Turning a described situation into an equation: the verbal model with units, and the three strategies the lesson names — use a formula, look for a pattern, and draw a diagram — plus the two-category problems that need only one letter.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 1 — Equations and Inequalities
Use Problem Solving Strategies and Models
Objectives
Five outcomes. None of them is about solving equations; all of them are about producing one worth solving.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.5 Use Problem Solving Strategies and Models §1.5, pp. 34-39 — the lesson these objectives are drawn from
Warm-up
Lessons 1.2 and 1.3 both ended with a verbal model. This lesson is about choosing which one to build.
Discussion prompt
A train covers 457 miles in 6.5 hours. Before writing anything: which known formula already describes this, and which of its three letters is the unknown?
Hint: There is only one formula in the Lesson 1.4 table that mentions time.
Answer:
\[ d = rt \;\Longrightarrow\; 457 = r \cdot 6.5 \]
Distance equals rate times time. The distance and the time are given, so the rate is the unknown. Recognising that a standard formula already is the verbal model saves you from inventing one, and it is the first of the three strategies this lesson names.
Concept
By now you can solve any linear equation you are handed. The remaining difficulty is producing the right one from a paragraph of English. A verbal model does that in two stages: first the relationship in words with units, then the same relationship in symbols.
verbal model — A word equation describing a real situation, with each quantity labelled by its unit, written before the algebraic equation.
Three strategies produce a verbal model: use a formula you already have, look for a pattern in given values, or draw a diagram. Most problems in this book yield to one of the three.
Figure (svg): A three-row diagram: the relationship in words, then the same relationship with units, then the equation in symbols
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.5 Use Problem Solving Strategies and Models §1.5, pp. 34-34
Section
Section 1
Concept
Write the relationship as a sentence, and put a unit on every quantity in it. The units are not decoration: they are what tells you whether the quantities should be added or multiplied, and whether the sides of your equation match.
\[ \text{distance (mi)} = \text{rate (mi/h)} \times \text{time (h)} \]
If both sides of the model do not carry the same unit, the model is wrong and no amount of correct algebra will rescue it.
Figure (svg): A three-row diagram: the relationship in words, then the same relationship with units, then the equation in symbols
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.5 Use Problem Solving Strategies and Models §1.5, pp. 34-34
Picture it
The middle row is the one students skip, and it is the one doing the work.
Figure (svg): A three-row diagram: the relationship in words, then the same relationship with units, then the equation in symbols
Only the third row can be solved, but only the second row can be checked. Writing both is what makes an unfamiliar problem safe.
Worked example
Example 1. The Acela covers 457 miles between Boston and Washington in 6.5 hours.
\[ \text{Find the average speed, in miles per hour.} \]
Choose the strategy: a formula already exists
Why: Distance, rate and time is one of the eight standard formulas from Lesson 1.4, so no model needs inventing.
\[ d = r t \]
Write the verbal model with units
Why: Miles equals miles per hour times hours, and the hours cancel to leave miles. The shape is confirmed before any number appears.
\[ \text{miles } = (\text{mi} / h) \times(h) \]
Substitute the known values
Why: Four hundred and fifty-seven miles, six and a half hours, and an unknown rate.
\[ 457 = 6.5 r \]
Divide each side by 6.5
Why: The division property of equality, undoing the coefficient.
\[ 70.3 = r\text{ approx} \]
Figure (svg): The solution to Worked example the average speed of a train shown as a ladder of expressions, one row per algebraic move
\[ r \approx 70.3 \; \text{mi/h} \]
Verify: run the units forward through the original formula
Why: Seventy point three miles per hour times 6.5 hours is about 457 miles, and the hours cancel to leave miles — the unit the question asked for. Notice this is a unit check and an arithmetic check at once, which is why the textbook offers it as the check for this example.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.5 Use Problem Solving Strategies and Models §1.5, pp. 34-34
Matching
The distance formula, asked four different ways.
Match the pairs
Why: The same formula produces three different operations depending on which letter is unknown, and in every case the units name the operation before you have to remember any rule. Two of these are the same operation applied to different numbers, which is worth noticing: the shape of the question, not its wording, decides the arithmetic.
