Systems in Two Variables: Free Response

Line each system up as $a_1x + b_1y = c_1$ with $a_2x + b_2y = c_2$ and compare matching numbers, coefficients before constants.

All three open identically, because the first equation never changes:

$\frac{a_1}{a_2} = \frac{4}{6} = \frac{2}{3}.$

In systems (i) and (ii) the $y$ coefficients give $\frac{-6}{-9} = \frac{2}{3}$, matching. In system (iii) they give $\frac{-6}{-8} = \frac{3}{4}$, which is not $\frac{2}{3}$.

That single mismatch finishes system (iii). Unequal coefficient ratios mean the two lines point in different directions, so they cross once and there is exactly one solution. The constants are never consulted, and could be replaced by any numbers at all without disturbing that verdict.

Systems (i) and (ii) are not finished, because agreeing coefficient ratios are compatible with two different endings. Only the constants separate them:

$\frac{22}{33} = \frac{2}{3}, \qquad \frac{22}{24} = \frac{11}{12}.$

In (i) the constant keeps the ratio the coefficients set, so the partner is the first equation rescaled and the two describe one line: infinitely many solutions. In (ii) the constant breaks that pattern, so the two lines point the same way and sit in different places: no solution.

Take system (iii) first. Multiply $4x - 6y = 22$ by $3$ and $6x - 8y = 24$ by $2$, so both carry $12x$:

$12x - 18y = 66, \qquad 12x - 16y = 48.$

Equal coefficients call for subtraction:

$-2y = 66 - 48 = 18, \qquad y = -9.$

Back into $4x - 6y = 22$: $4x + 54 = 22$, so $4x = -32$ and $x = -8$. Test the pair in the two equations as they were given:

$4(-8) - 6(-9) = 22, \qquad 6(-8) - 8(-9) = 24.$

Both hold, so the solution set of (iii) is the single pair $(-8, -9)$.

System (ii) needs no pair at all. Tripling $4x - 6y = 22$ and doubling $6x - 9y = 24$ gives

$12x - 18y = 66, \qquad 12x - 18y = 48,$

and one quantity cannot take two values at once, so nothing satisfies both equations. Its solution set is the empty set, which is a set and worth reporting as one; there is simply nothing inside it to test.

System (i) is different in kind, because there is nothing to pin down. Dividing $4x - 6y = 22$ by $2$ and dividing $6x - 9y = 33$ by $3$ turn both into

$2x - 3y = 11,$

so the system asks one question rather than two. Solve that shared equation for $y$:

$y = \frac{2x - 11}{3}.$

Now $x$ may take any value at all and $y$ is forced to match it, so the solutions are exactly the pairs $\left(x,\ \frac{2x - 11}{3}\right)$ as $x$ runs over all real numbers. Two of them: $x = 4$ gives $(4, -1)$ and $x = 7$ gives $(7, 1)$. Testing both against the originals:

$4(4) - 6(-1) = 22, \qquad 6(4) - 9(-1) = 33,$

$4(7) - 6(1) = 22, \qquad 6(7) - 9(1) = 33.$

Counting is not describing. The count says how many solutions there are; the rule says which pairs they are, and only the rule lets you produce one on demand.

Put both equations in standard form, with $a_2$ and $b_2$ nonzero so that both of the ratios the test compares are numbers at all, and with $b_1$ nonzero as well, as all three systems here are. Eliminate $y$ by multiplying the first equation by $b_2$, the second by $b_1$, and subtracting:

$(a_1b_2 - a_2b_1)\,x = c_1b_2 - c_2b_1.$

Now read the test's first half against that line. The ratios $\frac{a_1}{a_2}$ and $\frac{b_1}{b_2}$ differ exactly when $a_1b_2 \ne a_2b_1$, which is exactly when the coefficient on the left is not zero. If it is not zero, $x$ is forced to one value, back-substitution forces one matching $y$, and the system has exactly one solution. If it is zero, the left side is zero whatever $x$ may be, so the line either fails for every $x$ or holds for every $x$, and neither of those pins $x$ down to a single value.

So, wherever both of those ratios exist, the two conditions travel together. Differing ratios give exactly one solution, and exactly one solution can only have come from differing ratios, since agreeing ratios were just shown never to produce it. The first half is an 'exactly when', and reading it backwards is as safe as reading it forwards.

The proviso is not idle. Take $x + y = 1$ beside $y = 1$, where $a_2 = 0$, so $\frac{a_1}{a_2}$ is not a number and the two ratios cannot be compared at all. That system still has exactly one solution, $(0, 1)$. The test is silent there rather than wrong, which is why the nonzero coefficients belong in its statement.

