Quarry School

Turn a real situation into a linear model

Explain it like I am five

Think of a jar that already holds one coin. Each day you put in two more. The total is the coins already there plus the coins added over the days. A linear model is an equation that describes a situation with the same change for each input step. It has a starting amount and an amount per step. Money, songs and pay can follow this pattern too. Name what the input counts and what the output measures. Then ask which inputs make sense. A whole number of items or policies makes sense; a negative number of items does not.

x input unitsstarting amount +rate × xtotal outputinputoutput
A linear model adds the accumulated change to an amount present before the input begins.
Reminder
  • Order of operations. Multiply before adding: 520 + 80(3) = 520 + 240 = 760.
  • Solving an equation. In 1160 = 80n + 520, subtract 520 from both sides and divide both sides by 80. The check is 80(8) + 520 = 1160.
  • Function notation. C(100) asks for the cost output at input 100; replace x with 100.
  • Slope from two pairs. 920−7605−3 = 1602 = 80 dollars per policy.
  • Decimal multiplication. 37.5 × 100 = 3750; moving two place values corresponds to multiplying by 100.
  • Units. Dollars per item times items gives dollars; add the starting amount in dollars.
  • Domain. A whole-policy count can be 3 but cannot be 3.5 or −1.
Why it works. If each step adds m, then x steps add mx. Adding the initial amount b gives the total mx + b. The units explain the multiplication: dollars per item times items gives dollars. If two totals are supplied instead, their difference reveals the change earned or spent across the extra inputs. Dividing by that input change finds m, and subtracting mx from either total finds b. This describes the situation only while the constant-rate assumption and its allowed inputs remain valid.
RuleRule: A constant-rate model has output = starting amount + rate × input. Write f(x) = mx + b with units for m and b and a domain appropriate to the situation.
The same idea, five ways
Say it

Start with what is already there, then add the same amount for every step.

Write it

A linear model combines a fixed starting amount with change at a constant rate.

In math
  • f(x) = mx + b
  • f(0) = b
  • m = outputchangeinputchange
Like

A coin jar has a starting collection plus the same deposit each day.

See it
d daysstart with 1, add 2dJ(d) = 1 + 2d coinsinputoutput
The original jar model separates the starting coin from the coins added later.
The same idea, other ways
As a small story

One coin is in the jar before you begin. One day adds 2 coins, giving 3. Two days add 4 coins, giving 5. The same daily deposit explains J(d) = 1 + 2d.

input d daysoutput J(d) coins011325
Each one-day interval in this original table adds the same two coins.
By following the units

A charge of $2 per item times 3 items gives $6. If a $5 starting charge also applies, the total is $11. The item units cancel in the multiplication, leaving dollars.

$2 per item × 3 items = $6
$5 starting charge + $6 = $11
C(x) = 5 + 2x
Units show why the rate multiplies the input before you add the fixed amount.
.1Fixed cost plus marginal cost

A business can pay rent even when it makes no items. That starting expense is a fixed cost. Overhead means expenses of keeping the business running, such as office rent. Making each item adds a production cost. When each extra item costs the same amount, that amount is the marginal cost. Add the fixed cost to the variable cost from the items made.

  • Rule: C(x) = 1250 + 37.5x, because Ben pays $1,250 in fixed monthly overhead plus $37.50 for every item made that month.
  • The fixed cost is $1,250 per month and the marginal cost is $37.50 per item, because one is paid without production and the other is added by each extra item.
  • The variable cost is 37.5x dollars, because it changes with the number of items. The variable production cost per item is constant in this model.
  • The domain is x = 0, 1, 2, … items per month, because whole items cannot be negative or fractional. Continue only while the stated cost arrangement applies.
1224364860728496108120600120018002400300036004200480054006000(0, 1.25e+03)(−33.3, 0)fixed cost100 items
The line displays the trend, and only whole-number item inputs are used for actual production.
Reminder
  • Multiplication with a decimal. Multiplying by 100 shifts the decimal two places: 37.5 × 100 = 3750.
The same idea, five ways
Say it

Pay the monthly overhead, then add $37.50 for every item.

