Equations with Two Variables: Free Response

Put the first number of each pair in for $x$ and the second in for $y$, then compare the left side with $12$.

For $(4, 0)$:

$3(4) - 2(0) = 12 - 0 = 12.$

The two sides agree, so this pair is a solution. Notice that the term $2y$ did not disappear when $y$ was zero; it contributed $0$ and was still subtracted.

For $(0, 6)$:

$3(0) - 2(6) = 0 - 12 = -12.$

That is not $12$, so this pair is not a solution.

For $(2, -3)$:

$3(2) - 2(-3) = 6 + 6 = 12.$

Subtracting a negative adds, which is where this one is won or lost. The two sides agree, so it is a solution.

Choose the two coefficients first and let the constant fall out of the pair you were given.

Take $x + y$ on the left. Substituting the pair fixes what has to be on the right:

$5 + (-2) = 3,$

so $x + y = 3$ has $(5, -2)$ among its solutions. Take $2x + y$ instead and the same substitution gives

$2(5) + (-2) = 8,$

so $2x + y = 8$ works too. Neither is a multiple of the other: doubling the first gives $2x + 2y = 6$, which is not the second.

Each verification is the ordinary test from part A. The pair goes in, and the left side has to come out equal to the right.

Substituting the pair into $ax + by = c$ turns the requirement into one condition on the three numbers:

$5a - 2b = c.$

Read that as a recipe rather than an obstacle. Choose any $a$ and $b$ you like, provided they are not both zero, and the condition hands back the only constant that will do. Every such choice produces an equation with the pair among its solutions, and choices that are not proportional to one another produce equations that are not multiples of one another, so there is no end to them.

The condition runs the other way too: if $ax + by = c$ has the pair as a solution, substituting gives exactly that same relation, so no equation with this solution escapes it.

So one known solution pins down nothing about $a$ and $b$ separately, and pins $c$ down only once $a$ and $b$ have been settled. One pair is one condition on three numbers, which is why part B had so much room in it.

Test each pair separately, first number for $x$ and second for $y$.

For $(2, 5)$:

$4(2) - 5 = 8 - 5 = 3.$

The left side matches the right, so this pair is a solution.

For $(5, 2)$:

$4(5) - 2 = 20 - 2 = 18.$

That is not $3$, so this pair is not a solution.

Naomi's reason treats a pair as a bag holding two numbers. It is not: it is an ordered pair, and the order is what says which number does which job. Here the two numbers enter through different terms, one multiplied by $4$ and the other subtracted, so exchanging them changes the arithmetic entirely. That is why a solution is written with brackets and a comma and not as a pair of loose values.

Take the pairs her formula produces, then test them where it matters, in the original equation.

At $x = 0$ her formula gives $y = 3$, and at $x = 2$ it gives $y = 3 - 8 = -5$. Testing both:

$4(0) - 3 = -3, \qquad 4(2) - (-5) = 13.$

Neither is $3$, so the formula is not producing solutions.

Rearrange the equation properly. Add $y$ to both sides, then subtract $3$ from both sides:

$4x - y = 3 \quad\Rightarrow\quad 4x = y + 3 \quad\Rightarrow\quad y = 4x - 3.$

Check it at the same two values. At $x = 0$ it gives $(0, -3)$, and $4(0) - (-3) = 3$. At $x = 2$ it gives $(2, 5)$, the pair that passed in part A.

One test would have been thin. The two formulas agree at $x = \tfrac{3}{4}$ and nowhere else, so a single unlucky choice of $x$ would have produced a genuine solution from her formula and hidden the fault. Two values settle it.

A generating formula is worth having only if it is faithful in both directions, and each direction has its own reason.

Nothing it produces is spurious. The steps from $4x - y = 3$ to $y = 4x - 3$ are adding $y$ to both sides and subtracting $3$ from both sides, and each is undone by the opposite move. So a pair built by the formula satisfies the original equation:

$4x - (4x - 3) = 3.$

The left side collapses to $3$ whatever $x$ was chosen, which is the promise in full.

Nothing is missed either. Suppose some pair solves $4x - y = 3$. Feeding that pair's first coordinate into the formula returns $4x - 3$, and the original equation says that this is exactly the pair's second coordinate. So the formula reproduces every solution there is, one for each choice of the first coordinate, which is also why the solutions never run out.

