Building a linear model of a real situation from a rate of change and one observation, choosing a variable that counts from a stated zero, predicting with the model both algebraically and graphically, and writing a model from a verbal description of a fixed total. Includes what a prediction assumes and where a model stops being trustworthy.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 5 — Writing Linear Equations
Modeling with Linear Equations
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.5 Modeling with Linear Equations §5.5, pp. 298-304 — the lesson these objectives are drawn from
Warm-up
Chapter 5 has been writing equations from mathematical information. This lesson does the same job with information given in words.
Discussion prompt
The number of movie theatres grew by about 750 a year, and in 1993 there were about 26,000. Which of the two forms from this chapter does that information fit, and why?
Hint: Count what you were given: a rate and what else?
Answer:
\[ m = 750, \quad (t_1, y_1) = (8, 26\,000) \]
A rate of change is a slope, and one year with its count is a point. A slope and a point is exactly what point-slope form accepts, so Lesson 5.2 handles this — once the year 1993 has been turned into a number.
Concept
A linear model is a linear function used to describe a real situation. A rate of change compares two quantities that are changing, and slope is how a rate of change is usually written.
rate of change — A comparison of two changing quantities, such as 750 theatres per year. When a situation is modelled by a line, the rate of change is its slope.
Naming the units of the rate is what identifies it as a slope rather than a plain number.
Figure (svg): A rate of change written as a slope with its units
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.5 Modeling with Linear Equations §5.5, pp. 298-298
Section
Section 1
Concept
A rate of change compares a change in one quantity with a change in another. Written as a fraction, that comparison is a slope, and its units are the units of the two quantities divided.
\[ m = \dfrac{\text{change in output}}{\text{change in input}} \]
Figure (svg): A rate of change written as a slope with its units
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.5 Modeling with Linear Equations §5.5, pp. 298-298 — the definitions of linear model and rate of change
Picture it
The word per is the division line.
Figure (svg): A rate of change written as a slope with its units
Writing the units under and over the line makes the slope's meaning visible, and it also catches the commonest modelling error — using a quantity where a rate belongs.
Worked example
Three descriptions, three slopes.
\[ \text{Name the slope and its units: } 750 \text{ theatres a year}, \; 25 \text{ million admissions a year}, \; 40 \text{ dollars a month.} \]
Take the first
Why: The output is theatres and the input is years.
\[ m = 750,\text{ theatres per year} \]
Take the second
Why: Admissions in millions per year.
\[ m = 25,\text{ millions per year} \]
Take the third
Why: Dollars per month, and a charge, so it may be negative in a balance model.
\[ m = 40,\text{ dollars per month} \]
Note what they share
Why: Each has an output unit over an input unit.
Figure (svg): The solution to Worked example identify the rate and its units shown as a ladder of expressions, one row per algebraic move
\[ 750 \tfrac{\text{theatres}}{\text{year}}, \quad 25 \tfrac{\text{million}}{\text{year}}, \quad 40 \tfrac{\text{dollars}}{\text{month}} \]
Verify: multiply each rate by an input and check the units
Why: Seven hundred and fifty theatres per year times eight years gives six thousand theatres — the years cancel and theatres remain. If the units of the product are not the units of the output, the rate has been misidentified.
Sorting
Look for a per, or ask whether it changes.
Sort into buckets
Sort each number by its role in a model.
The word per, or the phrase a year, is the giveaway. Any number without one cannot be a slope, whatever role the problem seems to give it.
Worked example
Distinguishing the two is the whole of the modelling setup.
\[ \text{In 1993 there were } 26\,000 \text{ theatres, growing by } 750 \text{ a year. Which is the slope?} \]
Look at the units of the first
Why: Theatres, with no per in it.
Say what it is in the model
Why: It is one output value, so it belongs to a point.
Look at the units of the second
Why: Theatres per year.
