Systems with More Variables: Free Response

Choose $z$ as the letter to remove, because equation $(2)$ carries $-z$ and equation $(3)$ carries $+z$: those two cancel with no preparation at all. Add them:

$(x + 2y - z) + (2x + y + z) = 8 + 7, \qquad 3x + 3y = 15, \qquad x + y = 5.$

Now take a different pair, and it must bring in equation $(1)$, the one not yet used. Equation $(1)$ carries $2z$ while $(2)$ carries $-z$, so double $(2)$ first, both sides included, and add:

$(3x - y + 2z) + 2(x + 2y - z) = 5 + 16, \qquad 5x + 3y = 21.$

The two results, $x + y = 5$ and $5x + 3y = 21$, mention the same two unknowns, which is the whole point of removing the same letter twice.

From $x + y = 5$, write $y = 5 - x$ and substitute into the other equation:

$5x + 3(5 - x) = 21, \qquad 2x + 15 = 21, \qquad x = 3,$

so $y = 5 - 3 = 2$. The letter $z$ was only set aside, not lost. Put the two known values into any original equation that still carries it, say equation $(2)$:

$3 + 2(2) - z = 8, \qquad 7 - z = 8, \qquad z = -1.$

Check the candidate $(3, 2, -1)$ in all three, including the equations that were combined:

$3(3) - 2 + 2(-1) = 5, \qquad 3 + 2(2) - (-1) = 8, \qquad 2(3) + 2 + (-1) = 7.$

All three hold, so the solution is $(3, 2, -1)$.

Clearing $z$ from the pair $(2)$ and $(3)$ gives $x + y = 5$, an equation about $x$ and $y$. Clearing $x$ instead from the pair $(1)$ and $(2)$, by subtracting three times $(2)$ from $(1)$, gives an equation about $y$ and $z$:

$(3x - y + 2z) - 3(x + 2y - z) = 5 - 24, \qquad -7y + 5z = -19.$

Between them the two leftovers mention $x$, $y$ and $z$, three unknowns carried by two equations, so they cannot single out one triple. Solutions of the pair are easy to manufacture: choose $y = 7$, then $5z = -19 + 49$ gives $z = 6$, and $x = 5 - 7 = -2$. Test $(-2, 7, 6)$ against the originals:

$3(-2) - 7 + 2(6) = -1, \qquad -2 + 2(7) - 6 = 6, \qquad 2(-2) + 7 + 6 = 9,$

against the required $5$, $8$ and $7$. It fails every one.

Nothing here is false: both leftover equations are genuine consequences of the system, and you could keep combining and reach the answer eventually. The mixed pair is simply not finished, while two equations in the same two unknowns are.

Write the kept equation as $E$ and the replaced one as $F$, each of them a linear equation in $x$, $y$ and $z$. The round replaces $F$ by $F + kE$ for some number $k$.

One direction. Suppose a triple satisfies both $E$ and $F$. The multiplication property of equality lets you multiply the equation $E$ through by $k$, and the addition property lets you add the two true equations, so the triple satisfies $F + kE$. Nothing was lost.

The other direction, which is the half that is easy to forget. Suppose a triple satisfies $E$ and $F + kE$. Multiply $E$ by $-k$ and add it on:

$(F + kE) + (-k)E = F.$

So the triple satisfies $F$ as well, and no new solution was invented.

Both halves are needed, because a step that only preserved solutions in one direction could still hand you an answer the original system rejects, which is exactly the kind of thing the final check catches.

Neither half ever counted the unknowns. The letters appear only inside $E$ and $F$, and the two properties of equality apply to whole equations, so the argument reads the same with three unknowns, or with thirty.

The quantity to watch is the number of unknowns in the system still waiting to be solved, not the number of equations on the page.

Start with $n$ equations in $n$ unknowns. Keep one equation, which still carries every unknown, and use it to remove one chosen unknown from each of the other $n - 1$. Those $n - 1$ equations now mention only $n - 1$ unknowns, and by part A they have exactly the solutions they had before. The parked equation is set aside for later.

