Word Problems with Systems: Free Response

Two quantities are unknown, the number of bicycles and the number of tricycles, so each gets a letter of its own: let $b$ be the number of bicycles and $t$ the number of tricycles, both counts of machines.

The description states two separate facts, so expect two equations. The first counts machines, where a bicycle and a tricycle each contribute exactly one:

$b + t = 24.$

The second counts wheels, where a bicycle contributes $2$ and a tricycle contributes $3$:

$2b + 3t = 55.$

The two equations count different things, machines in one and wheels in the other, and that is what makes them two conditions rather than one condition written twice.

The machine equation is the easier one to rearrange, so use it to write one letter in terms of the other:

$b = 24 - t.$

Put that into the wheel equation, which then has a single unknown in it:

$2(24 - t) + 3t = 55, \qquad 48 - 2t + 3t = 55, \qquad 48 + t = 55, \qquad t = 7.$

Back-substitute into the machine equation: $b = 24 - 7 = 17$.

Now test the pair against both facts as they were stated, not just against the equation it was substituted into. Machines: $17 + 7 = 24$. Wheels: $2(17) + 3(7) = 34 + 21 = 55$. Both hold, so the rack holds $17$ bicycles and $7$ tricycles.

Take any whole number of bicycles from $0$ to $24$ and let the tricycles make the count up to $24$. Every one of those $25$ pairs satisfies the machine count. For instance $20$ bicycles and $4$ tricycles gives $20 + 4 = 24$ machines, yet its wheels come to $2(20) + 3(4) = 40 + 12 = 52$, three short of the $55$ counted on the rack.

So the machine count alone cannot answer the question: it leaves $25$ candidates standing. What the wheel count adds is a second, different measurement of the same rack. Writing $t = 24 - b$, the wheels of a candidate pair come to

$2b + 3(24 - b) = 72 - b.$

That expression takes a different value for every one of the $25$ candidates, running from $72$ wheels when there are no bicycles down to $48$ when they all are. Exactly one candidate therefore carries $55$ wheels, which is why the second fact narrows $25$ possibilities to one and why a problem about two unknown counts needs two conditions rather than one.

Let $x$ be the kilograms of house tea in the batch and $y$ the kilograms of jasmine tea. One fact fixes the weight of the batch:

$x + y = 10.$

The other fact is money, and the price given is per kilogram, so turn it into the cost of the whole batch first: $10$ kilograms at $15$ dollars each is $150$ dollars. Each tea contributes its own cost:

$12x + 20y = 150.$

From the weight equation $x = 10 - y$. Substitute:

$12(10 - y) + 20y = 150, \qquad 120 + 8y = 150, \qquad 8y = 30, \qquad y = 3.75,$

and then $x = 10 - 3.75 = 6.25$. Test both facts: the weight is $6.25 + 3.75 = 10$ kilograms, and the leaves cost $12(6.25) + 20(3.75) = 75 + 75 = 150$ dollars, which is $15$ dollars for each of the ten kilograms. Tea is weighed rather than counted, so a quarter of a kilogram is a perfectly good amount to scoop out.

Only the target changes. Ten kilograms at $22$ dollars each is $220$ dollars, so the system is

$x + y = 10, \qquad 12x + 20y = 220.$

Substituting $x = 10 - y$ as before:

$120 + 8y = 220, \qquad 8y = 100, \qquad y = 12.5,$

and then $x = 10 - 12.5 = -2.5$.

Nothing has gone wrong with the algebra, and it is worth checking that: $-2.5 + 12.5 = 10$, and $12(-2.5) + 20(12.5) = -30 + 250 = 220$. The pair satisfies both equations exactly. What rejects it is the shop, not the arithmetic. A mass of tea cannot be negative, and in any case $12.5$ kilograms of jasmine would already overflow a ten-kilogram batch on its own. So the equations faithfully describe an order that cannot be filled, and the honest report is that this batch is impossible rather than that the answer is $-2.5$ kilograms.

Write the cost of a batch in terms of a single mass. With $x + y = 10$, so $x = 10 - y$, the leaves cost

$12(10 - y) + 20y = 120 + 8y.$

Nothing outside the range is possible. A batch cannot hold a negative mass of either tea, so $y$ runs only from $0$ to $10$, and $120 + 8y$ therefore runs only from $120$ dollars to $200$ dollars. Divided by the ten kilograms, that is $12$ to $20$ dollars per kilogram, and no order outside those bounds can be met.

Everything inside the range is possible. Given a target of $P$ dollars per kilogram the batch must cost $10P$, so

$120 + 8y = 10P, \qquad y = \frac{10P - 120}{8}.$

At $P = 12$ this gives $y = 0$ and at $P = 20$ it gives $y = 10$, and larger targets give larger $y$, so every $P$ between the two bounds returns a $y$ between $0$ and $10$, with $x = 10 - y$ also between $0$ and $10$. Both masses are then amounts a batch can actually hold. The two endpoints are the unmixed batches, all house tea at the bottom and all jasmine at the top.