Worked example
Guided Practice 1. A jet averaging 540 miles per hour flies 6760 miles from New York to Tokyo.
\[ \text{How long does the flight take?} \]
Use the same formula, with time as the unknown now
Why: The formula does not change when the unknown moves; only which letter you solve for does.
\[ d = r t \]
Write the verbal model with units
Why: Miles equals miles per hour times hours, and the unknown is the hours.
\[ 6760 = 540 \times t \]
Divide each side by 540
Why: Miles divided by miles per hour gives hours, which is the unit the question wants.
\[ t = \frac{6760}{540} \]
Evaluate
Why: Six thousand seven hundred and sixty divided by five hundred and forty is about 12.5.
\[ t = 12.5\text{ approx} \]
Figure (svg): The solution to Worked example the same formula, a different unknown shown as a ladder of expressions, one row per algebraic move
\[ t \approx 12.5 \; \text{hours} \]
Verify: multiply back and check the size
Why: Five hundred and forty times twelve and a half is 6750 miles, within rounding of the 6760 given. The size is also plausible: at roughly 500 miles an hour, a 6760-mile flight has to take something over twelve hours, which matches what a transpacific flight actually takes.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.5 Use Problem Solving Strategies and Models §1.5, pp. 34-34
Trap
\[ \text{6760 miles at 540 miles per hour. How long?} \]
Multiply the two numbers, because both are given
Why: The operation is chosen from what is available rather than from what the units require.
\[ 6760 \times 540 = 3\,650\,400 \]
Miles times miles per hour gives miles squared per hour, which is not a time — and no journey takes three and a half million of anything.
\[ \text{6760 miles at 540 miles per hour. How long?} \]
Write the units and see which arrangement leaves hours
Why: Miles divided by miles per hour leaves hours, so the operation is division.
\[ \frac{6760 \text{ mi}}{540 \; \text{mi/h}} \approx 12.5 \text{ h} \]
The units decide the operation. When you cannot remember whether to multiply or divide, write them down and let them answer.
Sorting
Each of these claims to model a situation. Check only the units.
Sort into buckets
Sort each proposed model by whether its units make sense.
A model whose units do not balance is wrong before you check a single number, which is what makes this the cheapest check available.
Step zero
A car uses 14 gallons on a 550-mile trip, some of it on the highway at 40 miles per gallon and the rest in the city at 35.
Discussion prompt
Do not solve this. In plain English: how many unknowns are there really, how many letters do you need, and what unit does each side of your model carry? Say what forces the second quantity once you have named the first.
Hint: The two gallon amounts are not independent — something ties them together.
Answer:
There are two unknown gallon amounts but only one letter is needed, because the two must total 14. Naming the highway gallons g forces the city gallons to be 14 minus g.
\[ 550 = 40g + 35(14 - g) \]
Both sides carry miles: miles per gallon times gallons gives miles, twice over, and the two contributions add to the total distance.
Estimation
The Acela covers 457 miles in 6.5 hours.
Predict first
Roughly what is its average speed?
Correct: About 70 miles per hour.
\[ r = \frac{457}{6.5} \approx 70.3 \]
The size is also a sanity check on the situation: an express train averaging seventy miles an hour over a route with stops is entirely plausible, while three hundred would not be.
Why: Six and a half hours is close to seven, and seven times seventy is 490, a little above 457 — so the answer is a little under 70. The exact value is 70.3. Estimating first means that dividing the wrong way round, which would give about 0.014, could not survive, and neither could a slipped decimal point.
Section
Section 2
Concept
Many problems are a standard formula in disguise. Distance and rate, area, perimeter, temperature — if one of them already relates the quantities in the problem, it IS your verbal model and the work is just rearranging.
\[ d = rt, \quad A = lw, \quad P = 2l + 2w, \quad F = \tfrac{9}{5}C + 32 \]
Lesson 1.4 taught you to solve those formulas for any letter. This lesson is where that pays off, because the unknown is rarely the one the formula is written for.
Figure (svg): A unit cancellation showing miles per hour times hours giving miles, confirming the model
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.5 Use Problem Solving Strategies and Models §1.5, pp. 34-34
Picture it
Once the formula is chosen, the units confirm you chose the right one.
Figure (svg): A unit cancellation showing miles per hour times hours giving miles, confirming the model
The textbook offers exactly this as the check for Example 1. It is not an afterthought — it is the cheapest verification in the chapter.
Worked example
The same train problem, done the Lesson 1.4 way, so the two lessons connect.
\[ \text{Solve } d = rt \text{ for } r, \text{ then find the Acela's speed over 457 miles in 6.5 hours.} \]
Divide both sides by t
Why: The formula is a pure product, so one division isolates the rate — with the condition that the time is not zero.
\[ r = \frac{d}{t}, t\text{ not } 0 \]
Check the units of the rearranged formula
Why: Miles divided by hours gives miles per hour, which is what a rate should be.
\[ \text{mi} / h \]
Substitute the two known values
Why: Four hundred and fifty-seven miles over six and a half hours.
\[ r = \frac{457}{6.5} \]
Evaluate and round sensibly
Why: The given data has three significant figures, so an answer to three is honest.
\[ r = 70.3\text{ approx} \]
Figure (svg): The solution to Worked example rearrange the formula, then substitute shown as a ladder of expressions, one row per algebraic move
\[ r = \frac{d}{t} \approx 70.3 \; \text{mi/h} \]
Verify: substitute into the ORIGINAL formula
Why: Seventy point three times 6.5 is 456.95 miles, matching the 457 given to within rounding. Working forward through d equals rt rather than backward through the rearranged form is what makes this a genuine check on the rearrangement itself.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.5 Use Problem Solving Strategies and Models §1.5, pp. 34-34
Discrimination
Do not compute anything. Choose the formula from what the answer's unit must be.