Say what it is in the model
Why: It is the slope.
\[ m = 750 \]
Figure (svg): The solution to Worked example a rate against a quantity shown as a ladder of expressions, one row per algebraic move
\[ m = 750, \quad y_1 = 26\,000 \]
Verify: ask what happens to each number over two years
Why: The count of theatres changes from year to year, so it cannot be the slope, which is fixed. The growth of 750 a year is the same in every year, which is exactly what a constant slope means. Asking which number stays constant identifies the slope without any units at all.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.5 Modeling with Linear Equations §5.5, pp. 298-298
Trap
In 1993 there were 26,000 theatres, growing by 750 a year.
Write m = 26,000, since that is the larger and more prominent number
Why: The count is what the problem is about, so it looks like the main number.
That would say the number of theatres grows by 26,000 every year. Its units are theatres, not theatres per year, so it cannot be a rate at all.
\[ m = 750 \text{ theatres per year}, \quad (8, 26\,000) \text{ a point} \]
Look for the per, or ask which number stays the same from year to year
Why: Only a rate has two units in it, and only a rate is constant across the whole model.
Checking units before writing anything down is the habit that prevents this, and it costs one line.
Faded example
Output unit over input unit.
Fill in the blanks
The rate 750 theatres a year is a slope of 750, in theatres per year, and 26,000 theatres is a quantity rather than a rate.
Why: A rate has two units and a quantity has one, which is the fastest way to tell them apart. The slope is the number that stays the same across the whole model, while the count of theatres changes from year to year.
Elimination
A phone plan charges 40 dollars a month and had a 90 dollar handset.
Eliminate the wrong options
Which number is the slope of the total-cost model?
Survives elimination: A
Why: Only the forty carries two units, dollars over months, which is what makes it a slope. Option C is worth naming because adding quantities with different units is a common instinct and produces a number with no meaning at all — a units check rejects it before any algebra.
Socratic
The word linear is doing real work here.
Discussion prompt
Explain what choosing a linear model assumes about the situation, and give one real situation where that assumption would be wrong. Then say how you might notice the problem from data.
Hint: Think about what a constant slope means.
Answer:
A line has one slope everywhere, so a linear model assumes the quantity changes by the same amount in every unit of input — 750 theatres in each and every year. That is an assumption about the world, not a fact about the arithmetic, and it is the thing most worth stating when reporting a model.
It fails for anything growing by a percentage rather than an amount: a savings account earning interest gains more each year as the balance grows, so its steps increase. From data you would notice it as differences that grow steadily rather than staying put — which is the same constant-step test used in Lesson 4.2, run in reverse to detect that a line is the wrong shape.
Section
Section 2
Concept
Years and other large inputs are usually replaced by a count from a chosen starting point. The choice is yours, and it must be stated alongside the model.
\[ t = 0 \text{ represents } 1985 \;\Longrightarrow\; 1993 \text{ is } t = 8 \]
Figure (svg): A timeline showing years converted into a variable counted from a chosen zero
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.5 Modeling with Linear Equations §5.5, pp. 298-298 — Example 1, where t = 0 represents 1985
Picture it
Subtract the chosen zero year.
Figure (svg): A timeline showing years converted into a variable counted from a chosen zero
Using the years themselves would work and would give an intercept describing the year zero — a number with no meaning and an enormous size. Counting from a nearby year keeps every number readable.
Worked example
With t equal to zero at 1985.
\[ \text{Convert } 1993, \; 1997 \text{ and } 2005 \text{ into values of } t. \]
Subtract the zero year from 1993
Why: 1993 minus 1985.
\[ t = 8 \]
Do the same for 1997
Why: 1997 minus 1985.
\[ t = 12 \]
And for 2005
Why: 2005 minus 1985.
\[ t = 20 \]
State the convention
Why: Every model using t must say that t equals zero is 1985.
Figure (svg): A timeline showing years converted into a variable counted from a chosen zero
\[ 1993 \to 8, \quad 1997 \to 12, \quad 2005 \to 20 \]
Verify: convert one back
Why: t equal to twelve gives 1985 plus twelve, which is 1997 — the last year of the data. Converting in both directions confirms the convention was applied consistently, which is where this kind of error usually hides.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.5 Modeling with Linear Equations §5.5, pp. 298-299
Translation
t equal to zero represents 1985.