The second half claims the remaining case is always empty, and system (i) refutes it outright: its coefficient ratios agree, and part B produced a whole line of solutions rather than none. What agreement does establish is that exactly one is off the table. Choosing between the two survivors takes the constants, which is why (i) and (ii) can share all four coefficients and still end differently.

The two halves therefore stand in quite different relations to the truth. The first needs nothing added beyond the proviso that both of its ratios exist. The second is not a weak claim but a false one.

With $k = 5$ the system reads $5x + 6y = 18$ together with $2x + 3y = c$. Double the second equation, the letter on the right included, so both carry $6y$:

$4x + 6y = 2c.$

Subtracting that from $5x + 6y = 18$ removes $y$ in one step:

$x = 18 - 2c.$

Nothing has been decided about $c$, and nothing needs to be: whatever number it stands for, this line hands back a single value of $x$. Back-substitute into $2x + 3y = c$:

$3y = c - 2(18 - 2c) = 5c - 36, \qquad y = \frac{5c - 36}{3}.$

So the solution is the single pair $\left(18 - 2c,\ \frac{5c - 36}{3}\right)$. Check it at one value to be sure the algebra is sound. At $c = 3$ it gives $(12, -7)$, and

$5(12) + 6(-7) = 18, \qquad 2(12) + 3(-7) = 3.$

Both hold. The count is one for every $c$, because the elimination ended at a genuine value of $x$ rather than at a bare numerical statement, and it did so without ever asking what $c$ was.

At $k = 4$ the first equation is $4x + 6y = 18$, and halving it gives

$2x + 3y = 9.$

So the whole system is $2x + 3y = 9$ set beside $2x + 3y = c$, which makes the constant the only thing left in play.

With $c = 1$, subtracting one equation from the other leaves

$0 = 9 - 1 = 8,$

a false statement with no letters in it, so no pair satisfies both and there is no solution. That is the case the claim has in mind.

With $c = 9$ the same subtraction leaves

$0 = 0,$

which is true for every pair. The two equations are one equation written twice, and the solution set is the whole line: every pair $\left(x,\ \frac{9 - 2x}{3}\right)$ as $x$ runs over all real numbers. Two of them are $(3, 1)$ and $(0, 3)$, and both fit the originals:

$4(3) + 6(1) = 18, \qquad 2(3) + 3(1) = 9,$

$4(0) + 6(3) = 18, \qquad 2(0) + 3(3) = 9.$

So the arithmetic in the claim is right and the reasoning is half right. Matching the ratios does find the only value of $k$ at which a single solution becomes impossible, and everywhere else there really is exactly one. What matching cannot do is name which of the two remaining outcomes occurs, because it never looked at the right-hand sides. The claim treats one comparison as though it had answered two questions.

Split the plane of possible pairs at the one value of $k$ that makes the coefficient ratios agree. Those ratios are $\frac{k}{2}$ and $\frac{6}{3} = 2$, so they agree exactly when $k = 4$.

Take $k \ne 4$ first. The coefficient ratios then differ, the elimination ends at a genuine value of $x$, and the system has exactly one solution. Part A is the instance $k = 5$, where the pair came out with $c$ still inside it: the constant moved the crossing point around the plane and never once threatened the count.

Take $k = 4$. The first equation becomes $2x + 3y = 9$ and the second is $2x + 3y = c$, so subtracting gives

$0 = 9 - c.$

At $c = 9$ that reads $0 = 0$, true for every pair, and the system has infinitely many solutions. At any other $c$ it reads $0 = (\text{something nonzero})$, false for every pair, and the system has none.

Those three conditions are exclusive and between them cover every pair $(k, c)$, so each one is not merely sufficient but necessary: a system with no solution must have $k = 4$ and $c \ne 9$, since the other two branches deliver different counts. That is what makes the list a classification rather than three examples.

So the two letters have unequal powers. The letter $k$ controls the direction of the first line, and direction alone decides whether there is a crossing. The letter $c$ only slides the second line without turning it, which changes nothing while the directions differ, and decides everything once they agree.

Exactly two is unreachable, for a reason that has nothing to do with these particular numbers. Two different solutions would be two different points lying on both lines. But two different points determine exactly one line, so the two equations would have to describe the same line, and then every one of its points is a solution as well. Two solutions force infinitely many, so no choice of $k$ and $c$ can stop at two.

Two counts are unknown, so give each one a letter and be exact about what it counts. Let $s$ be the number of 200 gram bags sold and $b$ the number of 500 gram bags sold.

The mass record adds the grams: each small bag contributes 200 grams and each big bag 500, and the day came to 3600.