Sort into buckets
Sort each question by the formula it needs.
Worked example
The temperature formula from the Lesson 1.4 table, used as a model rather than rearranged for its own sake.
\[ \text{A thermostat reads } 68\degree\text{F. What is that in degrees Celsius?} \]
Choose the formula that relates the two scales
Why: It already exists, so it is the verbal model and nothing needs inventing.
\[ F = (\frac{9}{5}) C + 32 \]
Substitute the known Fahrenheit reading
Why: Sixty-eight degrees Fahrenheit is the given quantity; Celsius is the unknown.
\[ 68 = (\frac{9}{5}) C + 32 \]
Subtract 32 from each side
Why: The added constant is the outermost operation on C, so it comes off first.
\[ 36 = (\frac{9}{5}) C \]
Multiply each side by 5/9
Why: The reciprocal of nine fifths, which leaves C alone.
\[ C = (\frac{5}{9}) (36) = 20 \]
Figure (svg): The solution to Worked example a formula with an addition in it shown as a ladder of expressions, one row per algebraic move
\[ C = 20\degree\text{C} \]
Verify: convert back through the original formula
Why: Nine fifths of 20 is 36, and 36 plus 32 is 68 degrees Fahrenheit — the reading we started from. The answer is also plausible: 20 degrees Celsius is comfortable room temperature, which is what a thermostat set to 68 is meant to be.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.5 Use Problem Solving Strategies and Models §1.5, pp. 34-34
Trap
\[ \text{A rectangular field is 65 m by 45 m. How much fencing surrounds it?} \]
Use the area formula, because both dimensions are given
Why: The formula is chosen from which numbers are available rather than from what the question asks for.
\[ A = lw = 65 \times 45 = 2925 \]
The answer carries square metres, but fencing is measured in metres. The unit alone shows the wrong formula was used.
\[ \text{A rectangular field is 65 m by 45 m. How much fencing surrounds it?} \]
Ask what the question wants, then pick the formula whose unit matches
Why: Fencing is a length around the outside, which is the perimeter.
\[ P = 2l + 2w = 2(65) + 2(45) = 220 \text{ m} \]
Area and perimeter use exactly the same two numbers, which is precisely why the unit of the answer is the thing to check first.
Fill the middle
A cyclist rides for 2.5 hours at 18 kilometres per hour.
Fill in the blanks
d = rt \;\Longrightarrow\; d = 18 \cdot 2.5 = 45 \text___
Why: Kilometres per hour times hours gives kilometres, so the two given numbers multiply. Eighteen times two and a half is forty-five. The unit cancellation is what confirms multiplication rather than division: had the question asked for a time, the same two units would have divided instead.
Commit first
Answer, then rate your confidence honestly.
Predict first
A journey is made at 60 km/h out and 30 km/h back, over the same road. What is the average speed for the round trip?
Correct: 40 km/h.
\[ \frac{2d}{\tfrac{d}{60} + \tfrac{d}{30}} = \frac{2d}{\tfrac{3d}{60}} = 40 \]
Why: Averaging the two rates gives 45, which is wrong, because the trip spends twice as long at the slower speed. Take a 60 km road: out takes 1 hour, back takes 2 hours, so 120 km in 3 hours, which is 40 km/h. The answer does not depend on the distance — try 120 km and it comes out 40 again — because the distances cancel. Rates average by time, not by rate, and this is the classic demonstration of it.
Explain it
A classmate says checking units is a waste of time because they can just look at the answer.
Discussion prompt
In three sentences, explain what a unit check catches that looking at the answer does not, and give them one concrete example where the number looks fine but the units reveal a wrong setup.
Hint: Area and perimeter of a rectangle are a good source of examples.
Answer:
A unit check catches the error before the arithmetic, so it costs nothing when the setup is right and saves the whole computation when it is not. Looking at the answer can only catch errors you already have an expectation about.
Concretely: a 65 by 45 field gives 2925 for area and 220 for perimeter. Both are perfectly reasonable-looking numbers, and only the units — square metres against metres — say which one answers the fencing question.
Section
Section 3
Concept
When the problem gives a table of values, subtract consecutive entries. If every difference is the same, the situation has a single constant rate and the model is a starting value plus or minus that rate times the count.
\[ h = 2000 - 250t \]
Check at least three differences, not one. A single matching gap can be a coincidence; three in a row is what licenses the model.
Figure (svg): A table of paramotor heights with the constant drop of 250 feet marked between consecutive entries
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.5 Use Problem Solving Strategies and Models §1.5, pp. 35-35
Picture it
Example 2: the height of a paramotorist at each minute of a descent.