Match the pairs
Why: Each value of t is the year minus 1985. The last one is outside the range 1985 to 1997 that the data covered, which is worth noticing before any prediction made at it is reported.
Worked example
The model changes and the predictions do not.
\[ \text{Rewrite the theatre model with } t = 0 \text{ at } 1993 \text{ instead of } 1985. \]
Re-express the known point
Why: 1993 is now t equal to zero.
\[ (0, 26000) \]
Keep the slope
Why: The rate of growth has not changed.
\[ m = 750 \]
Write the model
Why: The intercept is now the 1993 count.
\[ y = 750 t + 26000 \]
Check a prediction
Why: 2005 is now t equal to twelve, giving nine thousand plus twenty-six thousand.
\[ 35, 000 \]
Figure (svg): The solution to Worked example a different choice of zero shown as a ladder of expressions, one row per algebraic move
\[ y = 750t + 26\,000 \quad (t = 0 \text{ is } 1993) \]
Verify: compare the two models' predictions
Why: The first model gave 35,000 for 2005 at t equal to twenty and this one gives 35,000 at t equal to twelve. Different equations, different intercepts, identical predictions — because the two are describing the same situation with different labels on the axis.
Error analysis
The student modelled the theatres with t equal to zero at 1985.
Annotate
On: \( \begin{aligned} m &= 750 \\ (t_1, y_1) &= (1993, 26\,000) \\ y - 26\,000 &= 750(t - 1993) \end{aligned} \)
The fix is to convert every year the moment it appears, and to write the convention at the top of the working where it can be seen at every step.
Faded example
Subtract the chosen zero year.
Fill in the blanks
t = 0 \text8 1985: \quad 1993 \rightarrow t = 20, \quad 2005 \rightarrow t = ___
Why: Each conversion is a subtraction from the chosen zero year. Substituting the year itself instead is the standard error, and it produces an intercept describing the year zero — a number both meaningless and enormous.
Prediction
The model is y equals 750t plus 20,000, with t equal to zero at 1985.
Predict first
What does the 20,000 represent?
Correct: The number of theatres in 1985.
\[ t = 0: \; y = 750(0) + 20\,000 = 20\,000 \]
Why: The intercept is the value of the model when t is zero, and t equal to zero was defined to be 1985. So the model says there were about 20,000 theatres that year, a figure that was not measured but reconstructed by the algebra. The third option is what you would get by substituting years directly instead of converting them, which is exactly why the convention has to be stated.
Socratic
Nothing stops you substituting 1993 for the variable.
Discussion prompt
Explain what goes wrong if the year is used as the variable directly. Then say whether the model would be wrong, or merely awkward.
Hint: Work out where the graph would cross the vertical axis.
Answer:
The model would read y equals 750 times the year minus 1,469,750, so its intercept describes the year zero and is about negative one and a half million theatres. Every number in the equation becomes huge, the graph's interesting part sits two thousand units from the axis, and the intercept means nothing at all.
It would not be wrong. Substituting 2005 gives 35,000, the same prediction, because it is the same line with a different label on the horizontal axis. It is purely awkward — and the awkwardness is the reason for the convention, not any error it prevents.
Section
Section 3
Concept
With a rate of change and one data point, point-slope form gives the model directly. Converting to slope-intercept form then reveals the value at the chosen zero.
The intercept that appears was not among the given information.
Figure (svg): A rate and one data point assembled into a model
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.5 Modeling with Linear Equations §5.5, pp. 298-298 — Example 1, Write a Linear Model, and its Study Tip
Picture it
A slope and a point, as in Lesson 5.2.
Figure (svg): A rate and one data point assembled into a model
The Study Tip in the textbook says exactly this: because you are given the slope and a point on the line, use point-slope form. The modelling is in the setup, not in the algebra.