Write it

The total monthly cost is the fixed cost plus the variable cost.

In math
  • C(x) = 1250 + 37.5x
  • C(0) = 1250
  • C(x + 1) − C(x) = 37.5
  • x ≥ 0, with x a whole number
Like

Rent is the cost of opening the workshop; materials add cost each time you make another item.

See it
100 items$1,250 + $37.50 ×items$5,000 that monthinputoutput
The monthly overhead is added once, while the per-item cost is multiplied by 100.
Worked exampleBen's cost, from no production to 100 items

Finding C(x) means describing the monthly cost for x items. Finding C(100) means putting in 100 items and reading the dollars out. Ben pays $1,250 monthly overhead and $37.50 per item. Find costs for zero, one and 100 items.

input x itemsoutput C(x) dollars0125011287.51005000↓ evaluate: input given, read the output below it
The output under zero is the fixed cost; the highlighted column gives the cost of 100 items.
  1. Let x count items made in one month and C(x) measure that month's cost in dollars.This makes the units and the nonnegative whole-number domain explicit.
  2. Write C(x) = 1250 + 37.5x.The overhead is present even at zero items, and every item adds $37.50.
  3. C(0) = 1250 + 37.5(0) = 1250. C(1) = 1250 + 37.5 = 1287.5.Zero items test the fixed cost; one item tests a single added production cost.
  4. C(100) = 1250 + 37.5(100) = 1250 + 3750 = 5000.One hundred items contribute $3,750 above the fixed overhead.
Answer
  • C(x) = 1250 + 37.5x
  • C(0) = $1,250.
  • C(1) = $1,287.50.
  • C(100) = $5,000.
Check Compute 100 groups of $37.50 as 10 × $375 = $3,750. Adding the $1,250 overhead gives $5,000 again.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: C(x) = (1250 + 37.5)x.
This charges the entire monthly overhead again for every item and gives zero cost when no items are made.
✓ Instead: C(x) = 1250 + 37.5x, so C(0) = 1250.
Tips and tricks
  • Tip: Test zero production. A fixed monthly cost must still appear.
.2A music collection growing each month

Marcus has a collection before he starts adding new songs. That collection is the starting amount. Each completed month adds the same new batch. This works like adding the same stack of books to a shelf every month. Count the starting collection once, then add the number of monthly batches times the size of a batch.

  • Rule: N(t) = 200 + 15t, because Marcus starts with 200 songs and adds 15 songs each month.
  • The initial value is 200 songs and the slope is 15 songs per month, because t counts months since the starting collection was recorded.
  • A year corresponds to input t = 12, because there are 12 months in a year.
  • Use nonnegative whole-number t for completed monthly batches. A fractional-month value on the drawn line is an estimate, because the description does not specify when individual songs are added.
24681012198220242264286308330352374396(0, 200)(−13.3, 0)starting collectionone year
The rising line shows the trend; the stated monthly batches give exact totals at whole months.
Reminder
  • Function notation. N(12) means use input 12 in the rule: N(12) = 200 + 15 × 12. It does not mean N times 12.
The same idea, five ways
Say it

Start with 200 songs and add 15 each completed month.

Write it

The collection contains its starting 200 songs plus 15 songs for each elapsed month.

In math
  • N(t) = 200 + 15t
  • N(0) = 200
  • N(12) = 380
  • t ≥ 0, with t a whole number for monthly batches
Like

A bookshelf begins with a stack of books and gains the same new stack every month.