$200s + 500b = 3600.$

The money record adds the dollars in the same way, each bag contributing its own price.

$12s + 30b = 216.$

Now classify before solving, by comparing matching numbers. The coefficients first:

$\frac{12}{200} = \frac{3}{50}, \qquad \frac{30}{500} = \frac{3}{50},$

and then the constants:

$\frac{216}{3600} = \frac{3}{50}.$

All three agree, so the second equation is the first one multiplied through by $\frac{3}{50}$, constant included. Two names for one line: the system is consistent, since solutions exist, and dependent, since the second equation restates the first.

That is the answer to the owner's question, and it is not the answer the owner wanted. The takings record is not a second constraint at all. Anything the mass record allows, the money record allows too, so the sheet as it stands cannot single out one pair of counts, and no amount of algebra will make it.

Strip both equations down before describing anything. Dividing the mass equation by 100 and the money equation by 6 turns both into the same thing:

$2s + 5b = 36.$

One equation in two letters leaves one of them free. Choose $b$ as the free one and solve for $s$:

$s = \frac{36 - 5b}{2}.$

So the solution set is every pair $\left(\frac{36 - 5b}{2},\ b\right)$ as $b$ runs over all real numbers.

Two of those pairs are days a shop could actually have had. At $b = 4$ the rule gives $s = 8$, and at $b = 6$ it gives $s = 3$. Test both against the two records as they were written, not against the reduced equation:

$200(8) + 500(4) = 3600, \qquad 12(8) + 30(4) = 216,$

$200(3) + 500(6) = 3600, \qquad 12(3) + 30(6) = 216.$

Both Saturdays fit the sheet perfectly, and they are not the same Saturday. That is what an infinite solution set looks like from inside the shop: two genuinely different days, indistinguishable from the records kept.

The rule describes more pairs than the shop can use, since a count of bags cannot be negative and cannot be a fraction. Those restrictions come from the situation and not from the equations, which is why they narrow the list without changing the classification.

The system would now be $200s + 500b = 3600$ together with $12s + 30b = 200$. Compare the same three quantities. The coefficients are untouched:

$\frac{12}{200} = \frac{3}{50}, \qquad \frac{30}{500} = \frac{3}{50},$

but the constants now give

$\frac{200}{3600} = \frac{1}{18}.$

The coefficients still keep step and the constants no longer do, which is the signature of parallel lines. Elimination says the same thing out loud. Multiplying the mass equation by $\frac{3}{50}$ gives $12s + 30b = 216$, and subtracting the new money record leaves

$0 = 216 - 200 = 16,$

a false statement carrying no letters, so no pair of counts satisfies both. The solution set is empty.

There is a plainer route to the same place. Whatever the counts were, the mass record already fixes $2s + 5b = 36$, and the takings are six times that quantity:

$12s + 30b = 6(2s + 5b) = 6(36) = 216.$

So 216 dollars was never in doubt once 3600 grams was recorded.

What the owner should conclude is about the sheet, not about the coffee. Saturday happened, so some pair of counts is true of it, and any honest pair of records must both hold of that pair. A system with no solution therefore reports that the two entries cannot both be honest: one of the totals was miscounted or mistyped. An empty solution set in a real situation is a finding about the data, not a failure of the method.

Work out what a single gram costs in each size:

$\frac{12}{200} = 0.06, \qquad \frac{30}{500} = 0.06.$

The two sizes sell at the same rate. So for any Saturday whatever, the takings are that rate times the grams sold, and the money equation is $0.06$ times the mass equation term for term. It cannot be a second constraint, because it is the first constraint measured in different units.

The claim runs both ways, which is what makes it worth stating. If the two per-gram prices agree, the coefficient ratios of the two records agree, so the count is either none or infinitely many and never one. If the two per-gram prices differ, then $\frac{12}{200} \ne \frac{30}{500}$, the coefficient ratios differ too, and the two records pin the day to exactly one pair of counts. So a takings record adds information exactly when the sizes are not priced at the same rate per gram.

Agreeing rates leave one thing still undecided, and part C is that thing: whether the recorded takings are the rate times the recorded mass. If they are, the second record is redundant, and if they are not, it contradicts the first.

Now the extra note. 'Twice as many small bags as big ones' is $s = 2b$, that is

$s - 2b = 0.$

Set it beside $2s + 5b = 36$ and run the first comparison: $\frac{1}{2}$ against $\frac{-2}{5}$. They differ, so this record is not a rescaling of the mass record, and the two together have exactly one solution. Substituting $s = 2b$:

$2(2b) + 5b = 36, \qquad 9b = 36, \qquad b = 4,$

so $s = 8$. Check all three statements about the day: $200(8) + 500(4) = 3600$ grams, $12(8) + 30(4) = 216$ dollars, and $8$ is twice $4$. The day was 8 small bags and 4 big ones.