Two directions, two guarantees: nothing spurious comes in, nothing genuine is left out. A formula with only the first property would be safe but incomplete, and one with only the second would hand you pairs you still had to check one at a time.

Two quantities are unknown, so give each a letter, and be precise that the letter counts items rather than naming a price. Let $m$ be the number of muffins sold and $j$ the number of juice cartons sold.

Muffins bring in $3$ dollars each, contributing $3m$ dollars, and cartons bring in $2$ dollars each, contributing $2j$ dollars. Together those make up the whole afternoon:

$3m + 2j = 24.$

The coefficients are the prices and the letters are the counts. Swapping those two roles is the usual slip, and it produces an equation that says nothing about the afternoon.

Work through the muffin counts in order, since each one leaves a definite amount for the juice.

Rearranging shows what is left once the muffins are paid for:

$2j = 24 - 3m.$

The muffins cannot bring in more than the whole takings, so $3m \le 24$ and the muffin count is at most $8$. That is what stops the search: past $8$ muffins the carton count would have to be negative.

Inside that range, the money left over has to be an exact number of $2$ dollar cartons, so $24 - 3m$ must be even. Since $24$ is even, that happens exactly when $3m$ is even, and so exactly when $m$ is even. Testing the even counts:

$m = 0, 2, 4, 6, 8 \quad \text{give} \quad j = 12, 9, 6, 3, 0.$

An odd muffin count leaves an odd number of dollars, which no whole number of cartons can make up. So exactly five afternoons are possible.

Check both pairs first, because the point of this part depends on their passing.

$3(2.5) + 2(8.25) = 7.5 + 16.5 = 24,$

$3(10) + 2(-3) = 30 - 6 = 24.$

Both are genuine solutions of the equation. What rules them out is not the equation but the situation: half a muffin cannot be sold, and a stall cannot sell a negative number of cartons. The counts have to be whole numbers, and they cannot be negative.

Two different things are therefore at work. The equation is one condition, and by itself it leaves a whole line's worth of pairs. The situation carries restrictions the equation knows nothing about, and those cut the endless family down to the afternoons of part B.

So the takings do say something: the muffin count is even and at most $8$. What they cannot do is name it. Several afternoons remain consistent with $24$ dollars, and no further work on this one equation will separate them, because nothing recorded so far distinguishes one from another.

Look at what one equation does to the other, rather than at particular pairs.

Doubling both sides of $2x + 3y = 12$ gives

$4x + 6y = 24,$

and halving both sides of that returns the first. Both moves are legal whatever numbers $x$ and $y$ stand for, and each undoes the other.

That settles both directions at once. If a pair makes $2x + 3y$ equal to $12$, it makes twice that expression equal to twice $12$, so it satisfies the second equation. If a pair makes $4x + 6y$ equal to $24$, halving gives $2x + 3y = 12$, so it satisfies the first. No pair can satisfy one and fail the other, so the two collections of solutions are one collection.

A list of tested pairs could never have reached that. Each successful test settles one pair and leaves infinitely many untested, so agreement across a list is evidence and not a verdict. A single failed test would have been decisive in the other direction, and that asymmetry is worth holding on to.

One pair is wanted, and it has to be produced rather than described.

Take $(0, 4)$, found by putting $x = 0$ into $2x + 3y = 12$ and solving $3y = 12$. Test it in both equations:

$2(0) + 3(4) = 12, \qquad 0 + 4 = 4.$

The first is satisfied and the second is not, since $4$ is not $5$. So a pair belongs to one collection of solutions and not the other, and the two equations are not one equation rewritten.

What the shared pair $(3, 2)$ establishes is smaller than it looks: this single pair meets both conditions at once. It says nothing about any other pair. Part A shows what a genuine rewriting looks like instead, an operation that carries every solution of one equation to a solution of the other and back again.

Attack it from one known solution rather than by trying values of $k$.

The pair $(0, 4)$ satisfies $2x + 3y = 12$. If $6x + ky = 36$ is to hold the same solutions, it must hold that one, so

$6(0) + 4k = 36,$

which forces $k = 9$. That is one direction finished: no other value has any chance, because every other value loses this single pair straight away.