Figure (svg): A table of paramotor heights with the constant drop of 250 feet marked between consecutive entries
The starting value 2000 is the entry at time zero, and the 250 is the common difference. Those two numbers are the whole model.
Worked example
Example 2. The table gives heights of 2000, 1750, 1500, 1250 and 1000 feet at minutes 0 through 4.
\[ \text{Find the height after 7 minutes.} \]
Subtract consecutive entries
Why: Two thousand minus 1750 is 250; 1750 minus 1500 is 250; and so on. Four identical differences.
\[ -250\text{ each minute} \]
Identify the starting value
Why: The entry at t equal to zero, before any descent has happened.
\[ 2000\text{ feet at } t = 0 \]
Write the verbal model with units
Why: Height equals initial height minus rate of descent times time. Feet equals feet minus feet per minute times minutes, and the minutes cancel.
\[ h = 2000 - 250 t \]
Substitute 7 for t
Why: Two hundred and fifty times seven is 1750.
\[ h = 2000 - 1750 \]
Evaluate
Why: Two thousand minus 1750 is 250 feet.
\[ h = 250 \]
Figure (svg): The solution to Worked example read the model off the table shown as a ladder of expressions, one row per algebraic move
\[ h = 2000 - 250t \qquad \text{and at } t = 7, \; h = 250 \text{ ft} \]
Verify: extend the table by hand
Why: Continuing the table past t equal to 4: at 5 minutes 750 feet, at 6 minutes 500, at 7 minutes 250. Counting down by 250 three more times reaches the same answer without using the formula at all, which is what confirms the formula matches the table.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.5 Use Problem Solving Strategies and Models §1.5, pp. 35-35
Pattern
A tank is draining. The table gives the volume in litres at each minute.
Step through it
How much is left after 8 minutes, and when does the tank run dry?
At eight minutes there are 180 minus 120, which is 60 litres. The model reaches zero at t equal to 12, and beyond that it predicts negative volumes — so 12 minutes is where the model stops describing anything real.
Worked example
Guided Practice 2. Heights of 2400, 2190, 1980, 1770 and 1560 feet at minutes 0 through 4.
\[ \text{Find the height after 8 minutes.} \]
Subtract consecutive entries to find the common difference
Why: Twenty-four hundred minus 2190 is 210, and every subsequent gap is 210 as well.
\[ -210\text{ each minute} \]
Write the model from the starting value and the rate
Why: The entry at time zero is 2400, and the descent is 210 feet per minute.
\[ h = 2400 - 210 t \]
Substitute 8 for t
Why: Two hundred and ten times eight is 1680.
\[ h = 2400 - 1680 \]
Evaluate
Why: Twenty-four hundred minus 1680 is 720 feet.
\[ h = 720 \]
Figure (svg): The solution to Worked example a different descent rate shown as a ladder of expressions, one row per algebraic move
\[ h = 2400 - 210t \qquad \text{and at } t = 8, \; h = 720 \text{ ft} \]
Verify: check the model against a value the table already gives
Why: At t equal to 3 the model gives 2400 minus 630, which is 1770 — exactly the table's fourth entry. A model that reproduces a value it was not built from is one you can trust to extend past the table.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.5 Use Problem Solving Strategies and Models §1.5, pp. 36-36
Error analysis
A student reads the first paramotor table and writes a model. One number is wrong.
Annotate
On: \( h = 250 - 2000t \;\Longrightarrow\; \text{at } t = 7, \; h = 250 - 14\,000 = -13\,750 \)
Always test a pattern model at t equal to zero. It must reproduce the starting value, and that single substitution catches a swap immediately.
Sorting
Each row is a sequence of values at times 0, 1, 2, 3. Subtract consecutive entries.
Sort into buckets
Sort each sequence by whether a constant-rate model fits.
Constant difference means add or subtract; constant ratio means multiply. Checking which one you have is the first thing to do with any table.
Edge cases
The paramotor model is h equals 2000 minus 250t.
Discussion prompt
What does the model predict at t equal to 10, and is that prediction meaningful? Find the exact time the model gives zero height, and say what the model can and cannot tell you about what happens after that moment.
Hint: Set the height to zero and solve for t.
Answer:
\[ 2000 - 250t = 0 \;\Longrightarrow\; t = 8 \]
At eight minutes the model gives zero, which is the landing. At ten minutes it gives negative 500 feet, which is not a height at all — the paramotorist is on the ground, not underneath it.
Every model built from a pattern has a domain: a range of inputs over which the pattern was observed and can be trusted. Here it is zero to eight minutes, and the model says nothing at all about what happens afterwards.
Reverse engineer
Here is a model. The table it came from has been lost.
