Chapter Summary and Practice

Problem
Solve −2 ≤ 3x + 1 < 10, first over real numbers and then over integers.

1. 1.Subtract 1 everywhere: −3 ≤ 3x < 9.
2. 2.Divide by positive 3: −1 ≤ x < 3. The real solution set is [−1, 3).
3. 3.For integer x retain only {−1, 0, 1, 2}. For natural x retain only {1, 2}.
4. 4.At x = −1 the original middle value is −2, which is allowed. At x = 3 it is 10, which is excluded.
5. 5.The real interval includes values such as 0.5; the integer set does not. A domain restriction changes the final set, not the solving rules.

Problem
A student solves −2x + 7 ≤ 1 and writes x ≤ 3. Identify and fix the mistake.

1. 1.Subtract 7 correctly: −2x ≤ −6.
2. 2.The final division is by −2, so ≤ must reverse to ≥. The correct answer is x ≥ 3.
3. 3.Check the disputed sides: x = 4 gives −1 ≤ 1, true; x = 2 gives 3 ≤ 1, false.
4. 4.The boundary x = 3 gives equality and belongs to the answer. Checking values on each side reveals the incorrect direction.

Problem
Solve 2x + 3 ≥ 7 and 5 − x > 3.

1. 1.The first gives 2x ≥ 4, so x ≥ 2.
2. 2.The second gives −x > −2, so x < 2.
3. 3.No value satisfies both. In particular, 2 passes the first but fails the strict second condition. The solution set is empty.
4. 4.If the second comparison were ≥ instead of >, it would give x ≤ 2, and the only common solution would be {2}. One symbol can therefore change the entire intersection.

Problem
Find consecutive odd positive integers, both smaller than 10, whose sum exceeds 11.

1. 1.Let the smaller odd integer be x; the larger is x + 2. Positivity gives x ≥ 1 and both being below 10 gives x + 2 < 10, hence x < 8.
2. 2.The sum requirement is x + x + 2 > 11. Thus 2x > 9 and x > 4.5.
3. 3.The odd positive integers satisfying 4.5 < x < 8 are 5 and 7.
4. 4.The pairs are (5, 7) and (7, 9). Both meet the size and sum requirements. Do not retain even integers or every real number in the interval.

Problem
Pieces x, x + 4, and 2x cm must fit in a 100 cm board. The third must be at least 6 cm longer than the second. Find x.

1. 1.Available length gives x + (x + 4) + 2x ≤ 100, so 4x + 4 ≤ 100 and x ≤ 24.
2. 2.The difference requirement gives 2x − (x + 4) ≥ 6, hence x − 4 ≥ 6 and x ≥ 10.
3. 3.Combine: 10 ≤ x ≤ 24 cm. These values make all piece lengths positive and keep x the shortest.
4. 4.At 10 the difference requirement is exactly met; at 24 the whole board is used. If complete use is required rather than merely fitting, only x = 24 remains.

Problem
A process permits 20 ≤ C ≤ 25. Give its Fahrenheit limits.

1. 1.Substitute F = (9/5)C + 32. Multiply the Celsius bounds by positive 9/5 to get 36 ≤ (9/5)C ≤ 45.
2. 2.Add 32 to all parts: 68 ≤ F ≤ 77.
3. 3.The interval is [68, 77]°F. Equality remains allowed at both ends because the original bounds were inclusive.
4. 4.The conversion preserves the interval order; it does not preserve the numerical interval width because the scales use different-sized degrees.

Problem
200 L of 10% solution is mixed with x L of 30% solution. Require concentration more than 15% and no more than 20%.

1. 1.Let x ≥ 0. Substance amount is 20 + 0.30x L and total volume is 200 + x L.
2. 2.Write 0.15 < (20 + 0.30x)/(200 + x) ≤ 0.20. The denominator is positive.
3. 3.For the lower bound: 20 + 0.30x > 30 + 0.15x, so 0.15x > 10 and x > 200/3.
4. 4.For the upper bound: 20 + 0.30x ≤ 40 + 0.20x, so 0.10x ≤ 20 and x ≤ 200.
5. 5.Therefore 200/3 < x ≤ 200 L. The lower endpoint produces exactly 15% and is excluded; the upper produces exactly 20% and is included.
6. 6.Check x = 100: 50/300 = 1/6, approximately 16.67%, which satisfies both bounds. A simple average of 10% and 30% would not account for unequal volumes.