See it
12 months200 + 15t380 songsinputoutput
Twelve batches add 180 songs to the starting collection of 200.
Worked exampleMarcus's songs after zero months, one month and a year

Finding N(12) means using 12 months as the input to find the number of songs. Marcus starts with 200 songs and adds 15 songs each month. Write the model and find the totals after zero months, one month and one year.

input t monthsoutput N(t) songs0200121512380↓ evaluate: input given, read the output below it
The highlighted input is 12 months, not one month.
  1. Let t measure completed months and N(t) count songs. Write N(t) = 200 + 15t.The starting collection contributes 200, and t monthly additions contribute 15t songs.
  2. N(0) = 200 and N(1) = 200 + 15 = 215.The smallest inputs check the starting collection and one monthly batch.
  3. One year is 12 months, so calculate N(12) = 200 + 15(12).The rate is per month, so the input must also be measured in months.
  4. 15(12) = 180, and 200 + 180 = 380.Add the 12 new batches to the collection that was already there.
Answer
  • N(t) = 200 + 15t
  • N(0) = 200 songs.
  • N(1) = 215 songs.
  • N(12) = 380 songs.
Check Six months add 90 songs. Two six-month groups add 180, and 200 + 180 = 380. The positive rate also explains why the collection increases.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: After one year, N(1) = 215 songs.
The input is measured in months. Input 1 means one month rather than one year.
✓ Instead: Convert one year to 12 months and use N(12) = 380 songs.
Tips and tricks
  • Tip: Write the input unit next to the number before substituting it.
.3Salary plus commission

A salesperson's pay can have two pieces. The base salary is the amount earned even with no sales. A commission is extra pay for each sale. This is like getting a regular allowance plus the same bonus for each completed job. If neither piece is stated, compare two weeks. Extra earnings divided by extra policies reveals the commission. Remove that week's commissions to uncover the base salary.

  • Rule: I(n) = 520 + 80n, because Ilya earns a $520 weekly base salary plus $80 commission per new policy.
  • The commission rate is $80 per policy, because an earnings increase of $160 accompanied an increase of 2 policies.
  • The base salary is $520 per week, because removing 3 commissions of $80 from $760 leaves $520.
  • The domain is n = 0, 1, 2, … policies sold in one week, because sold policies are whole objects and their count is nonnegative. The pay arrangement is assumed unchanged between the two weeks.
3 policies$520 + $80 ×policies$760 for the weekinputoutput
A whole-number policy count determines the commissions added to the weekly base salary.
Reminder
  • Interpreting a rate. $160 more for 2 more policies means $160 ÷ 2 = $80 per policy.
The same idea, five ways
Say it

Keep the weekly base pay, then add the same bonus for each new policy.

Write it

Weekly income equals base salary plus commission per policy times policies sold.

In math
  • I(n) = 520 + 80n
  • I(3) = 760
  • I(5) = 920
  • m = 920−7605−3 = 80
  • n ≥ 0, with n a whole number
Like

A regular allowance is paid before any job bonuses are added.

See it
input n policiesoutput I(n) dollars per week37605920
Two more policies correspond to $160 more earnings, revealing $80 per policy.
Worked exampleEvaluate the pay rule, then recover it from earnings

Finding I(n) means finding weekly pay for a given policy count. First use a $520 base salary and $80 per policy to find the pay for zero, one, three and five policies. Then recover that same model from the source earnings of $760 for three policies and $920 for five policies.

input n policiesoutput I(n) dollars37605920
Use the two observed weeks to infer the constant commission.
760240 + b=do the same thing to both sides
Subtract the commissions on both sides to find the weekly base salary.
  1. With the stated ingredients, write I(n) = 520 + 80n. I(0) = 520 and I(1) = 600.Zero policies give base pay; one policy adds one commission.
  2. I(3) = 520 + 80(3) = 760. I(5) = 520 + 80(5) = 920.Multiply the commission by the number of policies before adding the base salary.
  3. Now treat only the two earnings as given: m = 920−7605−3 = 1602 = 80 dollars per policy.The change in pay divided by the change in policy count finds the commission under the fixed-pay arrangement.
  4. Use three policies to find the base salary: 760 = 80(3) + b = 240 + b, so b = 760 − 240 = 520.Removing the commissions finds the pay that would remain with no policies.
  5. Plug the base salary back into that week's equation: 240 + 520 = 760. Write I(n) = 520 + 80n.The recovered base salary must reproduce the known earnings before it is used.
Answer
  • I(n) = 520 + 80n
  • Base salary: $520 per week.
  • Commission: $80 per policy.
  • I(0) = $520.
  • I(1) = $600.
  • I(3) = $760.
  • I(5) = $920.
Check The other week gives 80(5) + 520 = 920. Removing its $400 commission also leaves the same $520 base salary.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: The $760 earned for three policies is the base salary.
That total includes three commissions. Base salary means earnings at zero policies.
✓ Instead: Subtract 3 × $80 = $240 from $760 to find the $520 base salary.
Tips and tricks
  • Tip: For a base salary, remove all commissions from either known week's total and compare the results.
.4A target output asks for an input