Fill in the blanks
V = 180 - 15\,t
Why: The coefficient of t is the common difference of the original table, so the blank is 15 and the table read 180, 165, 150, 135 at times 0 through 3. Reading a model backwards into its table is the fastest way to check that the two numbers have not been swapped: substituting t equal to zero must return the first entry.
Section
Section 4
Concept
Some problems are about arrangement rather than arithmetic, and for those a sketch is not a study aid — it is the step that produces the correct model. The classic case is counting things placed along a line, where the count of objects and the count of gaps differ by one.
off-by-one — The error of counting the gaps between objects when the question asked for the objects, or the reverse. A drawing makes the difference visible immediately.
The two problems in this section are written for this deck rather than quoted from the textbook, because the book's own diagram example is set in artwork. The strategy is the lesson's; the numbers are ours.
Figure (svg): A straight fence 96 feet long with posts every 8 feet, drawn so that the posts can be counted including both ends
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.5 Use Problem Solving Strategies and Models §1.5, pp. 34-34 — the lesson names draw a diagram as one of its three strategies
Picture it
A straight fence 96 feet long, with a post every 8 feet including both ends.
Figure (svg): A straight fence 96 feet long with posts every 8 feet, drawn so that the posts can be counted including both ends
The division gives 12, and 12 is the right answer to a question nobody asked. The drawing is what shows that the posts number one more than the gaps.
Worked example
Written for this deck. The arithmetic is trivial; the modelling is not.
\[ \text{A straight fence is 96 ft long with a post every 8 ft, including both ends. How many posts?} \]
Draw the fence and mark the first two posts
Why: Even a rough sketch fixes that the first post sits at zero feet, not at eight.
\[ \text{post at } 0,\text{ post at } 8 \]
Count what the division actually gives
Why: Ninety-six divided by eight is twelve, and twelve counts the eight-foot GAPS between posts.
\[ 12\text{ gaps} \]
Read the drawing for the relationship
Why: Each gap has a post at its right end, and there is one extra post at the very start.
\[ \text{posts } =\text{ gaps } +1 \]
Add one
Why: Twelve gaps means thirteen posts.
\[ 13\text{ posts} \]
Figure (svg): The solution to Worked example counting posts along a fence shown as a ladder of expressions, one row per algebraic move
\[ \text{posts} = \frac{96}{8} + 1 = 13 \]
Verify: check the relationship on a tiny case you can count
Why: A 16-foot fence with posts every 8 feet has posts at 0, 8 and 16 — three posts and two gaps. The plus-one holds, and it holds for the same reason at any length: the starting post has no gap before it.
Prediction
Commit before reasoning it through.
Predict first
Trees are planted every 5 metres along a straight 100-metre path, with a tree at each end. How many trees?
Correct: 21 trees.
\[ \text{gaps} = \frac{100}{5} = 20 \qquad \text{trees} = 21 \]
The same reasoning reverses if the ends are not included, in which case there would be 19. Which version applies is decided by the wording, not by the arithmetic, which is why the sketch has to be drawn from the words.
Why: One hundred divided by five is twenty, and that counts the five-metre gaps. Since there is a tree at the start of the path as well as at the end of every gap, the trees number one more than the gaps. Drawing a 10-metre path with trees every 5 metres settles it: trees at 0, 5 and 10 — three trees, two gaps.
Worked example
Written for this deck. Two dimensions, one letter, because the words tie them together.
\[ \text{A rectangular field's length is 20 m more than its width, and 220 m of fencing surrounds it. Find both dimensions.} \]
Draw the rectangle and label the width w
Why: The drawing is what makes the next step obvious rather than a guess.
\[ \text{width } = w \]
Write the length in terms of w
Why: Twenty more than the width, which is what ties the two unknowns into one letter.
\[ \text{length } = w + 20 \]
Use the perimeter formula as the verbal model
Why: Perimeter equals twice the length plus twice the width, and both sides carry metres.
\[ 2(w + 20) + 2 w = 220 \]
Distribute and combine
Why: Two w plus forty plus two w is four w plus forty.
\[ 4 w + 40 = 220 \]
Solve for w, then find the length
Why: Four w is 180, so w is 45, and the length is 45 plus 20.
\[ w = 45, l = 65 \]
Figure (svg): The solution to Worked example a field described by a relationship shown as a ladder of expressions, one row per algebraic move
\[ w = 45 \text{ m}, \quad l = 65 \text{ m} \]
Verify: compute the perimeter from both dimensions
Why: Twice 65 plus twice 45 is 130 plus 90, which is 220 metres — the fencing given. The relationship also checks: 65 really is 20 more than 45, which confirms the labelling step and not just the arithmetic.