Worked example
This is Example 1 from the textbook.
\[ \text{Theatres grew by } 750 \text{ a year; in } 1993 \text{ there were } 26\,000. \text{ Model } y \text{ with } t = 0 \text{ at } 1985. \]
Identify the slope
Why: The rate of increase is 750 per year.
\[ m = 750 \]
Convert the observation
Why: 1993 is eight years after 1985.
\[ (8, 26000) \]
Write point-slope form
Why: Substitute all three numbers.
\[ y - 26000 = 750(t - 8) \]
Distribute and isolate y
Why: 750 times 8 is 6000, and adding 26,000 gives 20,000.
\[ y = 750 t + 20000 \]
Figure (svg): A rate and one data point assembled into a model
\[ y = 750t + 20\,000 \]
Verify: check the given observation
Why: At t equal to eight the model gives six thousand plus twenty thousand, which is twenty-six thousand — the figure for 1993. And the intercept of twenty thousand is the model's estimate for 1985, which was never given and came out of the algebra.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.5 Modeling with Linear Equations §5.5, pp. 298-298
Faded example
The slope and the converted point.
Fill in the blanks
m = 750, \; (8, 26\,000): \quad y - 26\,000 = 750(t - 8) \;\Longrightarrow\; y = 750t + 20000
Why: Distributing gives 750t minus 6000, and adding 26,000 to both sides leaves an intercept of 20,000. That intercept is the model's figure for 1985, produced by the algebra rather than measured.
Worked example
Guided Practice 1. The same structure, different numbers.
\[ \text{Attendance grew by about } 25 \text{ million a year; in } 1994 \text{ it was about } 1300 \text{ million.} \]
Identify the slope
Why: Twenty-five million admissions per year.
\[ m = 25 \]
Convert the observation
Why: 1994 is nine years after 1985.
\[ (9, 1300) \]
Write point-slope form
Why: Substitute.
\[ y - 1300 = 25(t - 9) \]
Convert
Why: Twenty-five times nine is 225, so the intercept is 1075.
\[ y = 25 t + 1075 \]
Figure (svg): The solution to Worked example the attendance model shown as a ladder of expressions, one row per algebraic move
\[ y = 25t + 1075 \]
Verify: check the observation and the units
Why: At t equal to nine the model gives 225 plus 1075, which is 1300 million — the given figure. The output is in millions throughout, so the intercept of 1075 means about 1075 million admissions in 1985, and stating the unit is part of the answer.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.5 Modeling with Linear Equations §5.5, pp. 298-298
Trap
In 1993 there were 26,000 theatres, growing by 750 a year.
Write y = 750t + 26,000 directly
Why: The observation is a starting figure in the sense that it is where you began, so it looks like b.
That model says there were 26,000 theatres in 1985, since t equal to zero is 1985. The observation belongs to 1993, which is t equal to eight, so it is a point rather than the intercept.
\[ y - 26\,000 = 750(t - 8) \;\Longrightarrow\; y = 750t + 20\,000 \]
Ask what value of t the observation belongs to before using it
Why: Only an observation at t equal to zero is the intercept.
Substituting t equal to eight into the finished model and checking that 26,000 comes back catches this immediately.
Elimination
Growth of 750 a year, with 26,000 in 1993 and t equal to zero at 1985.
Eliminate the wrong options
Which model is right?
Survives elimination: A
Why: Only the first satisfies both given facts: its slope is 750 and it gives 26,000 at t equal to eight. Each distractor fails one specific step — the conversion, the roles of the two numbers, or the choice of zero — and substituting t equal to eight rejects all three in one line.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| t = 0 at 1985 | t = 0 at 1993 | |
|---|---|---|
| The model | y = 750t + 20,000 | y = 750t + 26,000 |
| The slope | 750 | 750 |
| Prediction for 2005 | 35,000 at t = 20 | 35,000 at t = 12 |
The slopes and the predictions match and the intercepts do not, because the intercept is the only part of the model that depends on where the zero was put.
Socratic
Nobody measured 20,000 theatres in 1985.
Discussion prompt
Explain what the intercept of the theatre model actually is, given that it was not among the data. Then say how much confidence it deserves.
Hint: Ask what the model claims about 1985.
Answer:
It is the model's own estimate of the 1985 count, obtained by running the assumed constant growth backwards eight years from the single observation. It is a consequence of the two given facts rather than a third fact, so it inherits whatever error is in them.
