Word Problems with Linear Equations: Free Response

Say what the letter stands for before using it: let $n$ be the smaller number.

The words tie the other number to that one. "The larger is $14$ more than the smaller" is an instruction to add $14$, so the larger number is $n + 14$. Nothing else in the problem needs a letter of its own.

The sentence that states an equality is the one about the sum, and it becomes

$n + (n + 14) = 60.$

Let $m$ be the larger number. The tie between the two numbers now runs the other way: the smaller is $14$ LESS than the larger, so it is $m - 14$. The sum is still $60$:

$m + (m - 14) = 60.$

Combine the like terms and solve:

$2m - 14 = 60, \qquad 2m = 74, \qquad m = 37.$

The larger number is $37$, so the smaller is $m - 14 = 23$. Check against the words: $23 + 37 = 60$, and $37$ really is $14$ more than $23$.

A solution is not a bare number. It is a value OF something, and the two routes named different somethings.

In the first route the letter was the smaller number, so its value belongs to the smaller number. In the second the letter was the larger, so its value belongs to the larger. Set the two solved values side by side:

$n = 23, \qquad m = 37.$

Those are not two answers to one question. They are one answer each to two different questions, and the pair of numbers, which is what the problem actually asks about, comes out the same either way.

The contradiction appears only if the meanings are dropped and the two values are compared as though they measured the same thing. That is what step two of the method is for: write down what the letter stands for, with units wherever there are any. It costs one line, and it is the line that lets a solved value be read back into the words.

Only one piece is described on its own terms, and the other two are described by their relation to it, so the letter has to be that one: let $x$ be the length of the first piece in centimetres.

Now read the left side term by term. "Twice as long as the first" doubles it, giving $2x$. "$8$ cm longer than the first" adds $8$, giving $x + 8$. So the three pieces are

$x, \qquad 2x, \qquad x + 8.$

The right side is the board those pieces were cut from. Cutting loses no length, so the pieces add up to the whole $96$ cm, and that is the only equality the problem states.

Combine the like terms, then undo the addition and the multiplication:

$4x + 8 = 96, \qquad 4x = 88, \qquad x = 22.$

The letter was the FIRST piece, so $22$ cm is that piece and not yet an answer to the question. Write the other two out before deciding anything: the second is $2x = 44$ cm and the third is $x + 8 = 30$ cm.

Check against the words: $22 + 44 + 30 = 96$, the second is twice the first, and the third is $8$ more than the first. Of the three, the second piece is the longest, at $44$ cm.

The wording never describes the second or third piece by itself. It describes each one BY the first: "twice as long as the first", "$8$ cm longer than the first". Each of those phrases is an instruction for turning the first piece's length into another piece's length, so once the first piece has a letter, the other two have expressions:

$x, \qquad 2x, \qquad x + 8.$

One letter, three lengths, one equation.

Now break that chain. Suppose the problem said only that the board is $96$ cm long and that the second piece is $8$ cm longer than the third, with nothing linking either of them to the first. Then a cut of $68$ cm, $18$ cm and $10$ cm fits every word, and so does a cut of $48$ cm, $28$ cm and $20$ cm:

$68 + 18 + 10 = 96, \qquad 48 + 28 + 20 = 96.$

In each the second piece is $8$ cm longer than the third and the pieces make up the board, so the words do not pin the cut down, and there is nothing left for a single equation in one unknown to determine. That is the whole weight of step three: every other quantity has to be expressible in terms of the one you named.

Name the quantity that settles everything else: let $p$ be the pounds of house bean in the blend. The blend weighs $20$ pounds, so whatever is not house bean is premium bean, which is $20 - p$ pounds. That subtraction is what keeps the problem to one letter.

Both sides of the equation are amounts of money, each an amount times a price per pound. The beans used cost $6p$ dollars and $14(20 - p)$ dollars. The finished blend is to be worth $11$ dollars a pound for $20$ pounds, which is $11(20) = 220$ dollars:

$6p + 14(20 - p) = 220.$

Distribute, then collect the like terms:

$6p + 280 - 14p = 220, \qquad -8p = -60, \qquad p = 7.5.$

So the blend uses $7.5$ pounds of house bean, and the rest of the $20$ pounds, $20 - 7.5 = 12.5$ pounds, is premium bean.

Check against the words, in money: the house bean costs $6(7.5) = 45$ dollars and the premium costs $14(12.5) = 175$ dollars, together $220$ dollars for $20$ pounds, which is $11$ dollars a pound. A half pound is no obstacle here, because beans are weighed rather than counted, so a fractional amount is something the shop can actually scoop.

Only the target changes, so the setup is the one from part A with $5(20) = 100$ dollars on the right:

$6p + 14(20 - p) = 100, \qquad 280 - 8p = 100, \qquad p = 22.5.$

The algebra is faultless and the answer is still impossible. Notice where the impossibility shows up: the unknown came out positive, so a student watching only for a negative value sees nothing wrong. But $22.5$ pounds of one bean will not fit in a $20$ pound blend at all, and the expression written for the premium bean says the same thing more loudly:

$20 - p = 20 - 22.5 = -2.5.$

A blend cannot contain $-2.5$ pounds of anything, so no blend of these beans is worth $5$ dollars a pound. Writing out every quantity in the model at the solved value, and not just the letter, is what exposes it.

The general reason needs no algebra. Every pound of the blend is either a pound of house bean, costing $6$ dollars, or a pound of premium bean, costing $14$ dollars, so every pound costs at least $6$ dollars and the $20$ pounds cost at least $120$ dollars. A blend worth $5$ dollars a pound would have to cost

$20 \times 5 = 100 < 120 = 20 \times 6.$

Mixing can never take the price below the cheaper of the two things mixed, and the same argument the other way up says it can never take the price above the dearer.

