The algebra in this lesson is “Variables on Both Sides”'s, unchanged. The work is everything around it: naming the unknown as a number, finding the one quantity the words describe twice, and deciding whether the number the equation produces is an answer the situation will accept.
Core ideas
Three habits separate a model that works from an equation that merely has the right numbers in it.
Write let m = the number of months, not let m = months. If m is a thing rather than a count then 20 + 0.15m means nothing, and no step afterwards can repair it. Every unknown gets a noun phrase and a unit.
The equals sign has to be earned. Find a single quantity — a cost, a total, a length — that the situation describes twice, write both descriptions, and set them equal. Without that sentence there is nothing to write.
m = 300 is what the algebra produces. Whether the situation accepts 300, and whether 300 is what was asked for, are two further questions. A model that skips them reports nonsense with confidence.
The material
Naming the unknown, building both sides, and the two judgements at the end that the algebra cannot make for you.
Nothing here is a new manipulation. Every equation in this lesson is solved by the steps of “One- and Two-Step Equations”, “Variables on Both Sides” and “Equations with Fractions”. What is new is that the equation has to be produced from a paragraph of English, and then handed back to that paragraph at the end.
A model is an equation whose letters stand for named quantities in a described situation, together with a statement of what each letter means and what unit it is measured in.
The statement of meaning is part of the model, not a courtesy. An equation whose letters are undefined cannot be checked against the situation, and that check is the only thing that makes it a model rather than an exercise.
Choose as the unknown the quantity that the others are described relative to. In the length is 3 cm more than twice the width, the width is the one to name: the length is then 2w + 3 and nothing else has to be invented. Naming the length instead is legal and gives a width of (l − 3)/2, which is more to carry.
An equation says two expressions have the same value. So the sentence you are looking for in the problem is the one that says two quantities are equal, or asks when they will be. Everything before that sentence builds the two expressions; the sentence itself supplies the =.
Plan A charges 20 up front and 0.15 for each unit. Plan B charges 35 up front and 0.10 for each unit. Let m be the number of units. Plan A's total cost is 20 + 0.15m and plan B's is 35 + 0.10m. The question — at how many units do they cost the same? — is what supplies the equals sign.
The decimals are fractions: 0.15 = 3/20 and 0.10 = 1/10 = 2/20, so the coefficient of m on the left exceeds the one on the right by 1/20. “Equations with Fractions”'s clearing move applies unchanged — multiply every term by 20 and the equation becomes 400 + 3m = 700 + 2m, with no decimals in it at all.
Call the first n. The three are n, n + 1 and n + 2, and their sum is 3n + 3. So 3n + 3 = 48, giving n = 15 and the numbers 15, 16, 17. If the question asked for the largest, the answer is 17 — n was defined as the first, and the definition is what you read at the end.
Change the total to 50 and the same model gives 3n + 3 = 50, so n = 47/3. The equation is solved and the model is not: there is no run of three consecutive whole numbers with sum 50. The fraction is the algebra reporting that fact, not failing to produce an answer.
Interactive
Built from the numbers, solved, then judged
Method
The equation is written in step 3 and solved in step 4. Steps 1 and 2 are where models are won, and step 4's second half is where they are lost.
let x = the number of litres added, not let x = litres. Pick the quantity the others are described relative to, so that the rest of the problem can be written down without inventing a second letter.
One at a time, in words first if it helps. Twice the width plus 3 becomes 2w + 3. Do this before you look for the equation; half the errors in modelling are translation errors made in a hurry.
This is the only place an equals sign can legitimately come from. If you cannot say in English which quantity is being measured twice, you do not yet have an equation.
Substituting back proves the arithmetic. It says nothing about whether −100 litres can be poured or whether the question wanted the first number or the last. Those are separate, and they are the point of modelling.
This is the transfer example. The conserved quantity is pure substance: amount before plus amount added must equal the target fraction of the final volume. You have 20 litres at 10% and add x litres at 50%; first aim at 25%, then at 60%.
The second solve is correct algebra. Every step is legal and the substitution confirms the value: x = −100 really does satisfy the equation. What it does not do is answer the question, because there is no such thing as pouring −100 litres into a tank.
The reason is visible without algebra. Mixing a 10% solution with a 50% solution always lands somewhere between 10% and 50%, and 60% is outside that range. The negative answer is the model reporting an impossible request. A write-up that stopped at x = −100 would have reported it as a recipe.
For a faded model, let m be the number of months. Two club plans cost 50 + 8m and 20 + 11m. The unknown and both expressions have been supplied; you must say which question earns an equals sign, write and solve that equation, check both costs, and decide whether the resulting number of months is usable before opening the plans preset.
Model two venues that cost 120 + 4p and 60 + 6p, where p is the number of people. State the equality question, solve it, check both totals and judge the value against the situation. Then explain why a negative or fractional p could pass the equation and still fail this model.
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