It deserves the same confidence as the assumption of constant growth over those eight years. Since 1985 is inside the range the problem describes as growing at a constant rate, the estimate is reasonable — unlike an intercept extrapolated far outside the data, which would be a guess dressed as a number.
Section
Section 4
Concept
Once a model is written it can be used to predict unknown values, either by substituting algebraically or by reading a graph. Both assume the past pattern continues.
The two routes share no arithmetic, so agreeing is a real check.
Figure (svg): A prediction made both algebraically and by reading a graph
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.5 Modeling with Linear Equations §5.5, pp. 299-299 — Example 2, Use a Linear Model to Predict, with both methods
Picture it
An exact route and a fast one.
Figure (svg): A prediction made both algebraically and by reading a graph
The graph gives about 35,000 and the algebra gives exactly 35,000. When a prediction is being reported as approximate anyway, the graph is often enough.
Worked example
This is Example 2, Method 1, from the textbook.
\[ \text{Use } y = 750t + 20\,000 \text{ to predict the number of theatres in } 2005. \]
Convert the target year
Why: 2005 is twenty years after 1985.
\[ t = 20 \]
Substitute
Why: 750 times 20 is 15,000.
\[ y = 15000 + 20000 \]
Simplify
Why: The total is 35,000.
\[ y = 35000 \]
State the prediction
Why: About 35,000 theatres in 2005.
\[ \text{about } 35, 000 \]
Figure (svg): A prediction made both algebraically and by reading a graph
\[ y(20) = 35\,000 \]
Verify: check the prediction against the rate
Why: From 1993's 26,000 the model adds twelve years of growth at 750 a year, which is 9,000, giving 35,000. Reaching the same answer from the observation rather than the intercept is an independent route through the same model.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.5 Modeling with Linear Equations §5.5, pp. 299-299
Faded example
Convert the year, then substitute.
Fill in the blanks
2005: \; t = 20, \quad y = 750(20) + 20\,000 = 35000
Why: Converting the year first is the step that is most often skipped. Substituting twenty gives 15,000 plus 20,000, which is 35,000 theatres — and the answer should be reported as about 35,000, since the model is an approximation.
Worked example
This is Example 2, Method 2. The same question, read off a picture.
\[ \text{Read the prediction for } t = 20 \text{ off the graph of } y = 750t + 20\,000. \]
Plot the intercept
Why: At t equal to zero the model gives 20,000.
\[ (0, 20000) \]
Plot a second point
Why: At t equal to twenty it gives 35,000.
\[ (20, 35000) \]
Draw the line and find t equal to 20
Why: Read the height there.
\[ \text{about } 35, 000 \]
Compare with the algebra
Why: The two agree.
Figure (svg): A prediction made both algebraically and by reading a graph
\[ \text{at } t = 20, \; y \approx 35\,000 \]
Verify: say what each method is better at
Why: The algebra gives an exact figure and the graph shows the whole trend at once, including how far past the data the prediction sits. Since the answer is being reported as about 35,000 either way, the graph loses nothing — and it makes the extrapolation visible in a way the substitution does not.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.5 Modeling with Linear Equations §5.5, pp. 299-299
Trap
\[ y = 750t + 20\,000 \text{ at } t = 60 \]
Report 65,000 theatres in 2045
Why: The substitution is straightforward and the model gives a definite answer.
The model was built from data covering 1985 to 1997, so 2045 is nearly fifty years beyond it. The model has no evidence at all about that period, and the confident number hides that.
Report the prediction with the range the model came from.
State where the model was fitted whenever a prediction sits outside it
Why: A prediction inside the data is supported; one far outside is an assumption written as a number.
Reporting the range costs one clause and is the difference between a model and a claim.
Elimination
The model was built from data covering 1985 to 1997.
Eliminate the wrong options
What does predicting 2005 from it assume?
Survives elimination: A
Why: The textbook states this directly: when you predict, you assume the pattern established in the past will continue. That assumption is the whole content of the prediction, and it is what should be reported alongside the number.
Prediction
You need a figure accurate to the nearest hundred.