Trap
\[ \text{96 ft of fence, a post every 8 ft including both ends.} \]
Divide the length by the spacing
Why: The division is treated as the answer because it is the only arithmetic in sight.
\[ \frac{96}{8} = 12 \text{ posts} \quad \text{(wrong)} \]
Twelve is the number of gaps. Nobody asked for the gaps.
\[ \text{96 ft of fence, a post every 8 ft including both ends.} \]
Draw it, then count posts and gaps separately
Why: The sketch shows a post at each end and one at each eight-foot mark between them.
\[ \text{gaps} = \frac{96}{8} = 12, \qquad \text{posts} = 12 + 1 = 13 \]
Test it on a fence short enough to count by hand and the plus-one is unmistakable.
Translation
Each sentence fixes one dimension in terms of another.
Match the pairs
Why: In every case one dimension is named with a letter and the other is written in terms of it, which is the move that turns two unknowns into one. Notice the third one names the LENGTH as the letter, because the sentence describes the width in terms of the length — always label whichever quantity the sentence treats as the reference.
Missing information
A problem arrives incomplete.
Discussion prompt
A rectangular garden is surrounded by 60 metres of fencing. What are its dimensions? Say exactly what extra piece of information you would have to ask for before this question has a single answer, and give two different rectangles that both satisfy what you have been told.
Hint: Try a long thin rectangle and a nearly square one.
Answer:
The perimeter alone does not fix a rectangle. A 25 by 5 rectangle and a 20 by 10 rectangle both have a perimeter of 60 metres, and so does a 29 by 1.
\[ 2(25) + 2(5) = 60 \qquad 2(20) + 2(10) = 60 \]
You would have to ask for a relationship between the sides — that the length is 20 more than the width, say, or that the length is twice the width. One equation cannot pin down two independent unknowns, which is exactly the point Chapter 3 develops into systems of equations.
Comparison
Fill the blanks. The relationship depends only on whether the ends are included.
Comparison matrix
| Arrangement | Gaps | Objects |
|---|---|---|
| Posts on a straight fence, both ends | length / spacing | gaps + 1 |
| Posts on a straight fence, neither end | length / spacing | gaps - 1 |
| Posts round a circular field | circumference / spacing | gaps (they are equal) |
| 96 ft fence, 8 ft spacing, both ends | 12 | 13 |
The circular case is the one worth remembering: going round a loop, the last gap closes back onto the first post, so objects and gaps are equal. The plus-one is a fact about lines, not about counting.
Section
Section 5
Concept
When a total is split between two categories, name one part with a letter and write the other as the total minus that letter. Two unknowns become one, and one equation is enough.
\[ 550 = 40g + 35(14 - g) \]
This is the same move you met in Lesson 1.2 with the fifteen prints. Recognising it as a pattern rather than a trick is what makes it available under pressure.
Figure (svg): A bar split into two parts labelled highway gallons g and city gallons 14 minus g, with the mileage rate written under each
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.5 Use Problem Solving Strategies and Models §1.5, pp. 36-36
Picture it
Example 4: fourteen gallons split between highway and city driving.
Figure (svg): A bar split into two parts labelled highway gallons g and city gallons 14 minus g, with the mileage rate written under each
The bar is fixed at fourteen. Once the left piece is called g, the right piece has no freedom left — it must be fourteen minus g.
Worked example
Example 4. A car used 14 gallons over 550 miles, at 40 miles per gallon on the highway and 35 in the city.
\[ \text{How many gallons were used on the highway?} \]
Name the highway gallons g
Why: The question asks for the highway amount, so naming that one directly avoids an extra step at the end.
\[ \text{highway } = g\text{ gallons} \]
Write the city gallons as the total minus g
Why: The two amounts must total fourteen, which forces the second once the first is named.
\[ \text{city } = 14 - g\text{ gallons} \]
Write the verbal model in units
Why: Miles per gallon times gallons gives miles, twice over, and the two contributions add to the total distance.
\[ 550 = 40 g + 35(14 - g) \]
Distribute the 35
Why: Thirty-five times fourteen is 490, and thirty-five times negative g is negative 35g.
\[ 550 = 40 g + 490 - 35 g \]
Combine and solve
Why: Forty g minus thirty-five g is five g; subtracting 490 gives 60 equals 5g, so g is 12.
\[ g = 12 \]
Figure (svg): The solution to Worked example gallons on the highway shown as a ladder of expressions, one row per algebraic move
\[ g = 12 \text{ gallons on the highway} \]
Verify: compute the two distances separately
Why: Twelve gallons at 40 miles per gallon is 480 miles, and the remaining 2 gallons at 35 miles per gallon is 70 miles. Four hundred and eighty plus seventy is 550 miles, and twelve plus two is fourteen gallons — both given totals recovered.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.5 Use Problem Solving Strategies and Models §1.5, pp. 36-36
Matching
Each situation splits a fixed total between two categories.
Match the pairs
Why: Every one has the same shape: a fixed total, a letter for one part, and the total minus that letter for the other. The units differ — gallons, prints, coins, hours — but the modelling move does not, which is what makes it worth recognising as a pattern rather than learning four times.
