Linear Inequalities — Important Questions

76

Step 1: Let x = score in third test. Average of three tests = (70 + 80 + x)/3. Step 2: We need average strictly greater than 75: (70 + 80 + x)/3 > 75. Step 3: Multiply both sides by 3: 70 + 80 + x > 225 → 150 + x > 225. Step 4: Subtract 150: x > 75. So x must be STRICTLY greater than 75. Since scores are integers, the MINIMUM integer value satisfying x > 75 is x = 76. Common mistake: Writing x ≥ 75 (which would be for average ≥ 75, not average > 75). The word 'strictly greater' means we cannot include 75.

x < -3 or x > 2

Step 1: For a fraction to be positive, numerator and denominator must have the SAME sign. Critical points are x = 2 (numerator = 0) and x = -3 (denominator = 0). Step 2: Case 1 — Both positive: (x-2) > 0 AND (x+3) > 0 → x > 2 AND x > -3 → x > 2. Step 3: Case 2 — Both negative: (x-2) < 0 AND (x+3) < 0 → x < 2 AND x < -3 → x < -3. Step 4: Combining: x < -3 OR x > 2. At x = 2, the expression equals 0, which does NOT satisfy > 0. At x = -3, the expression is undefined. Common mistake: Students often take -3 < x < 2 (sign between the roots), which actually makes the fraction negative.

68°F < F < 86°F

Step 1: We have 20 < C < 30 (strictly between, so both endpoints excluded). Step 2: Substitute C = (5/9)(F-32) in the inequality: 20 < (5/9)(F-32) < 30. Step 3: Multiply all parts by 9/5 (positive, no sign change): 20 × (9/5) < (F-32) < 30 × (9/5) → 36 < F - 32 < 54. Step 4: Add 32 to all parts: 36 + 32 < F < 54 + 32 → 68 < F < 86. So 68°F < F < 86°F. Note: Both endpoints are STRICTLY excluded because the original inequality used strict < signs. Common mistake: Using ≤ instead of < in the final answer.

3 pairs

Step 1: Let x be the smaller even integer. The consecutive even integers are x and x+2. Step 2: Conditions: (i) x > 5, so the smallest even integer x ≥ 6. (ii) Sum < 23: x + (x+2) < 23 → 2x + 2 < 23 → 2x < 21 → x < 10.5. Step 3: From (i) and (ii): 5 < x < 10.5, and x must be an even integer, so x ∈ {6, 8, 10}. Step 4: The pairs are: (6,8), (8,10), (10,12). Verify (10,12): 10 > 5 ✓, 12 > 5 ✓, sum = 22 < 23 ✓. Check (12,14): 12+14=26 ≮ 23 ✗. So exactly 3 pairs exist. Common mistake: Forgetting that BOTH integers must be greater than 5, or using odd numbers instead of even.