Predict first
Which prediction method is appropriate?
Correct: The algebraic one, since reading a graph is not that precise.
\[ y(20) = 35\,000 \text{ exactly, from the model} \]
Why: Reading a graph to the nearest hundred on an axis marked in five-thousands is not possible, so the substitution is the only route to that precision. Whether the model deserves that precision is a separate question worth asking — the fourth option raises a fair point about the model, but the question asked which method computes the figure, and the algebra does.
Socratic
One of them is exact.
Discussion prompt
Say what is gained by doing a prediction both algebraically and graphically, given that the algebra is more precise. Then say what the graph shows that the number does not.
Hint: Ask what each method could get wrong.
Answer:
The two share no arithmetic, so agreeing catches a slip that rechecking the substitution would not. A sign error or a mis-converted year would put the algebraic answer somewhere the drawn line plainly does not go, and the disagreement is visible at a glance.
The graph also shows how far past the data the prediction sits, which the number hides entirely. Seeing the point at t equal to twenty sitting well to the right of where the data ended is the clearest possible reminder that the prediction rests on an assumption — which is why the textbook shows both.
Section
Section 5
Concept
When a situation is described in words, write the relationship as a sentence first, then assign a label and a unit to each quantity, and only then write the equation.
This is the algebraic modelling procedure from Lesson 1.6.
Figure (svg): A verbal model turned into labels and then into an equation
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.5 Modeling with Linear Equations §5.5, pp. 300-300 — Example 3, with its verbal model, labels and algebraic model
Picture it
Words, labels, then algebra.
Figure (svg): A verbal model turned into labels and then into an equation
The middle stage is where the units get attached, and attaching them is what makes the final multiplication sensible: dollars per pound times pounds gives dollars.
Worked example
This is Example 3 from the textbook.
\[ \text{Hamburger is } 3 \text{ dollars a pound and chicken } 4, \text{ with } 60 \text{ dollars to spend. Model the combinations.} \]
Write the verbal model
Why: Price times weight, plus price times weight, equals total cost.
Label the quantities
Why: Three and four are prices per pound; x and y are weights in pounds.
Write the algebraic model
Why: Substitute the labels into the sentence.
\[ 3 x + 4 y = 60 \]
Note the form
Why: It arrived in standard form with no rearranging.
Figure (svg): A verbal model turned into labels and then into an equation
\[ 3x + 4y = 60 \]
Verify: check the units of every term
Why: Dollars per pound times pounds gives dollars, so each term on the left is in dollars and matches the sixty on the right. A term whose units did not come out in dollars would mean a label had been attached to the wrong quantity.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.5 Modeling with Linear Equations §5.5, pp. 300-300
Faded example
Price times weight, twice, adding to the total.
Fill in the blanks
\text4 4 \text5, \text___ 5, \text___ 60: \quad ___x + ___y = 60
Why: This is Guided Practice 4 from the textbook. Each coefficient is a price in dollars per pound and each variable a weight in pounds, so every term on the left is in dollars and can be compared with the sixty on the right.
Worked example
Part b of Example 3. Substituting several values of x.
\[ \text{For } 3x + 4y = 60, \text{ find } y \text{ at } x = 0, 4, 8, 12, 16, 20. \]
Substitute zero
Why: 4y equals 60, so y is fifteen.
\[ 15 \]
Substitute four
Why: Twelve plus 4y equals 60, so y is twelve.
\[ 12 \]
Continue
Why: Eight gives nine, twelve gives six, sixteen gives three.
\[ 9, 6, 3 \]
Substitute twenty
Why: Sixty plus 4y equals 60, so y is zero.
\[ 0 \]
Figure (svg): A table of trade-offs between two quantities under a fixed budget
\[ y = 15, 12, 9, 6, 3, 0 \]
Verify: check the step against the slope
Why: Each four extra pounds of hamburger costs three pounds of chicken, so the outputs fall by three at each stage — a constant step matching the slope of negative three quarters. The two ends of the table are the intercepts, twenty pounds of hamburger alone or fifteen pounds of chicken alone.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.5 Modeling with Linear Equations §5.5, pp. 300-300
Trap
Hamburger is three dollars a pound, chicken is four dollars a pound, and you have sixty dollars.