The two earlier orders agree with this. A target of $15$ dollars lies inside the range and gives $y = \frac{150 - 120}{8} = 3.75$ kilograms; a target of $22$ dollars lies outside it and gives $y = \frac{220 - 120}{8} = 12.5$ kilograms, more jasmine than the batch weighs.

Two things about the printer are unknown, so give each a letter: let $f$ be the setup fee in dollars and $p$ the price of one shirt in dollars. Each past order is one fact about that pair, the fee being paid once however many shirts are ordered:

$f + 20p = 145, \qquad f + 50p = 265.$

The fee appears with the same coefficient in both, so subtracting the smaller order from the larger removes it:

$30p = 120, \qquad p = 4.$

Then $f = 145 - 20(4) = 145 - 80 = 65$. Test both orders as they were stated: $65 + 20(4) = 145$ dollars, and $65 + 50(4) = 265$ dollars. The setup fee is $65$ dollars and each shirt costs $4$ dollars.

Let $x$ be the number of shirts and $y$ the number of dollars. The printer's bill for $x$ shirts is the fee plus $4$ dollars a shirt, and the club's income is $12$ dollars a shirt:

$y = 4x + 65, \qquad y = 12x.$

Both are already solved for $y$, so set the two expressions equal:

$12x = 4x + 65, \qquad 8x = 65, \qquad x = 8.125,$

and then $y = 12(8.125) = 97.50$. Check the point on both lines: $4(8.125) + 65 = 32.5 + 65 = 97.50$ dollars, and $12(8.125) = 97.50$ dollars.

Shirts are ordered whole, so test the two whole numbers on either side. At $8$ shirts the club takes $12(8) = 96$ dollars and pays $4(8) + 65 = 97$ dollars, so it is a dollar down. At $9$ shirts it takes $12(9) = 108$ dollars and pays $4(9) + 65 = 101$ dollars, so it is seven dollars up. The smallest whole order that does not lose money is therefore $9$ shirts.

The two lines are drawn over the same pair of quantities, shirts and dollars, so their meeting point reads as $8.125$ shirts and $97.50$ dollars: the size of order at which what the club takes and what it pays are the same amount of money.

The algebra is under no obligation to respect the situation. Both equations hold at that point, and it still describes nothing the club can do, because a printer does not print an eighth of a shirt. The system is a faithful model of the money and a careless model of the goods, and it is the reader who has to supply the missing requirement that $x$ be a whole number.

So the choice lies between the two whole orders on either side, and each has to be priced out separately. At $8$ shirts, income against cost:

$12(8) = 96 \ \text{ against } \ 4(8) + 65 = 97.$

At $9$ shirts:

$12(9) = 108 \ \text{ against } \ 4(9) + 65 = 101.$

So the club is a dollar down at $8$ shirts and seven dollars up at $9$. Every extra shirt brings in $12$ dollars and costs $4$, closing $8$ dollars of the gap, so below the meeting point every whole order loses money and above it every whole order makes some. That is why the right move is to take the first whole number past the meeting point rather than the whole number nearest to it: here the nearest one is $8$, and $8$ shirts leaves the club out of pocket.

Work on the sentences as arithmetic, with the first number $60$ and the second $10$, and take them one at a time.

Three times the first added to the second:

$3(60) + 10 = 180 + 10 = 190,$

which the worksheet says should be $90$. That single failure is enough: the pair does not answer the problem.

The first added to twice the second:

$60 + 2(10) = 60 + 20 = 80,$

which is exactly what the worksheet says. So one sentence holds and the other does not.

Put those together with the remark that came with the answer. The pair really does satisfy both of the equations that were written down, and it fails a sentence of the problem, so at least one of those equations is not a faithful translation of the sentence it came from. The check that was run tested the arithmetic, and the arithmetic was never the problem.

Let $x$ be the first number and $y$ the second, and keep each multiplier attached to the number its sentence attaches it to:

$3x + y = 90, \qquad x + 2y = 80.$

The first is already close to being solved for $y$, so write $y = 90 - 3x$ and substitute into the second:

$x + 2(90 - 3x) = 80, \qquad x + 180 - 6x = 80, \qquad -5x = -100, \qquad x = 20.$

Then $y = 90 - 3(20) = 30$.

Test against the sentences, not just the equations: three times $20$ added to $30$ is $60 + 30 = 90$, and $20$ added to twice $30$ is $20 + 60 = 80$. Both hold, so the first number is $20$ and the second is $30$.