Imagine a rental bill with a starting fee and the same charge for every hour. If you know the hours, you can calculate the bill. If you know the bill, you can work backward to the hours. Remove the starting fee first. What remains is the charge for the hours, so divide it by the nonzero hourly rate. A flat membership price works differently. Changing your visit count leaves the bill unchanged. An exact target either equals that flat price for every allowed count or cannot be reached at all.

  • Rule: For m ≠ 0, solving mx + b = T finds the input x = T−bm that produces target output T, because subtracting b and dividing by m undo the model operations.
  • Rule: For m = 0, output is always b. Target b is reached at every allowed input; another target is never reached.
  • Rule: Check the solved input against the domain, because an algebraic solution may require a negative or fractional count that the situation forbids.
target billhourly rate × hours + starting fee=do the same thing to both sides
A target-output question works backward through the cost model.
Reminder
  • Solve and substitute. For 62 = 7h + 13, subtract 13 and divide by 7 to get h = 7. Putting 7 back gives 62.
The same idea, five ways
Say it

Say: the output is known; find every allowed input that gives it.

Write it

Solve the model equal to the target, then check its domain.

In math
  • f(x) = T
  • mx + b = T
  • m ≠ 0: x = T−bm
  • m = 0 and T = b: every allowed input
  • m = 0 and T ≠ b: no input
Like

Read a bill backward by removing its starting fee and counting its hourly charges.

See it
target Tmx + b=do the same thing to both sides
Both sides must describe the same requested output.
Worked exampleRecover hours from a total bill

Find the hours means the bill is known and the hour input is unknown. A rental costs $13 to begin plus $7 per whole hour. How many hours give a $62 bill?

627h + 13=do the same thing to both sides
The remaining bill counts seven-dollar hour charges.
  1. Write 62 = 7h + 13.The requested dollar output replaces C(h) and leaves h unknown.
  2. Subtract 13: 49 = 7h.This removes the starting fee to isolate the hourly charges.
  3. Divide by 7: h = 7.Seven equal seven-dollar charges account for the remaining forty-nine dollars.
Answer
7 whole hours.
Check 7 × 7 + 13 = 49 + 13 = 62, and 7 is an allowed nonnegative whole hour count.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: Solve C(v) = 30 for C(v) = 29 by dividing (30 − 29) by slope zero.
Zero cannot be a divisor and the visit count cannot change the fixed price.
✓ Instead: No input gives $30. Every allowed input gives $29 instead.
✗ Not this: Counterexample: A computed count of 3.5 whole items is automatically an exact attainable target.
The domain permits only whole item counts.
✓ Instead: Report that no allowed whole count reaches the target exactly if the nonzero-rate solution is 3.5.
Tips and tricks
  • Tip: Ask whether you are given an input to evaluate or an output to solve for. Check the rate before dividing and the domain after solving.
Strategy: step by step
  1. 1. Define the input and output, because a number without its meaning can lead to the wrong units or domain.
  2. 2. Identify the starting amount b and rate m. If two totals are given, divide output change by input change to find m.
  3. 3. If b is missing, substitute a known pair into output = mx + b and solve for b to recover the starting amount. Plug the value back into that pair.
  4. 4. Write the model with its units and allowed inputs, because counting objects differs from measuring continuous time.
  5. 5. For a given input, substitute it to find the output. For a target output, solve for the input, then plug it back in and check that it belongs to the domain.
Strategy
Strategy: Build, evaluate or reverse a real-world model
1
Are the starting amount and amount per input unit stated?
YesUse those as b and m.
NoIf a rate and one pair are given, substitute that pair to find b and check it. Otherwise use two distinct input-output pairs to find m, then b. If neither kind of information is supplied, the model is not determined.
↓
2
Does the input count indivisible objects or completed monthly additions?
YesUse nonnegative whole-number inputs such as 0, 1 and 2.
NoFor continuous measurements such as elapsed time, allow nonnegative real inputs during the period when the model applies.
↓
3
Is the question giving a target output rather than an input?
YesFor nonzero m, set the model equal to the target, solve for the input, plug back and check the domain. For m = 0, compare the target with b: equality makes every allowed input a solution; a different target has no solution.
NoReplace the input symbol with the stated value and compute the output.
  1. 1. Name both variables and their units.
  2. 2. Find the rate and starting amount, directly or from two input-output pairs.
  3. 3. Write f(x) = mx + b and state the meaningful domain.
  4. 4. Substitute a known input, or solve for a missing input from a stated output.
  5. 5. Plug the answer back into the model and interpret it with units.
Worked exampleOne coin already there, two added per day