Write 3 + 4 = 60 or x + y = 60, since those are the numbers given
Why: Jumping straight to symbols skips the step where each number gets a meaning.
The first says nothing about weights and the second says the total weight is sixty pounds rather than the total cost being sixty dollars. Neither is the situation.
\[ 3x + 4y = 60 \quad x, y \text{ in pounds} \]
Write the sentence and the labels before any equation
Why: Each price multiplies its own weight, and only then do the two costs add to the total.
The units check is what confirms it: dollars per pound times pounds is dollars, and dollars added to dollars can equal sixty dollars.
Elimination
Two items at 3 and 4 dollars a pound, with 60 dollars to spend.
Eliminate the wrong options
Which equation models it?
Survives elimination: A
Why: Each price multiplies its own weight and the two costs add to the budget. A units check settles it immediately: only the first has dollars on both sides, and option D is not linear at all since it multiplies two variables together.
Hypothesis
Predict before you decide.
Predict first
Which values of x make sense in the barbecue model 3x plus 4y equals 60?
Correct: Only x from 0 to 20, since outside that one weight would be negative.
\[ 0 \le x \le 20 \quad \text{and} \quad 0 \le y \le 15 \]
Why: Beyond twenty pounds of hamburger the cost exceeds the budget, so the chicken weight would come out negative, which is meaningless. The equation does have solutions everywhere, and only the segment between the two intercepts describes a purchase. Fractional pounds are perfectly buyable, and the table's multiples of four were chosen for convenience rather than being a restriction.
Socratic
The equation is short and the sentence is long.
Discussion prompt
Explain what the verbal model and the labels stage protect against, given that the final equation has only three numbers in it. Then say which of the two stages you would keep if you could only keep one.
Hint: Think about what kind of error each stage catches.
Answer:
The verbal model fixes the structure — which quantities multiply and which add — before any symbol is chosen, so it prevents writing x plus y equals 60 when the prices should be doing the multiplying. The labels stage attaches units, which prevents a price being paired with the wrong weight and makes the final units check possible.
The labels are the more valuable of the two, because they make an independent check available: if the units on each side do not match, something is wrong regardless of how sensible the equation looks. The verbal model can be held in your head for a simple situation; units written down catch errors that reasoning alone will not.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| A rate over time | A fixed total | |
|---|---|---|
| Given | a rate and one observation | two unit prices and a budget |
| Natural form | point-slope, then slope-intercept | standard form |
| What the slope means | the rate of change | the trade-off between the two items |
The first kind grows and the second is fixed, and each arrives in the form that displays what it is about. Recognising which kind you have decides everything that follows.
Pattern
Whether the situation is a growing count or a fixed budget, the same five moves cover it.
Step five's second half is the one most often left out, and it is what separates a prediction that is supported from one that is merely computed.
OpenStax Elementary Algebra 2e, §4.6 Find the Equation of a Line §4.6
Check
Convert the year first.
Check your understanding
With t = 0 at 1990, what value of t represents 2004?
Answer: A
Why: The variable counts years from the chosen zero, so t is 2004 minus 1990, which is fourteen. Converting back gives 1990 plus fourteen, which is 2004.
Check
A rate and a point.
Check your understanding
A quantity grows by 40 a year and was 300 at t = 5. What is the model?
Answer: A
Why: Point-slope gives y minus 300 equals 40 times t minus 5, so y equals 40t plus 100. Substituting t equal to five gives 200 plus 100, which is 300, confirming the given observation.
Check
Prices multiply weights.
Check your understanding
Two items cost 5 and 2 dollars a pound, with 40 dollars to spend. Which models the combinations?
Answer: A
Why: Each price multiplies its own weight and the two costs add to the budget. Dollars per pound times pounds gives dollars on each term, matching the forty dollars on the right.
Real world
A city's recycling rate was 28 per cent in 2015 and has risen by about 1.5 percentage points a year since.