Worked example
Guided Practice 4. A truck used 28 gallons over 428 miles, at 16 miles per gallon on the highway and 12 in the city.
\[ \text{How many gallons were used in the city?} \]
Name the city gallons c, since that is what is asked for
Why: Naming the quantity the question wants saves converting at the end.
\[ \text{city } = c\text{ gallons} \]
Write the highway gallons as 28 minus c
Why: The two must total twenty-eight.
\[ \text{highway } = 28 - c \]
Write the model: miles from each part, added
Why: Twelve miles per gallon times c gallons, plus sixteen times the rest, gives the total 428 miles.
\[ 428 = 12 c + 16(28 - c) \]
Distribute and combine
Why: Sixteen times twenty-eight is 448, and 12c minus 16c is negative 4c.
\[ 428 = 448 - 4 c \]
Solve
Why: Subtracting 448 gives negative 20 equals negative 4c, so c is 5.
\[ c = 5 \]
Figure (svg): The solution to Worked example gallons in the city shown as a ladder of expressions, one row per algebraic move
\[ c = 5 \text{ gallons in the city} \]
Verify: compute the two distances separately
Why: Five gallons at 12 miles per gallon is 60 miles, and the other 23 gallons at 16 miles per gallon is 368 miles. Sixty plus 368 is 428 miles, and five plus twenty-three is twenty-eight gallons. Both totals check, and the answer is plausible: most of this trip was highway, which the mileage figures already suggested.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.5 Use Problem Solving Strategies and Models §1.5, pp. 36-36
Error analysis
A student sets up the fuel problem with a letter for each category.
Annotate
On: \( 550 = 40g + 35c \quad \text{with } g \text{ and } c \text{ both unknown} \)
This is not a wrong equation so much as an incomplete use of the information. Chapter 3 will teach you to keep both letters and both equations; for now, substitute one into the other at the modelling stage.
Fill the middle
Twenty coins, all nickels and dimes, worth 1.55 dollars in total.
Fill in the blanks
0.05k + 0.10(20 - k) = 1.55
Why: Naming the nickels k forces the dimes to be twenty minus k, since the two must total twenty coins. The equation then has one unknown: 0.05k plus 2 minus 0.10k equals 1.55, giving negative 0.05k equals negative 0.45, so k is 9 nickels and 11 dimes. Checking: 45 cents plus 110 cents is 155 cents.
Real world
A concert sold 400 tickets and took 7200 dollars. Adult tickets cost 24 dollars and student tickets cost 12.
Discussion prompt
Set this up with a single letter, write the equation, and say how many of each were sold. Then say what you would check to be confident the answer is right, and what would have told you immediately if you had mixed up which price went with which letter.
Hint: Name the adult tickets and let the students be what is left.
Answer:
\[ 24a + 12(400 - a) = 7200 \;\Longrightarrow\; 12a + 4800 = 7200 \;\Longrightarrow\; a = 200 \]
Two hundred adult and two hundred student tickets. Check both totals: 200 plus 200 is 400 tickets, and 4800 plus 2400 is 7200 dollars.
Had the prices been swapped the answer would have come out as 200 again by coincidence here — so the real safeguard is the average: 7200 over 400 is 18 dollars a ticket, exactly halfway between 12 and 24, which is only possible with an even split.
Two truths and a lie
All three claim to model the fuel problem.
Eliminate the wrong options
Two of these are correct. Knock those out and keep the broken one.
Survives elimination: C
Why: The survivor is the broken one. Using the same letter for both categories claims the car used g gallons on the highway AND g gallons in the city, which would make the total 2g rather than 14. It gives g equal to about 7.33, so a total of 14.67 gallons — contradicting the 14 given, which is the check that exposes it.
Comparison
Fill the blanks. The form the information arrives in chooses the strategy.
Comparison matrix
| The problem gives you | Strategy | What you produce |
|---|---|---|
| Quantities a known formula relates | use a formula | the formula, rearranged |
| A table of values | look for a pattern | start plus or minus rate times t |
| A shape, a layout, or things in a row | draw a diagram | a labelled sketch, then an equation |
| A total split between two categories | name one, subtract for the other | one equation in one letter |
| A rate and a fixed starting amount | look for a pattern, or use a formula | fixed plus rate times variable |
Notice the last row can be reached two ways. Strategies overlap, and that is fine — the goal is a correct model, not the officially intended route to it.
Pattern
One routine turns any of this chapter's word problems into an equation.
Step four is the one that gets skipped, and skipping it is why word problems feel like guessing. The unit check is what turns a guess into a decision.
OpenStax Algebra and Trigonometry 2e, §2.3 Models and Applications §2.3
Check
A formula problem. Watch which letter is unknown.
Check your understanding
A jet averages 540 miles per hour. How long does it take to fly 6760 miles?