Nobody ages faster than anybody else, so the one letter counts the years for both of them: in $x$ years the father is $45 + x$ and the son is $9 + x$.

"Three times as old" multiplies the SON's age, because he is the younger of the two:

$45 + x = 3(9 + x).$

The multiplier belongs to the smaller quantity. Attaching it to the father's age instead would be writing down that the son is the older of the two, which is not what the words say.

Three times as old:

$45 + x = 3(9 + x), \qquad 45 + x = 27 + 3x, \qquad 18 = 2x, \qquad x = 9.$

In $9$ years he is $54$ and the son is $18$, and $54 = 3 \times 18$.

Four times as old, the same setup with one symbol changed:

$45 + x = 4(9 + x), \qquad 45 + x = 36 + 4x, \qquad 9 = 3x, \qquad x = 3.$

In $3$ years he is $48$ and the son is $12$, and $48 = 4 \times 12$.

Five times as old:

$45 + x = 5(9 + x), \qquad 45 + x = 45 + 5x, \qquad 0 = 4x, \qquad x = 0.$

That one is today: $45 = 5 \times 9$ already. A solution of zero is not a failure to find a date; it is the equation reporting that the condition holds at this moment.

The larger the multiple asked for, the sooner it happens, and the three dates arrive in that order: $9$ years away, $3$ years away, and today.

Find the date first:

$45 + x = 6(9 + x), \qquad 45 + x = 54 + 6x, \qquad -9 = 5x, \qquad x = -1.8.$

A negative $x$ is a date in the past, $1.8$ years ago, when the father was $43.2$ and the son $7.2$, and indeed $43.2 = 6 \times 7.2$. So the equation has a perfectly good solution and the answer to the question asked is still no: no FUTURE date works. The algebra can only report a date; deciding whether that date is admissible is a question about the situation.

Why it had to come out that way is worth seeing without solving anything. The father is $36$ years older than the son today and on every other day, because both ages advance at the same rate. For him to be six times as old, his extra $36$ years must be five times the son's age at that moment, so the son must be $36 \div 5 = 7.2$ years old. But the son is already $9$ and only gets older, so that moment is behind them. Asking for a still bigger multiple demands a still younger son, which puts every multiple above today's five further into the past.

Run the same reasoning forwards. As the son ages, the fixed gap of $36$ years is a smaller and smaller multiple of his age, so the comparison slides steadily towards one, and it never arrives, because the father stays $36$ years ahead forever. The multiples still to come are exactly those between one and five.

Read the advertisement literally. "$30\%$ off every item" means the amount removed is $30\%$ of the price the coat HAD, which is the unknown, and not $30\%$ of the $91$ dollars that were left once the removal had happened. Rosa took her percentage of the wrong number, and the difficulty of the whole question sits there rather than in any arithmetic.

Name the unknown: let $p$ be the pre-sale price in dollars. The sale removes $0.30p$ dollars from it, and what remains is what the receipt says:

$p - 0.30p = 91, \qquad 0.70p = 91.$

Combining $p - 0.30p$ into $0.70p$ is the same like-terms step as always: one whole $p$ less three tenths of it leaves seven tenths of it.

Divide by the factor multiplying the unknown:

$0.70p = 91, \qquad p = \frac{91}{0.70} = 130.$

Check against the advertisement, which is the only thing worth checking against: $30\%$ of $130$ dollars is $39$ dollars, and $130 - 39 = 91$ dollars, the amount on the receipt.

Now put Rosa's figure through the identical test. Had the coat been $118.30$ dollars, the sale would have taken $0.30 \times 118.30 = 35.49$ dollars off it, leaving

$118.30 - 35.49 = 82.81$

dollars, which is not what she paid. Her answer does not survive the words of the problem. Substituting it into her own working would have confirmed nothing except her own working, which is why the final check is always made against the original sentences.

Her arithmetic was not careless. It was a correct answer to a question nobody asked. Adding $30\%$ of a number to that number is exactly what a $30\%$ increase does:

$91 + 0.30(91) = 1.30 \times 91 = 118.30.$

So $118.30$ dollars is what a $91$ dollar item sells for after a $30\%$ markup. The base of that percentage is the $91$ dollars, and that is precisely what makes it the wrong calculation here, where the percentage in the story belongs to the larger, unknown, pre-sale price.

A number carries no record of the question it came from, which is why the last step of the method is to read an answer back into the words instead of admiring it on the page.

Follow an unnamed price through both moves. Removing $30\%$ leaves $70\%$ of what it is applied to, and adding $30\%$ gives $130\%$ of what IT is applied to, so a price of $P$ dollars becomes

$P \to 0.70P \to 1.30 \times 0.70P = 0.91P.$

The round trip returns $91\%$ of the original price, never the whole of it, and the starting price was never named, so this is not a fact about one coat. Two of them make it concrete: a $200$ dollar coat falls to $140$ and rises to $182$, and Rosa's own coat, at $130$ dollars, falls to $91$ and rises to $118.30$. Both land $9\%$ short.

The reason is what a percentage IS. It is never an amount of money on its own; it is a fraction OF some quantity, and which quantity is part of the instruction. The $30\%$ that came off was $30\%$ of the larger pre-sale price, while the $30\%$ going back on is $30\%$ of the smaller sale price, so the two amounts of money are simply not the same one. To undo the cut you have to go back to the base it was taken from, which is what naming that price and leaving it a letter does.