Finding the number of coins after one day means using input d = 1 to find the output. A jar starts with one coin and receives two coins at the end of every day. Write J(d) and find J(1).

1 completed day1 + 2d3 coinsinputoutput
Multiply the daily addition by the completed days, then add the starting coin.
  1. Let d be the number of completed days and J(d) the number of coins.The input counts daily deposits, so d is a nonnegative whole number.
  2. Write J(d) = 1 + 2d.There is 1 starting coin and each completed day contributes 2 coins.
  3. J(1) = 1 + 2(1) = 3.An input of 1 means one daily deposit.
Answer
  • J(d) = 1 + 2d, with d a nonnegative whole number.
  • J(1) = 3 coins.
Check Count the starting coin and two new coins: 1 + 2 = 3. Also J(0) = 1, so the model keeps the correct starting amount.
Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.
Rung 1Rung 1: Starting amount plus a one-step rate

Writing B(w) means finding the total books after w completed weeks. A shelf starts with two books and gains three books each week. Find B(1) and B(4).

4 completed weeks2 + 3w14 booksinputoutput
The starting books are counted once; the weekly additions are counted four times.
  1. Let w count completed weeks and B(w) count books. The domain is nonnegative whole weeks.The books are added in weekly batches.
  2. Write B(w) = 2 + 3w.Start with 2 books and add 3 for each weekly batch.
  3. B(1) = 2 + 3(1) = 5. B(4) = 2 + 3(4) = 14.One batch adds 3 and four batches add 12.
Answer
  • B(w) = 2 + 3w
  • B(1) = 5 books.
  • B(4) = 14 books.
Check Four batches of 3 make 12 books added. Together with the starting 2, that is 14 books. B(0) = 2 checks the start.
Rung 2Rung 2: Evaluate the cost of 100 items

Finding C(100) means using 100 items as the input to find monthly cost in dollars. Use Ben's $1,250 fixed overhead and $37.50 cost per item.

100 items1250 + 37.5x$5,000inputoutput
Keep the fixed cost and the per-item cost in their separate roles.
  1. Write C(x) = 1250 + 37.5x for nonnegative whole-number item counts.The monthly starting cost is fixed while each item adds a constant cost.
  2. Substitute 100: C(100) = 1250 + 37.5(100).The question gives the item count and asks for the corresponding output.
  3. 37.5(100) = 3750, so C(100) = 1250 + 3750 = 5000.Add the variable production cost to the fixed overhead.
Answer
C(100) = $5,000 for the month.
Check The cost above overhead is $5,000 − $1,250 = $3,750. Dividing that by 100 items gives the stated $37.50 per item.
Rung 3Rung 3: Find salary from two earnings