Answer: A
Why: Miles divided by miles per hour gives hours: 6760 over 540 is about 12.5. Multiplying back, 540 times 12.5 is 6750 miles, matching to within rounding.
Check
A pattern problem. Find the common difference first.
Check your understanding
A paramotorist's height in feet is 2400, 2190, 1980, 1770 at minutes 0, 1, 2, 3. What is the height after 8 minutes?
Answer: A
Why: The common difference is 210 feet per minute, so h = 2400 - 210t. At t = 8 that is 2400 - 1680, which is 720 feet. Checking at t = 3 gives 1770, matching the table.
Check
A two-category problem. One letter is enough.
Check your understanding
A truck used 28 gallons over 428 miles, at 16 miles per gallon on the highway and 12 in the city. How many gallons were used in the city?
Answer: A
Why: Letting c be the city gallons, 12c + 16(28 - c) = 428 gives 448 - 4c = 428, so c = 5. Checking: 5 gallons at 12 mpg is 60 miles and 23 at 16 mpg is 368, totalling 428.
Real world
You are stocking a bake sale. Cookies cost 40 cents each to make and sell for 1.50; brownies cost 70 cents and sell for 2.25. You have 60 dollars of ingredients money and want to make exactly 100 items.
Discussion prompt
Set this up with one letter, write the equation, and find how many of each you can make. Then say which of the three strategies from this lesson you used, and what you would check before trusting the answer.
Hint: Name the cookies and let the brownies be what is left of the hundred.
Answer:
\[ 0.40c + 0.70(100 - c) = 60 \;\Longrightarrow\; 70 - 0.30c = 60 \;\Longrightarrow\; c = \tfrac{100}{3} \approx 33.3 \]
This is the two-category move, and the answer comes out fractional — about 33 cookies and 67 brownies, costing 59.90, or 34 and 66 costing 59.60. Neither hits 60 exactly.
That is worth noticing rather than rounding away: the model is continuous but the situation is not, so the honest answer is that no whole-number combination spends exactly 60 dollars. Checking whether the answer has to be a whole number is part of checking against the situation, not against the equation.
Commit first
Answer, then rate your confidence honestly.
Predict first
A table of values has a constant difference of 12 between consecutive entries. Does that guarantee the model is a starting value plus 12 times t?
Correct: Yes, but only if the times are equally spaced.
\[ \text{equally spaced } t: \quad \Delta h = 12 \;\Longrightarrow\; \text{rate} = 12 \text{ per unit} \]
Why: A constant difference gives a constant rate only when each difference covers the same amount of time. If the table listed heights at minutes 0, 1, 3 and 6, a difference of 12 between consecutive entries would mean 12 per minute, then 6 per minute, then 4 — not one rate at all. Checking the spacing of the input column is a step that gets skipped because textbook tables almost always space them evenly.
Explain it
They can solve any equation you hand them but freeze at word problems.
Discussion prompt
In four sentences or fewer, give them a procedure for turning a paragraph into an equation. Include the one check that tells them whether their model is wrong before they solve it.
Hint: The check is one sentence and it involves units.
Answer:
Read the question and write down what unit the answer must carry. Then write the relationship in words with a unit on every quantity, choosing whichever of the three strategies fits — a known formula, a pattern in a table, or a drawing.
Replace the words with symbols only after the words are written down. The check is that both sides of the word version must carry the same unit; if they do not, the model is wrong and solving it will waste your time.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For multiply-or-divide, write the units and see which arrangement leaves the unit the question wants. For tables, check three differences before committing and test the model at t equal to zero. For spatial problems, always sketch a small version you can count by hand to find the relationship. For two categories, name the one the question asks for and write the other as the total minus it. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Divide a page into three columns headed use a formula, look for a pattern, and draw a diagram. In each column write one problem from this lesson that the strategy solved, the verbal model with units underneath it, and the final equation beneath that. Across the bottom of the page, draw one bar split into two pieces, label the left piece with a letter and the right piece with the total minus that letter, and write out the fuel-efficiency equation beneath it. Finally, in the margin, write the single sentence you would say to yourself at the start of any word problem — the one about what unit the answer has to carry — and box it.
If the boxed sentence is not about units, write it again. That sentence is the whole lesson compressed to one line.
Recap
Five things, and none of them is solving an equation — that part you already had.
| If the problem gives you | Reach for |
|---|---|
| Distance, rate or time | The formula d equals rt |
| A table of values | The common difference |
| Objects in a row or a shape | A sketch you can count |
| A total split two ways | One letter, and total minus it |
| Anything at all | The units, before the symbols |
Lesson 1.6 takes the same modelling and changes the relationship from equals to at least or at most, which turns an equation into an inequality.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.5 Use Problem Solving Strategies and Models §1.5, pp. 34-39 — everything on these slides traces back here
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