Now confirm that $9$ really does the job, which does not follow from the argument above. With $k = 9$ the equation reads $6x + 9y = 36$, and each side is three times the matching side of $2x + 3y = 12$. Tripling and dividing by three are opposite moves, so by exactly the reasoning of part A the two hold the same pairs.

Both halves are needed. The first rules out every other candidate, the second establishes the survivor, and an argument with only the first half would have shown no more than that $9$ is the only value that could possibly work.

Deal with the two equations separately before comparing them.

$4x = 12$ holds one unknown. Dividing both sides by $4$ gives $x = 3$, and no other number survives, so there is exactly one solution and it is a single number.

$4x - 3y = 12$ holds two. Choosing any value for $x$ leaves a one-variable equation that fixes $y$, so its solutions come in ordered pairs, and since every choice of $x$ produces one, there is no end to them.

Now the last question. A solution of the second equation has to supply a value for each letter, and $x = 3$ supplies one, so it is not a solution; it is a condition that some solutions meet. Exactly one meets it, since putting $x = 3$ into the equation gives

$12 - 3y = 12,$

so $y = 0$ and the pair is $(3, 0)$. The number $3$ that solved the first equation outright reappears here as one coordinate of one pair among infinitely many, which is the whole change in a sentence.

Halfway between two pairs means halfway in each coordinate.

Between $(3, 0)$ and $(0, -4)$:

$\left(\tfrac{3 + 0}{2},\ \tfrac{0 + (-4)}{2}\right) = \left(\tfrac{3}{2}, -2\right).$

Test it in the equation:

$4\left(\tfrac{3}{2}\right) - 3(-2) = 6 + 6 = 12.$

It is a solution. Now take the pair halfway between $(3, 0)$ and that one:

$\left(\tfrac{9}{4}, -1\right), \qquad 4\left(\tfrac{9}{4}\right) - 3(-1) = 9 + 3 = 12.$

A solution again.

Nothing about the two starting pairs was special, and each round hands back a genuine solution lying strictly between the two it came from. So the halving can be applied to the new pair and one of the old ones, and then again, and it never runs out of room: between any two solutions there is another.

That is more than the fact that there are infinitely many. Infinitely many solutions could still have sat apart from one another, the way the whole numbers sit apart on a number line, and a table of tidy pairs quietly invites that picture. The halving rules it out. No two solutions have empty space between them, so the plotted points do not stay as separate dots but fill in, which is why the graph is drawn as an unbroken straight line rather than a row of marks.

Sort out what the table actually contains. Every row was made by choosing a value for one variable and computing the other from the equation. Choosing $x = 6$, for instance, leaves

$4(6) - 3y = 12, \qquad y = 4,$

so the row $(6, 4)$ begins life as a solution and is only afterwards plotted. Every row is like that, so the table offers instances of one direction only: these solutions turn out to be points of the line. It offers no instance at all of the other direction, because no point was ever taken off the line and tested in the equation. The evidence runs one way, and the two directions are genuinely separate claims.

Suppose the second claim failed, so that the line carried points that are not solutions. Reading off would then be unsafe. Drawing the line through two computed solutions and then taking the point above $x = 9$, a value never computed, would hand you a pair with no guarantee attached, and the habit of plotting a third solution to check that the first two were right would lose its meaning as well, since landing on the line would no longer say anything. Almost everything the picture is used for leans on the direction the table cannot supply.

Both directions do hold for a linear equation, which is what licenses drawing the line from two solutions and reading further solutions off it.

The instruction has to be answered on this equation's own terms. A solution here is an ordered pair and there is no end to them, so the only complete answer is a description of the whole collection. Solving for one variable produces one:

$y = \frac{4x - 12}{3},$

so the solutions are exactly the pairs $\left(x, \frac{4x - 12}{3}\right)$ as $x$ runs over every number, which is the line of part C.

The first response is true but partial. The pair $(3, 0)$ does satisfy the equation, and part A produced it, but it is one member of an endless family offered as though it were the answer. The habit comes from the previous chapter, where naming the number really was the whole job.

The second response is not partial but wrong, and it confuses two different things. Solutions are not scarce here; there are more of them than any one-unknown equation ever had. What one equation cannot do is single one out, and that is a statement about how much the equation determines, not about whether anything satisfies it. Reading 'I cannot pin it down' as 'there is nothing to pin down' is the mistake, and the rest of the chapter exists because pinning it down takes more than this single equation.