Algebraic Solution of Linear Inequalities in One Variable

Correct answer: B. 9

Solve the double inequality before counting integers. Start with −5 < 2x + 3 ≤ 13. Subtract 3 from all three parts: −8 < 2x ≤ 10. Divide all parts by positive 2, so the directions remain unchanged: −4 < x ≤ 5. The integers strictly greater than −4 and less than or equal to 5 are −3, −2, −1, 0, 1, 2, 3, 4, and 5. Counting them gives 9 solutions, so option B is correct. The endpoint −4 is excluded because the left inequality is strict, while 5 is included because the right inequality is non-strict. Option A misses one integer, option C adds an invalid endpoint or count, and option D overcounts the interval. The governing idea is intersection of an interval with the set of integers.

Correct answer: A. \(x<\frac{56}{11}\)

Clear the denominators by multiplying the entire inequality by 10, which is positive, so the inequality direction remains unchanged. We obtain \(40-2(3x+2)>10+5(x-6)\). Expanding gives \(40-6x-4>10+5x-30\), or \(36-6x>5x-20\). Adding \(6x+20\) to both sides yields \(56>11x\). Since 11 is positive, division gives \(x<56/11\). Thus option A is correct. The original version’s explanation incorrectly simplified the constant to 66 and the listed value \(29/11\) was not supported by the algebra. Options B and C reverse the direction or sign, while D uses the reciprocal incorrectly.

Correct answer: A. ∅

Clear all denominators by multiplying the inequality by 12, which is positive and therefore preserves the sign. The left side becomes 3(2x+7)=6x+21. The right side becomes 2(3x−1)+20=6x−2+20=6x+18. Thus the inequality reduces to 6x+21<6x+18. Subtracting 6x gives 21<18, a false statement independent of x. Consequently no real number can satisfy the original inequality, so the solution set is empty and option A is correct. Option B would follow from a true identity after cancellation, while options C and D incorrectly assume a restriction on x despite the variable disappearing and leaving a contradiction.

Correct answer: B. \(x\ge \frac{38}{7}\)

Multiply both sides by 6, the positive least common multiple of the denominators 2 and 3; therefore the inequality sign is preserved. This gives \(36-3(3x-4)\le 2(5-x)\). Expand: \(36-9x+12\le 10-2x\), so \(48-9x\le 10-2x\). Subtracting 10 and adding 9x gives \(38\le 7x\). Since 7 is positive, division yields \(x\ge 38/7\). Therefore option B is correct. The equality case is included because the original symbol is \(\le\), not a strict inequality. Options A and D reverse the direction, while C wrongly excludes the boundary value.

Correct answer: B. 4≤x<9

A compound inequality means that both component inequalities must hold simultaneously. First solve 3x−8<2x+1. Subtract 2x and add 8 to obtain x<9. Next solve 2x+1≤5x−11. Add 11, subtract 2x, and divide by 3: 12≤3x, so x≥4. The required solution is the intersection of x<9 and x≥4, namely 4≤x<9. Therefore option B is correct. The lower endpoint 4 is included because the second relation is ≤, while the upper endpoint 9 is excluded because the first relation is strict. Option A omits valid values from 4 onward, option C omits the valid interval, and option D incorrectly excludes 4 and includes 9.