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4.5 — May I Have More, Please?
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Use inequalities to describe budgets, averages, and competing costs, then interpret the solutions in their original situations.
1. In class: A Solidify Understanding Task
Write the condition before solving
Define a variable with its units, write expressions for the relevant quantities, and choose the comparison that matches the situation.
| Phrase | Comparison | Example | Includes the limit? |
|---|---|---|---|
| No more than | Less than or equal to: \(\le\) | x is no more than 7: \(x\le7\) | Yes |
| Greater than | Greater than: \(>\) | x is greater than 7: \(x>7\) | No |
After solving, interpret the answer: counts such as cartons, pizzas, and students must be whole numbers, and quantities such as time and cost cannot be negative. Check a boundary value in the original situation to see whether it works.
Worked example 1 — The milk-carton budget
Original situation — Main 1: Milk costs $0.35 per carton plus $75 for delivery. Elvira has a $500 budget.
(a) Model and solve. Let \(m\) be the number of cartons. The total cost must be less than or equal to $500:
(b) Interpret. The maximum is 1,214 cartons. Any whole-number count from 0 through 1,214 stays within the budget. Buying 1,214 costs $499.90; buying 1,215 costs $500.25.
(c) Represent the solution. The exact set of carton counts is \(\{m\in\mathbb Z\mid0\le m\le1214\}\), where \(\mathbb Z\) means the integers. The unrestricted real-number model gives \(\left[0,\frac{8500}{7}\right]\), but only its whole-number values represent carton counts.
The packet asks for interval notation even when the answer counts whole objects or people. The teacher notes write \(\left[0,1214\right]\) for the milk-carton answer, but that interval also includes fractions. An interval alone does not impose a whole-number restriction. Use set notation for the exact count, or explicitly state the integer restriction alongside the interval. This is the distinction asked about in Section 4.2, Go 25. For a notation refresher, see the set-builder explanation.
Worked example 2 — Keep enough ranch dressing in reserve
Original situation — Main 2: Elvira starts with 130 gallons, uses 2.25 gallons per day, and needs 20 gallons remaining when she reorders.
(a) Let \(d\) be elapsed days. The amount remaining must be greater than or equal to 20:
Dividing by −2.25 reverses the inequality.
(b) The model allows about 48.89 days before reaching the reserve. After 48 full days, 22 gallons remain; after 49 full days, only 19.75 remain. If ordering on whole-day boundaries, reorder by the end of day 48.
(c) For elapsed time, \(\left[0,\frac{440}{9}\right]\) or \(\{d\in\mathbb R\mid0\le d\le\frac{440}{9}\}\). If counting only completed days, use \(\{d\in\mathbb Z\mid0\le d\le48\}\).
Model a limit on an average
Main 3 asks about the fifth veggie-burger order. An average is the sum of the values divided by their number. Let \(v\) be the fifth order’s cost in dollars:
Multiply both sides by 5: \(327+v\le410\), so \(v\le83\). With the contextual lower bound, (b) the cost can be from $0 through $83, and (c) the interval is \(\left[0,83\right]\) or the set is \(\{v\in\mathbb R\mid0\le v\le83\}\).
Compare the two total costs
For “cheaper,” compare the complete costs using a strict \(<\). A delivery charge or other fixed charge belongs in the expression even though it does not depend on the quantity purchased.
Worked example 3 — Choose the cheaper lettuce supplier
Original situation — Main 5: Val’s charges $250 plus $1.50 per bag; Sally’s charges $100 plus $4.00 per bag.
(a) Let \(b\) be the number of bags. For Val’s to be cheaper:
(b) Buy 61 or more bags. At 60 bags, both suppliers charge $340, so equality does not satisfy “cheaper.”
(c) The real-number inequality gives \((60,\infty)\). The exact bag counts are \(\{b\in\mathbb Z\mid b\ge61\}\).
For Main 4, let \(p\) be the number of pizzas. (a) The comparison is \(44.20+6.25p<14.50p\), giving \(p>\frac{884}{165}\approx5.358\). (b) Making the pizzas is cheaper for 6 or more pizzas. (c) The real model gives \(\left(\frac{884}{165},\infty\right)\), while the exact counts are \(\{p\in\mathbb Z\mid p\ge6\}\).
Worked example 4 — Meet a lower and an upper bound
Original situation — Main 6: Each student pays $2.75; lunch revenue is between $1,168.75 and $1,438.25.