Discussion prompt
Build a model with a stated zero year, predict the rate for 2030, and say what the model implies about 2060 and whether you would report it.
Hint: The rate of increase is the slope; state your zero.
Answer:
\[ t = 0 \text{ at } 2015: \; (0, 28), \; m = 1.5 \;\Longrightarrow\; r = 1.5t + 28 \]
\[ 2030: \; t = 15, \; r = 22.5 + 28 = 50.5\% \]
The model predicts about fifty per cent by 2030, which is a plausible target for a city already at twenty-eight and improving steadily. The intercept here happens to be a measured value rather than a reconstructed one, because 2015 was chosen as the zero.
For 2060 the model gives 95.5 per cent, and by 2063 it exceeds one hundred — which is impossible, and is the clearest possible signal that the model has been pushed past where it means anything. Real recycling rates approach a ceiling rather than rising forever, so the linear model is a description of an early period rather than a law. Report the 2030 figure with the assumption stated; do not report the 2060 one at all.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
A count was 26,000 in 1993 and grows by 750 a year. With t = 0 at 1985, what is the y-intercept of the model?
Correct: 20,000, the model's value at t = 0.
\[ y - 26\,000 = 750(t - 8) \;\Longrightarrow\; y = 750t + 20\,000 \]
\[ t = 8: \; 6000 + 20\,000 = 26\,000 \;\checkmark \]
Why: The observation belongs to t equal to eight, not to t equal to zero, so it is a point rather than the intercept. Running the growth backwards eight years from 26,000 removes 6,000 and leaves 20,000 as the model's figure for 1985. The first option is the natural instinct and it is what makes this the standard error in the lesson: a given starting figure is only the intercept if it was measured at the chosen zero.
Explain it
They can write equations from numbers and freeze when handed a paragraph.
Discussion prompt
In no more than four sentences, explain how to turn a description of a steady change into an equation. Then tell them the one thing to check before writing any symbols.
Hint: Find the rate, find one observation.
Answer:
A usable answer: look for the number with a per in it — so much a year, so much a pound — because that is the slope. Then look for one moment where you know the actual amount, turn its date into a count from whatever year you decide to call zero, and you have a point. A slope and a point is point-slope form, which you already know.
Before writing anything, check the units of each number. A slope has two units and a quantity has one, so if you find yourself about to use a number with one unit as the slope, you have picked the wrong number.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Rate against quantity is fixed by checking units: a rate has two and a quantity has one. Year conversion is fixed by writing the convention at the top of the page and converting every year the moment it appears. Turning a total into an equation is fixed by writing the verbal model and the labels first. The assumption is fixed by naming the range the data covered and saying whether the prediction sits inside it. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page write a real situation with a steady rate of change, and beside the rate write its units as a fraction. Choose a zero for your input variable, state that choice in a sentence, and convert one observation into an ordered pair. Build the model in point-slope form and convert it, circling the intercept and writing beside it what that number means in your situation. To the right, graph the model and mark the region your data actually covered, shading it, and mark one prediction inside that region and one well outside. In the lower half, write a second situation with a fixed total split between two items, and produce its equation in three stages: the verbal sentence, the labelled quantities with units, and the algebra. Finally, in the margin, write one sentence saying what your first model assumes about the world.
The units in each term of your second model should all come out the same. If one term is in different units from the others, a price has been paired with the wrong quantity, and no amount of rechecking the arithmetic will find it.
Recap
Five things, and the last is what separates a model from an assertion.
| If the question says | Your first move is |
|---|---|
| Increases by so much per year | That is the slope |
| In 1993 there were this many | That is a point; convert the year |
| Let t = 0 represent 1985 | Subtract 1985 from every year |
| Predict the value in 2005 | Convert the year, then substitute |
| You have this much to spend | Write the verbal model, then standard form |
Lesson 5.6 closes the chapter with one more relationship between lines. Parallel lines shared a slope; perpendicular lines turn out to have slopes related in a way you can compute, which completes the geometry of Chapter 5.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 5 Writing Linear Equations — Lesson 5.5 Modeling with Linear Equations §5.5, pp. 298-304 — everything on these slides traces back here
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