Real-world inequalities

1. Let n = the number of notebooks. The total cost is 3n dollars.
2. She can spend no more than $40: 3n \le 40
3. Divide both sides by 3: n \le \frac{40}{3} = 13.\overline{3}
4. Notebooks come in whole numbers, and she wants the greatest number that fits, so round down to 13.

Answer: 13 notebooks

Check: 13 \times 3 = 39, and 39 \le 40 ✓. 14 \times 3 = 42, and 42 \le 40 is false — she cannot afford 14.

Tip: For "the most you can afford" questions, always round down, even if the decimal part is .5 or more.

1. Let h = hours worked. Earnings = 15h.
2. He needs at least $350: 15h \ge 350
3. Divide by 15: h \ge \frac{350}{15} = 23.\overline{3}
4. He must work whole hours, and 23 hours is less than 23.\overline{3}, so 23 is not enough. Round up to 24.

Answer: 24 hours

Check: 24 \times 15 = 360 \ge 350 ✓. 23 \times 15 = 345, which is less than $350 ✗.

Tip: For "the least you need to reach a goal", always round up — even when the decimal is small, like .3.

1. Let c = the number of crates. Each crate weighs 85 lb, so the cargo weighs 85c pounds.
2. The cargo can be at most 2,000 lb: 85c \le 2000
3. Divide both sides by 85: c \le \frac{2000}{85} \approx 23.53
4. Crates come in whole numbers, and a maximum must stay under the limit, so round down to 23.

Answer: 23 crates

Check: 23 \times 85 = 1{,}955 lb, and 1{,}955 \le 2{,}000 ✓. With 24 crates: 24 \times 85 = 2{,}040 lb, which is over the limit ✗.

Tip: For any "greatest number that fits" question, round down — even when the decimal is .5 or more.

1. Let w = the number of weeks. She volunteers 7w hours.
2. She needs at least 120 hours: 7w \ge 120
3. Divide both sides by 7: w \ge \frac{120}{7} \approx 17.14
4. She must reach the goal, and 17 weeks is not quite enough, so round up to 18.

Answer: 18 weeks

Check: 18 \times 7 = 126 hours, and 126 \ge 120 ✓. In 17 weeks she has only 17 \times 7 = 119 hours ✗ — one hour short.

Tip: For "at least / minimum to reach a goal" questions, round up. Rounding down always leaves you just short.

1. Let b = number of boxes. Total weight = 180 + 45b.
2. "At most 1,500 lb": 180 + 45b \le 1500
3. Subtract 180: 45b \le 1320
4. Divide by 45: b \le \frac{1320}{45} = 29.\overline{3}
5. Boxes are whole numbers; round down to 29.

Answer: 29 boxes

Check: 180 + 45(29) = 180 + 1305 = 1485 \le 1500 ✓. 180 + 45(30) = 1530 > 1500 ✗.

Tip: The fixed amount (the operator) uses up part of the limit before the first box is loaded — subtract it first.

1. Cost of Plan A = 25 + 0.15t. Cost of Plan B = 40.
2. Plan A is cheaper when 25 + 0.15t < 40
3. Subtract 25: 0.15t < 15
4. Divide by 0.15: t < 100

Answer: fewer than 100 texts (t < 100).

Check: At 80 texts, A costs 25 + 12 = 37 < 40 ✓. At 120 texts, A costs 25 + 18 = 43 > 40 ✓. At exactly 100 texts both cost $40 (a tie).

Tip: The break-even point (where the costs are equal) is the boundary. Test one number on each side to see which plan wins where.

1. "41°F or colder" means F \le 41. Substitute the formula: \frac{9}{5}C + 32 \le 41
2. Subtract 32: \frac{9}{5}C \le 9
3. Multiply by the reciprocal \dfrac{5}{9} (positive, no flip): C \le 9 \cdot \frac{5}{9} = 5

Answer: C \le 5

Check: C = 5: \tfrac{9}{5}(5) + 32 = 9 + 32 = 41 ✓ (exactly the limit). C = 0: 32 \le 41 ✓, which makes sense — 0 °C is freezing.

Tip: "Or colder" / "or less" / "at most" all mean \le. Substitute the formula, then solve like any inequality.

