Algebraic Solution of Linear Inequalities in One Variable

Correct answer: B. (x>\frac{1}{2})

Multiplying by positive (12) gives (4x-8+6x+3<10x-14). This reduces to (-5<-14), which is false, so there is no solution.

Correct answer: A. no solution

After multiplying by positive (12), variable terms cancel and (-5<-14) remains. A false final statement gives an empty solution set.

Correct answer: B. \({-2,-1,0,1,2}\)

The inequality gives (-6\le3x<12), hence (-2\le x<4). For integers, the correct list is ({-2,-1,0,1,2,3}).

Correct answer: C. (a=3)

We get ((a-2)x>a-4); for (x>3), (\frac{a-4}{a-2}=3) and (a-2>0) are needed. This gives (a=1), which violates the direction condition, so no option is correct.

Correct answer: A. \(a=1\)

Rearranging the inequality gives \((a-2)x>a-4\). For \(a=1\), this becomes \(-x>-3\). Dividing by a negative number reverses the inequality sign, so \(x<3\). Hence, \(a=1\) is correct. For instance, \(a=3\) gives \(x>-1\), not the required solution. Exam tip: Always reverse an inequality sign when multiplying or dividing by a negative quantity.

Correct answer: A. \(x>-\frac{7}{2}\)

Since \(p>0\), dividing by \(p\) does not reverse the inequality: \(-(4x+1)<2(3-x)\). Thus, \(-4x-1<6-2x\), so \(-7<2x\), giving \(x>-\frac{7}{2}\). Option B has the inequality direction reversed. Exam tip: reverse an inequality only when multiplying or dividing by a negative quantity.

Correct answer: A. (x\ge\frac{7}{2})

Multiplying by positive (10) gives (2(2x+3)-5(x-4)\le7-x). Simplification gives (26-x\le7-x), which is false, so no solution exists.

Correct answer: A. \(-\frac{7}{2}\le x<\frac{9}{2}\)

Multiply every part of the compound inequality by the positive number 4, so the inequality signs remain unchanged: \(-12\le 2x-5<4\). Adding 5 throughout gives \(-7\le 2x<9\). Dividing by 2 gives \(-\frac{7}{2}\le x<\frac{9}{2}\). Therefore, option A is correct. Option B incorrectly reverses the inclusive and exclusive endpoint signs. Exam tip: Inequality signs reverse only when multiplying or dividing by a negative number.

Correct answer: B. ( \frac{70}{3}\le s\le\frac{160}{3})

The direct answer is option B: \(\frac{70}{3}\le s\le\frac{160}{3}\). The compound inequality means both bounds must hold: \(40\le\frac{3s+10}{2}\le85\). Multiply every part by positive 2: \(80\le3s+10\le170\). Subtract 10 throughout: \(70\le3s\le160\). Divide by positive 3: \(\frac{70}{3}\le s\le\frac{160}{3}\). Option A, \(20\le s\le50\), does not come from the correct subtraction and division. Option B is exactly the derived interval. Option C has both endpoints shifted upward by \(10/3\), so it is wrong. Option D is a rough interval and fails the exact boundary calculation. Both endpoints are included because the original signs are \(\le\), not strict signs. Keep the order of a compound inequality while performing the same operation on all three parts.

Correct answer: A. \(x\leq 13\)

Multiply both sides by 15. Since 15 is positive, the inequality sign does not change: \(5x+10\geq 6x-3\). Rearranging gives \(10+3\geq 6x-5x\), so \(13\geq x\), i.e. \(x\leq 13\). The option \(x\geq 13\) incorrectly reverses the direction of the inequality. Exam tip: reverse an inequality sign only when multiplying or dividing by a negative number.

Correct answer: A. (x<\frac{39}{7})

Multiplying by positive (8) gives (64-2(3x-2)>16+x+6). This gives (68-6x>22+x), so (x<\frac{46}{7}).

Correct answer: B. \(x<\frac{46}{7}\)

Multiplying both sides by the positive number \(8\) does not change the inequality sign: \(68-6x>22+x\). Thus, \(46>7x\), so \(x<\frac{46}{7}\). Option A has the inequality sign reversed. Exam tip: the sign reverses only when multiplying or dividing an inequality by a negative number.

Correct answer: A. \(x\le -2\)

Opening the brackets gives \(10-5x\ge 12-3x+2\), or \(10-5x\ge14-3x\). Subtracting 10 from both sides and adding \(3x\) gives \(-2x\ge4\). On dividing by \(-2\), the inequality sign reverses, so \(x\le-2\). The option \(x\ge-2\) results from the common mistake of not reversing the sign. Exam tip: always reverse an inequality when multiplying or dividing by a negative number.

