10.3 Creating Linear Models from Context
Key Takeaways
- Define each variable with units before writing output = (rate)(quantity) + start; the slope is the changing piece per unit and the intercept is the amount when the input is 0.
- About 30% of PSAT 8/9 Math is in context: translate the science, social studies, or real-world sentence first, then solve.
- Swapping the rate with the starting amount is the usual trap—a $12 fee plus $3 per gigabyte is C = 3g + 12, not C = 12g + 3.
- At least and at most in a story produce inequalities (≥ or ≤), not equations; equality is included unless the sentence is strict.
- After solving, substitute, check units, and reject unreasonable results such as a negative number of tickets or a phone bill below the monthly fee when usage is zero or more.
College Board's Algebra domain is not only "solve for x." The same domain asks you to create linear equations and inequalities that model a situation. About 30% of PSAT 8/9 Math items are in context: a science measurement, a social-studies quantity, or a real-world cost. This independent OpenExamPrep section is the modeling habit that feeds 9.1 Linear Equations in One Variable, 9.2 Linear Equations in Two Variables, 10.1 Systems of Two Linear Equations, and 10.2 Linear Inequalities. OpenExamPrep does not claim official approval, review, partnership, or exact equivalence with College Board materials.
A linear model has a constant rate and, often, a starting amount. In slope-intercept language, y = mx + b:
- m is the rate (dollars per gigabyte, centimeters per week, people per year).
- b is the value when the independent variable is 0 (monthly fee, height at week 0, population now).
If the story is an exact total, you write an equation. If the story says at least or at most, you write an inequality. Units ride along so you do not add dollars to hours. Algebra is still about 16–18 operational questions on this test (about 42.5% of Math). A calculator and Desmos are allowed as checks after the model exists. This test does not assess trigonometry, circle theorems, or rational or radical equations, so the models here stay linear.
A five-step translation
- Name the variables with units. "Let g be gigabytes used this month. Let C be the bill in dollars."
- Separate start from rate. The start happens even at quantity 0. The rate happens per unit.
- Write one equation or inequality. Match "at least" to ≥ and "at most" to ≤.
- Solve with the algebra from earlier sections.
- Check reasonableness: substitute, confirm units, and ask whether the number could happen in the story.
The usual trap is swapping rate and start: a $12 fee plus $3 per gigabyte is C = 3g + 12, not C = 12g + 3. A second trap is writing an equation when the sentence was a bound ("at most $36").
Worked example: phone plan
A plan charges $12 per month plus $3 per gigabyte. Let g be gigabytes and C be cost in dollars.
C = 3g + 12
| g (GB) | Calculation | C (dollars) |
|---|---|---|
| 0 | 3(0) + 12 | 12 |
| 2 | 6 + 12 | 18 |
| 4 | 12 + 12 | 24 |
| 6 | 18 + 12 | 30 |
| 8 | 24 + 12 | 36 |
Four gigabytes: C = 3(4) + 12 = 12 + 12 = $24. Zero gigabytes still costs $12—that is how you know 12 is the intercept, not the slope. If the family wants the bill at most $36: 3g + 12 ≤ 36, so 3g ≤ 24, g ≤ 8. Eight gigabytes costs exactly 24 + 12 = 36 and is allowed. Nine gigabytes: 27 + 12 = 39, which exceeds 36.
Wrong model C = 12g + 3 at 4 GB gives 48 + 3 = 51, which is not the plan. Wrong model C = 3g forgets the fee and gives $12 at 4 GB instead of $24. Desmos check: graph y = 3x + 12 and confirm (4, 24) and (0, 12). For the inequality, graph y = 36 and see they meet at x = 8.
A second plan: $18 per month plus $4 per gigabyte. C = 4g + 18. Six gigabytes: 24 + 18 = $42. Budget at most $50: 4g + 18 ≤ 50, 4g ≤ 32, g ≤ 8. Check 8 GB: 32 + 18 = 50. Check 9 GB: 36 + 18 = 54, over budget.
Worked example: fundraiser
A club already has $40 and earns $6 per raffle ticket. Let t be tickets sold and M be money in dollars.
M = 6t + 40
Five tickets: 6(5) + 40 = 30 + 40 = $70. Goal: at least $100. 6t + 40 ≥ 100 6t ≥ 60 t ≥ 10
Ten tickets: 60 + 40 = 100, which meets "at least." Nine tickets: 54 + 40 = 94, which is short. Mixing the $6 rate with the $40 start as M = 40t + 6 would claim five tickets raise 200 + 6 = 206 dollars, which is not reasonable for a $6 ticket and a $40 till.
