in One Variable

Simplifying the inequality gives \(11-10+2x>3x+4\), or \(1+2x>3x+4\). Subtracting \(2x\) gives \(1>x+4\), and subtracting 4 gives \(-3>x\). Hence, \(x<-3\) is the correct solution. The close distractor \(x\leq -3\) is incorrect because the inequality is strict: at \(x=-3\), both sides are equal, so this value is not included. Exam tip: Reverse an inequality sign only when multiplying or dividing by a negative number.

Multiplying by 12 does not reverse the inequality because 12 is positive: \(4(x+7)-3(2x-1)\ge 0\). This gives \(4x+28-6x+3\ge 0\), or \(31-2x\ge 0\). Hence \(2x\le 31\), so \(x\le \frac{31}{2}\). The closest distractor is \(x\ge \frac{31}{2}\), but the negative coefficient of \(x\) must be handled carefully while rearranging. Exam tip: reverse an inequality sign only when multiplying or dividing by a negative number.

Expanding the right-hand side gives \(2(3x-4)=6x-8\). Thus, the inequality becomes \(6x-5<6x-8\). Subtracting \(6x\) from both sides gives \(-5<-8\), which is false. Hence, no real value of \(x\) satisfies the inequality, so the solution set is \(\varnothing\). Options such as \(x<3\) or \(x>3\) may result from not checking the remaining numerical inequality after the \(x\)-terms cancel. Exam tip: when identical variable terms cancel, always test whether the resulting numerical statement is true or false.

Multiplying both sides by the positive number 3 gives \(24-6x\le x+9\). Thus, \(15\le 7x\), so \(x\ge \frac{15}{7}\). Option C excludes the boundary value \(\frac{15}{7}\), but equality is allowed in the given inequality. Exam tip: Reverse an inequality sign only when multiplying or dividing by a negative number.

Multiplying all three parts by 2 gives \(-12\le 4-x<6\). From the left part, \(-12\le4-x\), we get \(x\le16\); from the right part, \(4-x<6\), we get \(x>-2\). Their common solution is \(-2<x\le16\). Option C incorrectly includes \(-2\): substituting \(x=-2\) makes the right-hand condition \(3<3\), which is false. Exam tip: split a compound inequality into two inequalities and take their intersection.

The direct answer is option B: \(x\le -13\). First solve each inequality separately. From \(\frac{x-2}{3}\ge\frac{2x+1}{5}\), multiply by the positive number 15, so the sign does not change: \(5(x-2)\ge3(2x+1)\). Thus \(5x-10\ge6x+3\), so \(-13\ge x\), or \(x\le-13\). The second inequality gives \(x+4\le10\), hence \(x\le6\). “And” means both conditions must hold, so we take their intersection. Every number at most \(-13\) is also at most 6; therefore the combined solution is \(x\le-13\). Option A, \(x\le6\), gives only the second condition and includes values that fail the first. Option B is correct because it satisfies both. Option C, \(-13<x\le6\), wrongly excludes \(-13\) and includes numbers greater than \(-13\). Option D, \(x\ge-13\), reverses the direction and includes unsuitable large values. Remember: for “and”, use the common overlap; multiplying an inequality by a positive number keeps its direction.

Multiplying both sides by the positive number 6 does not change the inequality sign: \(5x+2<3(x-3)+24=3x+15\). Hence \(2x<13\), so \(x<\frac{13}{2}\). Option B incorrectly includes equality; since the original inequality is strict, \(x=\frac{13}{2}\) is not a solution. Exam tip: Reverse an inequality sign only when multiplying or dividing by a negative number.

Simplify the left-hand side: \(8-3(2-x)=8-6+3x=2+3x\). Thus, \(2+3x\le 5x-10\), which gives \(12\le 2x\). Dividing by 2 gives \(x\ge 6\). The close distractor \(x>6\) wrongly excludes \(x=6\), which must be included because the inequality contains \(\le\). Exam tip: An inequality sign does not reverse when dividing by a positive number.

The first inequality gives (x>4), and the second gives (x<-7). For an OR condition, take the union of both solution sets.

Multiplying both sides by the positive number 6 does not change the inequality sign: \(2(7-2x)\le x+5\). Thus, \(14-4x\le x+5\), so \(9\le 5x\), giving \(x\ge \frac{9}{5}\). The option \(x>\frac{9}{5}\) is incorrect because at \(x=\frac{9}{5}\), both sides are equal, and \(\le\) includes equality. Exam tip: Reverse an inequality sign only when multiplying or dividing by a negative number.

Solve both parts of the compound inequality separately. From \(5\le 3x-1\), we get \(6\le 3x\), so \(x\ge 2\). From \(3x-1<2x+11\), we get \(x<12\). Applying both conditions together gives \(2\le x<12\). Option D incorrectly excludes 2 and includes 12. Exam tip: For a compound inequality, solve each condition and then write their common interval.