(a) Let \(n\) be the number of students. Treating the stated endpoints as included, the two conditions can be written together as a compound inequality:
Both comparisons must hold. Divide all three quantities by positive 2.75:
(b) From 425 through 523 students, inclusive.
(c) The interval \(\left[425,523\right]\) describes the real-number bounds. The student counts are \(\{n\in\mathbb Z\mid425\le n\le523\}\).
2. Ready — Translate inequality phrases
| Phrase | Comparison | Example | Includes the bound? |
|---|---|---|---|
| At least; no fewer than | Greater than or equal to: \(\ge\) | x is at least 7: \(x\ge7\) | Yes |
| At most; no more than; does not exceed | Less than or equal to: \(\le\) | x is at most 7: \(x\le7\) | Yes |
| Fewer than; less than | Less than: \(<\) | x is less than 7: \(x<7\) | No |
| Greater than; more than | Greater than: \(>\) | x is more than 7: \(x>7\) | No |
Try the translations, then check the key.
Check your work — Ready 1–10 answer key
- \(x\ge13\)
- \(x\ge13\)
- \(x\le13\)
- \(x\le13\)
- \(x\le13\)
- \(x<13\)
- \(x\le13\)
- \(x<13\)
- \(x\ge13\)
- \(x\le13\)
3. Set — Average and cost models
Use the same modeling methods as in the in-class examples. Define the variable, solve, and give an answer that fits the situation. Try the problems independently before checking each part.
Check your work — Set 11–13 answer key
11 — Ethan’s pumpkins
(a) Let \(w\) be the second pumpkin’s weight in pounds:
(b) More than 847 pounds. Exactly 847 gives an average of 875 and does not meet the requirement. If using whole pounds, the minimum is 848 pounds.
(c) \((847,\infty)\) or \(\{w\in\mathbb R\mid w>847\}\). For whole-pound measurements, use \(\{w\in\mathbb Z\mid w\ge848\}\).
12 — Aaron’s third test
(a) Let \(s\) be the third test score:
(b) Scores from 96 through 100, inclusive. The maximum possible score supplies the upper bound.
(c) \(\left[96,100\right]\) or \(\{s\in\mathbb R\mid96\le s\le100\}\). If scores must be whole numbers, the exact set is \(\{96,97,98,99,100\}\).
13 — Compare texting plans
(a) Let \(t\) be the number of texts. The second plan costs more when
(b) 65 or more texts. At 64 texts, the two plans cost the same.
(c) The real model gives \((64,\infty)\). The actual counts are \(\{t\in\mathbb Z\mid t\ge65\}\).
4. Go — Systems by substitution
Substitute the expression from one equation into the other, solve for one variable, then find the other variable. Give the answer as an ordered pair \((x,y)\), and check it in both equations. The packet supplies a worked example. Try the four systems before opening the key; numbering follows the packet.
Check your work — Go 21–24 answer key
Solution: \((-2,3)\)
Solution: \((1,2)\)
Solution: \((10,0)\)
Solution: \((2,1)\)
Additional learning resources
| Resource | What it supports |
|---|---|
| Khan Academy — Two-step inequality word problem: R&B | Translating a verbal lower bound into an inequality and solving for an unknown quantity. |
| Khan Academy — Two-step inequality word problems | Practice writing inequalities from situations and interpreting the solutions. |
Teacher’s review of the source packet
Main 1(c), 4–6(c): Whole-number counts need an explicit integer restriction. The teacher’s milk example gives \(\left[0,1214\right]\) and a set-builder expression without that restriction, although it rules out fractional cartons. The clarification above distinguishes the real interval from the contextual set of counts.
Coverage — Section 4.5
| Packet location | Where supported |
|---|---|
| Main 1–6 · pp. 19–21 | Budgets, reserves, averages, cost comparisons, compound inequalities, and contextual notation; visible solutions aligned with parts (a)–(c) |
| Ready 1–10 · p. 22 | Phrase reminder and compact checking answers |
| Set 11–13 · pp. 22–23 | Contextual models and lettered solutions with notation and endpoint explanations |
| Go 21–24 · p. 23 | Brief substitution reminder and ordered-pair answers; no additional practice |
Problems and adapted material from Mathematics Vision Project, Secondary Mathematics One, Module 4: Equations and Inequalities (May 2016), Section 4.5, printed pp. 19–23; teacher notes for 4.5. MVP material is licensed under CC BY 4.0. Explanations and solutions added; selected wording and presentation adapted. Original materials are linked above.
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