Pies. The club has $25 and sells pies for $8. They need at least $89. 8p + 25 ≥ 89 8p ≥ 64 p ≥ 8
Check 8: 64 + 25 = 89. Check 7: 56 + 25 = 81, which is below 89. The least whole number of pies is 8. Stopping at 8p ≥ 64 and bubbling 64 is an SPR/multiple-choice trap: 64 is 8p, not p.
Worked example: temperature and a plant (science)
A sample starts at 12°C and cools 3°C per hour. Let h be hours and T be temperature in °C.
T = 12 - 3h
After 2 hours: 12 - 6 = 6°C. When is T ≤ 0? 12 - 3h ≤ 0 -3h ≤ -12 Divide by -3 and reverse: h ≥ 4.
Check h = 4: 12 - 12 = 0 ≤ 0, true. Check h = 3: 12 - 9 = 3 ≤ 0, false. Cooling longer makes temperature smaller, so after dividing by a negative the inequality opens toward larger h. Forgetting the reverse yields h ≤ 4, which would say three hours of cooling already reaches 0°C even though the sample is still at 3°C.
A plant is 8 cm tall and grows 2 cm per week. Let w be weeks and H be height in centimeters.
H = 2w + 8
After 5 weeks: 10 + 8 = 18 cm. When is the plant at least 20 cm? 2w + 8 ≥ 20, so 2w ≥ 12, w ≥ 6. Week 6: 12 + 8 = 20. Week 5 is 18, which is not yet 20. Units matter: if you treat 2 as "per day" you will report 6 days instead of 6 weeks.
Worked example: population and a library (social / civic numbers)
A town has 8,400 people and grows by 120 people per year. Let y be years from now and P be population.
P = 120y + 8400
Five years: 120(5) + 8400 = 600 + 8400 = 9,000. When does the population reach 9,240? 120y + 8400 = 9240 120y = 840 y = 7
Check: 120(7) = 840, and 840 + 8400 = 9240. Swapping to P = 8400y + 120 at y = 5 gives a "population" in the tens of thousands after five years that ignores the 8,400 starting count as a start.
A library starts with 120 books and adds 15 each month: B = 15m + 120. After 4 months: 60 + 120 = 180. When are there 210 books? 15m + 120 = 210, so 15m = 90, m = 6. Check: 90 + 120 = 210.
Worked example: draining tank (negative rate)
A tank holds 40 liters and drains 5 liters per minute. Let m be minutes and W be liters left.
W = 40 - 5m
When is at most 10 liters left? 40 - 5m ≤ 10 -5m ≤ -30 Divide by -5 and reverse: m ≥ 6
Six minutes: 40 - 30 = 10. Five minutes: 40 - 25 = 15, which is still more than 10. More draining time means less water, so "at most 10 liters" is later, not earlier. The no-reverse trap m ≤ 6 would claim that time is running out of water while the tank is still at 15 liters after 5 minutes. Empty would be 40 - 5m = 0, m = 8, which is a different question.
Units, reasonableness, and two-equation stories
| Story piece | Typical unit | Role in y = mx + b |
|---|---|---|
| Monthly fee, starting height, money already in the till | dollars, cm, people | Intercept b |
| Per gigabyte, per week, per ticket, per year | dollars/GB, cm/week | Slope m |
| "At least $100" | dollars | Inequality on the output |
| "At most 8 GB" | GB | Inequality on the input |
After you get a number:
- Substitute back into the model.
- Confirm the unit matches the question (hours vs minutes, dollars vs number of tickets).
- Reject nonsense: a negative number of pies, 400 hours of work in one school day, or a phone bill smaller than the $12 fee when g ≥ 0.
If this is SPR, type the number only—no dollar sign, no comma, no percent, no unit. Mixed numbers as improper fractions or decimals. Entry rules are in 15.1 Student-Produced Response Questions.
Sometimes a story needs two linear equations, as in 10.1: two different purchases, two unknown prices. Sometimes it needs one inequality, as in 10.2. Read whether you have one unknown or two, and whether the constraint is exact or a bound. Taxi version of the rate/start trap: $3 to start and $2 per mile is C = 2m + 3. Three miles cost 6 + 3 = $9. The swapped model C = 3m + 2 gives 9 + 2 = 11, which is the wrong fare.
A phone plan costs $12 each month plus $3 per gigabyte. Which equation gives the monthly cost C, in dollars, after g gigabytes?
A tank starts with 40 liters and drains 5 liters per minute. After m minutes the amount of water is at most 10 liters. Which inequality for m is correct?
A club has $25 and sells pies for $8 each. What is the least number of pies they must sell to have at least $89?