1. Let x = the fourth score. The mean of four games is \dfrac{142 + 158 + 135 + x}{4}.
2. The mean must be greater than 150: \frac{435 + x}{4} > 150
3. Multiply both sides by 4: 435 + x > 600
4. Subtract 435: x > 165
5. The inequality is strict, so 165 itself does not work. The lowest whole score is 166.

Answer: 166

Check: \dfrac{435 + 166}{4} = \dfrac{601}{4} = 150.25 > 150 ✓. With 165: \dfrac{600}{4} = 150, which is not greater than 150 ✗.

Tip: For mean goals, multiply the target mean by the total number of values to get the needed sum, then subtract what you already have.

1. Cost of Gym A = 10 + 5c. Cost of Gym B = 45.
2. Gym A is cheaper: 10 + 5c < 45
3. Subtract 10 from both sides: 5c < 35
4. Divide by 5: c < 7

Answer: c < 7 (fewer than 7 classes a month)

Check: At 6 classes, Gym A costs 10 + 30 = \$40 < \$45 ✓. At 7 classes, Gym A costs 10 + 35 = \$45 — a tie, not cheaper. At 8 classes, Gym A costs $50 — more ✓.

Tip: Find the break-even number (where the costs are equal), then think: the plan with the lower fixed fee wins for small amounts of use.

1. Weight limit. Let j = jars. Box plus jars: 4 + 2.2j \le 50.
   • Subtract 4: 2.2j \le 46
   • Divide by 2.2: j \le 20.\overline{90}, so at most 20 whole jars (round down).
2. Space limit. At most 24 jars fit: j \le 24.
3. Both limits must be met, so use the smaller maximum: 20.

Answer: 20 jars

Check: 4 + 2.2(20) = 4 + 44 = 48 lb \le 50 ✓, and 20 jars fit in the box ✓. With 21 jars: 4 + 46.2 = 50.2 lb > 50 ✗.

Tip: When there are two limits, solve each one separately. The answer is the one that runs out first (the smaller maximum).

1. Let p = the price before tax. With 8% tax, she pays p + 0.08p = 1.08p.
2. She can spend at most $100: 1.08p \le 100
3. Divide both sides by 1.08: p \le \frac{100}{1.08} = 92.592\ldots
4. Prices are in cents. Rounding up to $92.60 would go over the limit, so round down to $92.59.

Answer: $92.59

Check: 92.59 \times 1.08 = 99.9972 \approx \$100.00, which is within $100 ✓. 92.60 \times 1.08 = 100.008, which is $100.01 after rounding — just over ✗.

Tip: "Price including tax" means multiply by (1 + \text{rate}). To work backwards, divide by (1 + \text{rate}) — do not subtract the tax rate from the total.

1. Perimeter of a rectangle: P = 2l + 2w. With l = 30: P = 2(30) + 2w = 60 + 2w
2. The fence can be at most 100 ft: 60 + 2w \le 100
3. Subtract 60 from both sides: 2w \le 40
4. Divide by 2: w \le 20

Answer: w \le 20 (the width can be at most 20 ft)

Check: With w = 20: P = 60 + 40 = 100 ft \le 100 ✓. With w = 21: P = 102 ft — not enough fence ✗.

Tip: A rectangle has two lengths and two widths. Subtract both lengths from the total, then split what is left between the two widths.

1. Temperature after t minutes: 14 - 1.5t.
2. It must be between 2 and 8, inclusive: 2 \le 14 - 1.5t \le 8
3. Subtract 14 from all three parts: -12 \le -1.5t \le -6
4. Divide all three parts by -1.5 and flip both signs: 8 \ge t \ge 4
5. Rewrite from smallest to largest: 4 \le t \le 8.

Answer: 4 \le t \le 8 (from 4 to 8 minutes)

Check: At t = 4: 14 - 6 = 8°C ✓. At t = 8: 14 - 12 = 2°C ✓. At t = 6: 14 - 9 = 5°C, inside the range ✓.

Tip: A compound inequality is solved like a regular one — just do every step to all three parts. Dividing by a negative flips both symbols.