Correct answer: A. \(x>-1.6\)

Multiply both sides of the inequality by 10 to remove the decimals: \(2x+15>7-3x\). Moving \(3x\) to the left and 15 to the right gives \(5x>-8\). Since 5 is positive, dividing by 5 does not reverse the inequality sign: \(x>-\frac{8}{5}=-1.6\). Therefore, option A is correct. Option B incorrectly reverses the inequality sign. Exam tip: An inequality sign stays unchanged when multiplying or dividing by a positive number.

Correct answer: A. (x\ge\frac{3}{5}) and (x<\frac{21}{2})

The left inequality gives (18-3x\le2x+9), hence (x\ge\frac{9}{5}); the right gives (x<\frac{21}{2}). The correct intersection is (x\ge\frac{9}{5}) and (x<\frac{21}{2}).

Correct answer: A. \(k=-1\)

Rearranging the inequality gives \((k-2)x\le 9\). For the solution to be \(x\ge -3\), the coefficient of \(x\) must be negative. Using the boundary value \(x=-3\), we get \((k-2)(-3)=9\), so \(k-2=-3\) and hence \(k=-1\). If \(k=1\), the solution is \(x\ge -9\), so it is not correct. Exam tip: when dividing an inequality by a negative number, reverse the inequality sign.

Correct answer: B. 4

Multiplying both sides by the positive number 2 does not reverse the inequality sign: \(x-a\le 6\). Hence, \(x\le a+6\). Since the given solution is \(x\le 10\), we get \(a+6=10\), so \(a=4\). If 6 were chosen, the boundary would be \(x\le 12\), not the given one. Exam tip: Multiplying or dividing an inequality by a positive number leaves its sign unchanged.

Correct answer: B. \(-3<x\le3\)

Multiplying by positive (2) gives (-8<1-3x\le10). From (-9<-3x\le9), reversing signs gives (-3\le x<3).

Correct answer: D. infinitely many

The first part gives (x>-\frac{2}{3}), and the second gives (x\le10). Integer solutions are (0) through (10), totaling (11).

Correct answer: C. (11)

Solving both parts gives (x>-\frac{2}{3}) and (x\le10). Thus integers are (x=0,1,\ldots,10), totaling (11).

Correct answer: B. (x\le-1)

Direct answer: option B, \(x\le-1\). Begin with \(6-2x\ge x+9\). Subtract \(x\): \(6-3x\ge9\). Subtract 6: \(-3x\ge3\). Divide by \(-3\), reversing the inequality: \(x\le-1\). Option B is correct and includes equality because the original sign is \(\ge\). Option A has the direction reversed. Option C has the wrong boundary and direction. Option D has the wrong boundary, although its direction is not the needed one either. Checking \(x=-1\) gives 8 on both sides, so the endpoint belongs. Remember: a negative divisor flips \(\ge\) into \(\le\).

Correct answer: A. \(x<-6\)

Given \(9x-4(2x+1)>3x+8\), expanding the bracket gives \(9x-8x-4>3x+8\), or \(x-4>3x+8\). Subtracting \(3x\) from both sides gives \(-2x-4>8\), so \(-2x>12\). On dividing by \(-2\), the inequality sign reverses, giving \(x<-6\). The close distractor \(x>-6\) results from forgetting to reverse the sign when dividing by a negative number. Exam tip: always reverse an inequality sign when multiplying or dividing both sides by a negative quantity.

Correct answer: A. \(x\ge 7\)

Multiplying both sides by the positive number \(9\) does not reverse the inequality: \(3(4x-1)-(x+2)\ge 9x+9\). Simplifying gives \(11x-5\ge 9x+9\), so \(2x\ge 14\). Hence, \(x\ge 7\). The close distractor \(x\le 7\) would result from incorrectly reversing the sign; no multiplication or division by a negative number occurs here. Exam tip: always check the sign of the number when multiplying or dividing an inequality.

Correct answer: A. \(x\le-\frac{1}{2}\)

Expanding the left-hand side gives \(2(1-3x)=2-6x\), and expanding the right-hand side gives \(5-4(2x+1)=1-8x\). Thus, \(2-6x\le1-8x\). Adding \(8x\) to both sides gives \(2+2x\le1\), so \(2x\le-1\). Dividing by 2 gives \(x\le-\frac{1}{2}\). The option \(x\ge-\frac{1}{2}\) is incorrect because no division or multiplication by a negative number occurs, so the inequality sign does not reverse. Exam tip: reverse an inequality sign only when multiplying or dividing by a negative number.

Correct answer: C. all real (x)

After simplification, a true statement like (1>0) remains. Therefore every real (x) satisfies the inequality.