Recovering the income rule means finding the pay per policy and the pay with no policies. Ilya earned $760 with three policies and $920 with five. Assume the same base salary and commission rate in both weeks.

input n policiesoutput I(n) dollars37605920
Compare the two input changes and output changes before finding the base salary.
  1. Read the pairs as (3, 760) and (5, 920).The input is policies and the output is weekly dollars.
  2. m = 920−7605−3 = 1602 = 80 dollars per policy.Two extra policies earned $160 more, so each contributes $80.
  3. Find b from the three-policy week: 760 = 80(3) + b = 240 + b, so subtract 240 to obtain b = 520.Removing the commissions isolates the base weekly salary.
  4. Plug back: 80(3) + 520 = 760. Write I(n) = 80n + 520 for nonnegative whole-number policy counts.The recovered salary must fit the observed week, and policies are counted as whole objects.
Answer
  • I(n) = 80n + 520
  • Commission: $80 per policy.
  • Base salary: $520 per week.
Check At n = 5, I(5) = 400 + 520 = 920, matching the other week. The positive commission makes income increase with sales.
Rung 4Rung 4: Solve for a target weekly income

Solving I(n) = 1160 means the weekly output is $1,160 and you must find the policy input that gives it. Under I(n) = 80n + 520, how many policies produce that income?

116080n + 520=do the same thing to both sides
Remove the base salary before dividing the remaining target earnings by the commission rate.
  1. Set 1160 = 80n + 520.The target output replaces I(n), leaving the number of policies unknown.
  2. Subtract 520 from both sides: 640 = 80n.Removing the base salary isolates the income that must come from commissions.
  3. Divide both sides by 80: n = 64080 = 8.Each policy earns $80 of commission, so this finds the needed number of policies.
  4. Plug back: I(8) = 80(8) + 520 = 640 + 520 = 1160.The found input must give the target output, and 8 is an allowed nonnegative whole-number count.
Answer
8 new policies give $1,160 for the week.
Check Five policies gave $920. Three more policies add 3 × $80 = $240, and $920 + $240 = $1,160.
Rung 5Rung 5: A zero-rate target has all or no inputs

Find an input means decide which allowed visit counts give the requested charge. A membership costs $29 for the month regardless of visits. For C(v) = 29 with nonnegative whole visit counts, solve C(v) = 29 and C(v) = 30.

C(v) = 0v + 29
Target 29: every allowed input
Target 30: no input
Compare a zero-rate model with the target instead of dividing by zero.
  1. Write the model as C(v) = 0v + 29.The visit count has zero effect on the charge.
  2. For target 29, the equation is 29 = 0v + 29, which is true for every allowed v.Any allowed count leaves the same output; no division is needed.
  3. For target 30, the equation is 30 = 0v + 29, which says 30 = 29 and has no solution.Changing the input cannot change the constant output.
Answer
  • Target $29: every nonnegative whole visit count works.
  • Target $30: no visit count works.
Check At v = 0, C(0) = 29; at v = 6, C(6) = 29. The output stays 29 for any other allowed count as well.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: Every money problem can be modeled by one constant-rate line.
A discount, a changed commission rate or a second monthly charge can change the amount per step. The fixed-rate assumption must come from the situation.
✓ Instead: Use one line only during the period and input range where the same starting cost and per-step rate apply.
✗ Not this: Counterexample: A solution of n = 3.5 is automatically a valid policy count.
This model counts whole policies. A fractional solution does not give an attainable exact target under that counting rule.
✓ Instead: Check the domain after solving. For a target that requires 3.5 policies, report that no whole-number count attains it exactly.
Tips and tricks
  • Tip: Put the units beside m and b: dollars per item versus dollars for the starting monthly cost.
  • Tip: For a target output with a nonzero rate, subtract the start, divide by the rate, then plug back and check the domain. With zero rate, compare the target with the constant output instead.
Trap. Using the first observed total as the initial value when its input is not zero. Remove its accumulated change to find the starting amount.