The first sentence has two numbers in it and a multiplier of $3$ to attach to one of them, so the natural misreading is to attach it to the other one. That gives

$x + 3y = 90, \qquad x + 2y = 80,$

with the second sentence translated correctly. Subtracting the second from the first leaves $y = 10$, and then $x = 80 - 20 = 60$: exactly the reported pair. It satisfies both of those equations, since $60 + 3(10) = 90$ and $60 + 2(10) = 80$, which is precisely the successful check that was reported.

Now rule out a rival. Suppose instead that the second sentence had been the one misread, as twice the first added to the second:

$3x + y = 90, \qquad 2x + y = 80.$

Subtracting gives $x = 10$ and then $y = 60$, the pair $10$ and $60$, which is not what was reported. So that misreading is not the one that happened.

One other explanation fits the numbers arithmetically: copying the first sentence's total as $190$ rather than $90$ would also be satisfied by the reported pair. But that is a slip of the pen rather than a misreading of the words, and it does not explain a first equation that the solver believed said $90$.

A check has to be against something the mistake cannot have touched. Substituting a pair into the equations you wrote yourself is a check against work that already contains the mistake: if the system solved is the misread one, its own solution satisfies it exactly, whatever the worksheet said. That check can catch a slip in the solving. It is structurally incapable of catching a slip in the translating, because the translation is the very thing it assumes.

The only thing untouched by the mistake is the wording of the problem. So the check that works is to read the pair back into each sentence and do the arithmetic on the words:

$3(60) + 10 = 190 \ \text{ against the required } \ 90.$

Every sentence has to be tried, and the reason is sharper than luck. The misreading changed one sentence and left the other alone, so the correctly translated equation belongs to the intended system and to the misread system alike. Any pair that solves the misread system is therefore bound to satisfy that sentence. A student who tests only that sentence is guaranteed a pass, no matter how badly the other one was translated, and the one sentence that could have raised the alarm is the one they did not try.

The sum clue and the difference clue are

$x + y = 40, \qquad x - y = 8.$

The $y$ terms are already opposites, so adding the two equations removes $y$:

$2x = 48, \qquad x = 24,$

and then $y = 40 - 24 = 16$.

Test the doubling clue, which took no part in the solving: twice the smaller less $8$ is $2(16) - 8 = 24$, which is the larger. It holds. All three clues are true of the pair $24$ and $16$.

Both of the named clues say what the larger number is in terms of the smaller:

$x = y + 8, \qquad x = 2y - 8.$

Setting the two expressions equal gives $y + 8 = 2y - 8$, so $y = 16$ and $x = 16 + 8 = 24$, the same pair as before.

Now the general reason, in two steps. First, a single clue does not name a pair. Each of these clues is one linear equation in two unknowns, and from the two-variable work its solutions fill a whole line; the sum clue alone is satisfied by $30$ and $10$, by $25$ and $15$, and by endlessly many other pairs. So no clue on its own settles anything, and it takes a second clue to cut the line down.

Second, write all three clues as the larger in terms of the smaller:

$x = 40 - y, \qquad x = y + 8, \qquad x = 2y - 8.$

As the smaller number grows by one, these change by $-1$, by $+1$ and by $+2$ respectively. Three different rates mean no two of the lines are parallel and no two are the same line, so any two of them meet in exactly one pair. And the pair $24$ and $16$ satisfies all three clues, so it lies on all three lines. Whichever two clues are chosen, their single meeting pair must be that one, since $24$ and $16$ already lies on both of them. Hence every choice of two clues returns the same answer, and it is only the count of clues, two, that is doing the work.

Two of the clues already leave exactly one candidate, the pair $24$ and $16$. Call it $P$, and let the fourth clue say whatever it likes.

If $P$ satisfies the fourth clue, then $P$ satisfies all four. It is also the only such pair, because anything satisfying all four in particular satisfies the first two, and those two admit only $P$:

$x + y = 40, \qquad x - y = 8 \quad \Rightarrow \quad (x, y) = (24, 16).$

So the puzzle still has exactly one answer, and the same answer as before.

If $P$ fails the fourth clue, then nothing satisfies all four, by the same argument run backwards: any pair satisfying all four would have to be $P$, and $P$ does not qualify. The puzzle has no answer at all.

So a fourth clue is printable exactly when the answer already satisfies it, and in that case it contributes nothing beyond reassurance: the number of unknowns is two, two independent clues have already used up all the freedom there was, and a clue the answer satisfies cannot cut the field below the single candidate left standing. The one thing a further clue can still do is empty the field altogether, which is the other case. A reader who works carefully and finds that no pair fits all four clues has not made an error; they have proved the setter's clues disagree, which is a fault in the puzzle. That is the same count read from the other end: two unknowns need two clues, fewer leaves a whole line of pretenders, and more can only agree or contradict.