Begin by reading the hypothesis as two statements about numbers rather than as two equations waiting to be solved:

$ap + bq = c, \qquad ar + bs = c.$

Now substitute the proposed pair into the left-hand side of the equation and expand, keeping $t$ visible:

$a\bigl(p + t(r - p)\bigr) + b\bigl(q + t(s - q)\bigr)$

$= ap + t\,a(r - p) + bq + t\,b(s - q).$

The letters $a$ and $b$ only multiply, so each distributes across its bracket and leaves $t$ standing as a plain factor. Gather the two terms carrying $t$:

$= (ap + bq) + t\bigl(a(r - p) + b(s - q)\bigr)$

$= (ap + bq) + t\bigl((ar + bs) - (ap + bq)\bigr).$

That inner bracket is now built entirely out of the two quantities the hypothesis has already evaluated. Both are $c$, so

$= c + t(c - c) = c + 0 = c.$

The left side comes out equal to the right side, so the pair satisfies the equation. Nothing was assumed about $t$ beyond its being a number, so the conclusion holds for every $t$ at once, negative values and zero included. At $t = 0$ the pair is $(p, q)$ and at $t = 1$ it is $(r, s)$, so the construction includes the two pairs it started from.

Suppose a system of two linear equations in two variables has two different solutions, $(p, q)$ and $(r, s)$, and follow where that leads.

A solution of the system satisfies each equation, so both pairs satisfy the first equation and both pairs satisfy the second. Part A therefore applies twice, once to each equation, with the same two pairs and the same $t$. For every number $t$, the pair

$\bigl(p + t(r - p),\ q + t(s - q)\bigr)$

satisfies the first equation and satisfies the second, which is to say it is a solution of the system.

It remains to check that different values of $t$ really do produce different pairs, since a construction that kept handing back the same pair would prove nothing. The two starting pairs are different, so $r - p$ and $s - q$ are not both zero. If $r - p \ne 0$, then two different values of $t$ give two different first coordinates, because $p + t(r - p)$ changes whenever $t$ does. If instead $r - p = 0$, then $s - q \ne 0$, and the same argument runs on the second coordinate. Either way, distinct values of $t$ give distinct pairs.

There are infinitely many numbers $t$, so the system has infinitely many solutions. Having two therefore forces having infinitely many, and the same conclusion arrives from any starting count of two or more: just pick two of the solutions and run the construction. So no system can stop at exactly two, or at exactly seventeen, or at any finite total above one.

With one, none and infinitely many all realisable, and every other count now forbidden, the list of three is complete.

Check the two pairs first, since the whole part depends on their passing:

$(-4)^2 + (-3)^2 = 16 + 9 = 25, \qquad 3^2 + 4^2 = 9 + 16 = 25.$

Now run the construction with $(p, q) = (-4, -3)$ and $(r, s) = (3, 4)$ at $t = \tfrac{1}{2}$. The coordinate differences are $r - p = 7$ and $s - q = 7$, so the pair is

$\left(-4 + \tfrac{7}{2},\ -3 + \tfrac{7}{2}\right) = \left(-\tfrac{1}{2},\ \tfrac{1}{2}\right).$

Test it:

$\left(-\tfrac{1}{2}\right)^2 + \left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4} + \tfrac{1}{4} = \tfrac{1}{2},$

which is nowhere near $25$. The construction has failed, and it is worth being precise about where.

In part A the letter $a$ multiplied its bracket, so $a\bigl(p + t(r - p)\bigr)$ came apart into $ap$ plus $t$ times something, with $t$ appearing to the first power and as a single factor. Squaring does not come apart that way:

$\bigl(p + t(r - p)\bigr)^2 = p^2 + 2tp(r - p) + t^2(r - p)^2.$

The term in $t^2$ has no partner to cancel against, so the substituted expression is not the value at $t = 0$ plus $t$ times a difference, and the telescoping that produced $c + t(c - c)$ never gets started. The step with no counterpart is the distribution of the coefficient across the bracket, and what was doing the work in part A was not the shape of the construction but the fact that each variable appeared to the first power multiplied by a constant.