Simplifying the left side gives \(-6x+8+7\ge x-6\), or \(-6x+15\ge x-6\). Adding \(6x\) and \(6\) to both sides gives \(21\ge 7x\). Dividing by 7, we get \(x\le 3\). Although \(x<3\) is a close distractor, \(x=3\) also satisfies the inequality, so equality must be included. Exam tip: Reverse an inequality sign only when multiplying or dividing by a negative number.

Multiplying both sides of the inequality by the positive number 4 gives \(3x+8-2(x-6)>20\). Thus, \(3x+8-2x+12>20\), so \(x+20>20\) and hence \(x>0\). \(x\geq 0\) is incorrect because at \(x=0\), the left-hand side equals 5, which is not greater than 5. Exam tip: multiplying or dividing an inequality by a positive number does not reverse its sign.

Simplify the right-hand side: \(3(3x-1)-2=9x-5\). The inequality becomes \(9x-4\le 9x-5\). Subtracting \(9x\) from both sides gives \(-4\le-5\), which is false. Hence, the inequality is not true for any real value of \(x\), so the solution set is \(\varnothing\). Options such as \(x\le1\) or \(x\ge1\) would arise only if an \(x\)-term remained after simplification. Exam tip: when variable terms cancel, check whether the remaining numerical statement is true or false.

Multiplying by 6 gives \(3x-(x-3)\le 24\). Thus, \(2x+3\le 24\), so \(2x\le 21\) and \(x\le \frac{21}{2}\). Option B excludes \(\frac{21}{2}\), but this boundary value must be included because the inequality uses \(\le\). Exam tip: multiplying an inequality by a positive LCM clears denominators without reversing the inequality sign.

The first inequality gives \(x\ge 4\), and the second gives (x<3). Both conditions cannot hold together.

From the first part, \(7-4x<3(2-x)\) becomes \(7-4x<6-3x\), which gives \(x>1\). The second part, \(3(2-x)\le12\), becomes \(6-3x\le12\), giving \(x\ge-2\). The intersection of these two conditions is \(x>1\), so option D is correct. Remember to reverse the inequality sign when multiplying or dividing by a negative number.

Since 7 and 3 are positive, multiply both sides by 21 without reversing the inequality sign: \(3(2x+1)\ge 7(x-5)\). Thus, \(6x+3\ge 7x-35\), which gives \(38\ge x\), or \(x\le 38\). The close distractor \(x<38\) is incorrect because at \(x=38\), both sides are equal, so 38 is included. Exam tip: An inequality sign changes direction only when multiplying or dividing by a negative number.

Simplifying the left-hand side gives \(3(1-x)-2(4-3x)=3-3x-8+6x=3x-5\). Thus, \(3x-5>5x-10\), which gives \(5>2x\). Since 2 is positive, dividing both sides by 2 does not reverse the inequality sign, so \(x<\frac{5}{2}\). The option \(x\le \frac{5}{2}\) is incorrect because equality is not included in the original inequality. Exam tip: while expanding brackets, multiply \(-2\) by every term in the second bracket.

Using the LCM of the denominators, 6, multiply the inequality by 6 to get \(3(x-1)+2(x-2)+(x-3)\ge 24\). This simplifies to \(6x-10\ge 24\), so \(6x\ge 34\), giving \(x\ge \frac{17}{3}\). The option \(x>\frac{17}{3}\) is incorrect because the boundary value \(x=\frac{17}{3}\) also satisfies the inequality. Exam tip: Include the boundary value when the inequality sign is \(\ge\) or \(\le\).

Solve the compound inequality in two parts. From \(-5<2-\frac{x}{3}\), we get \(-7<-\frac{x}{3}\); multiplying by \(-3\) reverses the inequality, giving \(x<21\). From \(2-\frac{x}{3}\le4\), we get \(-\frac{x}{3}\le2\), hence \(x\ge-6\). Their common solution is \(-6\le x<21\). Option C has the endpoint signs reversed. Exam tip: always reverse an inequality sign when multiplying or dividing by a negative number.

Multiplying both sides of the inequality by the positive number 4 gives \(8-(3x-1)\ge 2(x+5)\). Thus, \(9-3x\ge 2x+10\), so \(-1\ge 5x\). Dividing by 5 gives \(x\le -\frac{1}{5}\). Option B excludes \(-\frac{1}{5}\), but the original inequality contains \(\ge\). Exam tip: the inequality sign does not reverse when multiplying or dividing by a positive number.

Multiply both sides by the positive number 6 to clear the denominators, so the inequality sign remains unchanged: \(24x+24-2(2x-3)<3(5x+6)\). This simplifies to \(20x+30<15x+18\), giving \(5x<-12\), and hence \(x<-\frac{12}{5}\). Since the original inequality is strict, \(x=-\frac{12}{5}\) is not included. Exam tip: reverse the inequality sign only when multiplying or dividing by a negative number.

Clearing denominators gives (4x-22\le 1), so (x\le \frac{23}{4}). In fractional inequalities, multiply by the LCM first.