1. Let g = guests. Cost = 250 + 18g.
2. Budget limit: 250 + 18g \le 1500.
3. Subtract 250: 18g \le 1250.
4. Divide by 18: g \le 69.\overline{4} → at most 69 whole guests (round down).
5. Room limit: g \le 75.
6. Both conditions must hold, so take the smaller maximum: 69.

Answer: 69 guests

Check: 250 + 18(69) = 250 + 1242 = 1492 \le 1500 ✓. 70 guests: 250 + 1260 = 1510 > 1500 ✗.

Tip: When there are two limits, the stricter one decides. Check every condition in the story before choosing.

1. Rate: each extra hour adds 10 - 7 = 3 dollars.
2. Starting value: going back one hour from $7 gives 7 - 3 = 4 dollars. So cost = 4 + 3h. (Check: 4 + 3(1) = 7 ✓, 4 + 3(4) = 16 ✓.)
3. He can spend at most $30: 4 + 3h \le 30
4. Subtract 4: 3h \le 26.
5. Divide by 3: h \le 8.\overline{6}. Round down to 8 whole hours.

Answer: 8 hours

Check: 4 + 3(8) = 28 \le 30 ✓. 9 hours would cost 4 + 27 = 31 > 30 ✗.

Tip: From a table, the rate is the change between rows, and the fixed part is the value at 0 — not the first number in the table.

1. Read the graph. At t = 0 the tank holds 600 gallons. From (0, 600) to (5, 400) it loses 200 gallons in 5 hours: 200 \div 5 = 40 gallons per hour.
2. Water left after t hours: 600 - 40t.
3. It must be at least 200: 600 - 40t \ge 200
4. Subtract 600: -40t \ge -400.
5. Divide by -40 and flip the sign: t \le 10.
6. Time starts at 0, so 0 \le t \le 10.

Answer: 0 \le t \le 10

Check: On the graph, the line crosses the 200-gallon grid line at t = 10 hours ✓. After that, the line drops below 200 ✓.

Tip: On a graph you can often check your answer directly: find the target value on the vertical axis, go across to the line, and read the time below it.

1. To remove +25 from the left side, subtract 25 from both sides: 30m \le 200 - 25 = 175 Kim added 25 and got 225, so Step 1 is the first error.
2. Correct work from there: m \le \dfrac{175}{30} = 5.8\overline{3}.
3. Months are whole and this is a maximum, so round down: at most 5 months.
4. (Step 3 also rounds the wrong way, but the first mistake is in Step 1.)

Answer: Step 1 is the first error; the correct answer is at most 5 months.

Check: 5 months: 25 + 150 = \$175 \le \$200 ✓. 6 months: 25 + 180 = \$205 > \$200 ✗.

Tip: Use the opposite operation to move a number: +25 is undone by subtracting 25. Check your final answer in the original story every time.

1. Start: 1800 - 65f \ge 300.
2. Subtract 1,800 from both sides: -65f \ge -1500.
3. Divide by -65 and flip the sign: f \le \frac{1500}{65} \approx 23.08
4. Families come in whole numbers, and this is a maximum, so round down: at most 23 families.

Answer: The food bank can serve at most 23 families.

Check: 23 families: 1800 - 65(23) = 1800 - 1495 = 305 \ge 300 ✓. 24 families: 1800 - 1560 = 240 < 300 ✗ — the reserve is too small.

Tip: When the amount left must be at least something, the number of items you can give away has a maximum. If your answer says "at least," re-check the sign flip.

1. Let g = gallons used. Bill = 18.50 + 0.006g.
2. Bill must be no more than $60: 18.50 + 0.006g \le 60
3. Subtract 18.50: 0.006g \le 41.50.
4. Divide by 0.006: g \le \frac{41.50}{0.006} = 6916.\overline{6}
5. Round down to stay within budget: 6,916 gallons.

Answer: 6,916 gallons

Check: 18.50 + 0.006(6{,}916) = 18.50 + 41.496 = 59.996 \approx \$60.00 ✓. With 6,917 gallons: 18.50 + 41.502 = 60.002, which rounds to $60.00 but is over the limit ✗.

Tip: Estimate first: about $40 at less than a cent per gallon means several thousand gallons. If your answer is far off, recheck the decimal.