The consequence is not a technicality. The system $x^2 + y^2 = 25$ together with $y = x + 1$ has exactly the two solutions checked above, since substituting gives $x^2 + (x + 1)^2 = 25$, that is $2x^2 + 2x - 24 = 0$, and halving every term of that leaves $x^2 + x - 12 = 0$, whose factors $(x + 4)(x - 3)$ give $x = -4$ and $x = 3$ and nothing else. Exactly two solutions, the count part B forbade. So the list of three outcomes is a theorem about linear systems specifically, and the word linear in its statement is carrying weight rather than describing the examples.

Answer the first question for each system before reaching for a second word.

For $7x + 2y = 5$ with $-14x - 4y = 3$, the coefficient ratios are $\frac{7}{-14} = -\frac{1}{2}$ and $\frac{2}{-4} = -\frac{1}{2}$, agreeing, while the constants give $\frac{5}{3}$, which does not. Coefficients in step and constants out of step is the parallel case, so there is no solution and the system is inconsistent. Now the second question has nothing to work on: independent and dependent are two ways for a system to have solutions, so a system with none earns neither. It keeps one label, not two.

For $x - 4y = 6$ with $-3x + 12y = -18$, all three comparisons agree at $-\frac{1}{3}$, since $\frac{1}{-3}$, $\frac{-4}{12}$ and $\frac{6}{-18}$ are the same number. The partner is the first equation multiplied through by $-3$, so the two describe one line. Solutions exist, so the system is consistent, and the second equation restates the first, so it is dependent.

For $5x + y = 2$ with $x - 3y = 10$, the coefficient ratios are $\frac{5}{1} = 5$ and $\frac{1}{-3}$, which differ, so there is exactly one solution and the system is consistent and independent. Finding it: $y = 2 - 5x$ from the first equation gives

$x - 3(2 - 5x) = 10, \qquad 16x = 16, \qquad x = 1,$

so $y = -3$, and $(1, -3)$ fits both, since $5(1) + (-3) = 2$ and $1 - 3(-3) = 10$.

The four words are not four cases in a row. They are answers to two questions asked in order, and the second question is only reached when the first is answered yes.

The claim leans on the two systems failing to name one pair, which is true of both, and slides from there to their being the same situation, which is not. Find something one has and the other cannot.

Take $x - 4y = 6$ with $-3x + 12y = -18$ and produce a solution outright. Setting $y = 0$ in the first equation gives $x = 6$, so try $(6, 0)$:

$6 - 4(0) = 6, \qquad -3(6) + 12(0) = -18.$

Both hold, so this system has a solution, and part A said it has a whole line of them. Failing to name one pair here is not a shortage of answers but a surplus.

Now $7x + 2y = 5$ with $-14x - 4y = 3$, where no amount of searching would settle anything, since not finding a solution is not the same as there being none. Argue instead. Doubling the first equation gives $14x + 4y = 10$, and multiplying the second by $-1$ gives $14x + 4y = -3$. Both moves are reversible, so any pair satisfying the originals satisfies these. But then the one quantity $14x + 4y$ takes the value $10$ and the value $-3$ at once, and a number has one value. No such pair exists.

So the two situations are opposites dressed alike. One has too many solutions to name a single answer; the other has none at all. That is exactly why dependent sits under the heading consistent, beside the one-solution case, rather than beside the inconsistent one.

Build the first two partners together, since they differ only in one number. Multiplying $6x + 4y = 12$ by $\frac{3}{2}$, a factor that is not a whole number, gives the left side $9x + 6y$ and the constant $18$. So

$9x + 6y = 18$

is the same line as the first equation and produces infinitely many solutions, while

$9x + 6y = 20$

keeps that left side and breaks the constant, giving parallel lines and no solution. Check the ratios of the second one: $\frac{6}{9} = \frac{2}{3}$ and $\frac{4}{6} = \frac{2}{3}$ agree, and $\frac{12}{20} = \frac{3}{5}$ does not.

The third partner is built to a specification instead. It has to make $\frac{6}{a_2}$ equal $\frac{12}{c_2}$, which multiplies out to $6c_2 = 12a_2$, so $c_2$ has to be twice $a_2$. Take $a_2 = 3$ and $c_2 = 6$, making that shared ratio $2$, and then choose $b_2$ so the middle ratio breaks step: $b_2 = 1$ gives $\frac{4}{1} = 4$, which is not $2$. The partner is

$3x + y = 6.$

Since its coefficient ratios differ from the first equation's, the system has exactly one solution. Halving $6x + 4y = 12$ gives $3x + 2y = 6$, and subtracting $3x + y = 6$ leaves

$y = 0, \qquad 3x = 6, \qquad x = 2.$

Test $(2, 0)$: $6(2) + 4(0) = 12$ and $3(2) + 0 = 6$. So two of the three ratios agree here and the system is still as far from dependent as it could be.