So one round turns a pending system of size $n$ into a pending system of size $n - 1$. The size is a whole number, it strictly falls each round, and nothing in the round can put an unknown back:

$n \ \longrightarrow \ n - 1 \ \longrightarrow \ n - 2 \ \longrightarrow \ \cdots \ \longrightarrow \ 1.$

After $n - 1$ rounds the pending system is one equation in one unknown, which you solve outright. For $n = 3$ that is two rounds: three unknowns to two, two to one.

The argument is not finished when the eliminating stops. Each parked equation carries exactly one unknown that the values found so far do not cover, so back-substituting into them one at a time, in the reverse order they were parked, recovers every remaining value. The process both halts and delivers an answer for each unknown.

Both readings lean on part A. Every round preserves the solution set, so the collection of triples solving the collapsed system is the very collection solving the system you started from.

A false collapse settles the question. If a stage reads

$0 = 4,$

no triple can satisfy it, so the collapsed system has no solutions, so neither does the original. That verdict needs nothing else: one false collapsed statement anywhere in the work is enough.

An always-true collapse does not. A stage reading

$0 = 0$

says that the combination used produced no new information, so one equation was redundant. That alone does not count the solutions, because a redundant equation and a contradictory one can sit in the same system. Take

$x + y + z = 1, \qquad 2x + 2y + 2z = 2, \qquad x + y + z = 5.$

Combining $x + y + z = 1$ with $2x + 2y + 2z = 2$ gives $0 = 0$, while combining it with $x + y + z = 5$ gives

$(x + y + z) - (x + y + z) = 5 - 1, \qquad 0 = 4.$

The system has no solution despite the always-true collapse. So $0 = 0$ narrows the answer to infinitely many or none, and what remains after the redundant equation is set aside is what decides: if the survivors are consistent and leave an unknown free, the count is infinite.

Let $s$, $m$ and $l$ be the numbers of small, medium and large chairs built. Each is a count, so each is a whole number and none can be negative.

Each equation comes from one STATION, not from one chair. Read down a station's column of times, multiply each time by the number of chairs of that size, and set the total against the hours booked. Cutting takes $1$, $2$ and $3$ hours per small, medium and large chair:

$s + 2m + 3l = 29.$

Assembly takes $2$, $2$ and $1$ hours:

$2s + 2m + l = 22.$

Finishing takes $1$, $3$ and $2$ hours:

$s + 3m + 2l = 26.$

Three unknowns, and three conditions that are genuinely different, since no station's row of times is a copy or a multiple of another's.

Remove $s$, whose coefficient is $1$ in the cutting equation. Subtract twice the cutting equation from the assembly equation:

$(2s + 2m + l) - 2(s + 2m + 3l) = 22 - 58, \qquad -2m - 5l = -36, \qquad 2m + 5l = 36.$

Now remove the same unknown from the finishing equation, the one not yet used, by subtracting the cutting equation from it:

$(s + 3m + 2l) - (s + 2m + 3l) = 26 - 29, \qquad m - l = -3.$

The second gives $m = l - 3$. Substituting:

$2(l - 3) + 5l = 36, \qquad 7l = 42, \qquad l = 6,$

so $m = 3$, and the cutting equation returns $s + 6 + 18 = 29$, giving $s = 5$.

Check the schedule of $5$ small, $3$ medium and $6$ large chairs at every station:

$5 + 6 + 18 = 29, \qquad 10 + 6 + 6 = 22, \qquad 5 + 9 + 12 = 26.$

All three totals are met, and all three counts are whole and positive, as counts of chairs have to be.

Two equations carrying three unknowns cannot fix all three, so solve for two counts in terms of the third. Subtracting the cutting equation from the assembly equation clears $m$:

$(2s + 2m + l) - (s + 2m + 3l) = 22 - 29, \qquad s - 2l = -7, \qquad s = 2l - 7.$

Putting that into the cutting equation gives $2m = 36 - 5l$, so

$s = 2l - 7, \qquad m = 18 - \frac{5l}{2}.$

Now apply what a count of chairs must be. For $m$ to be a whole number, $l$ has to be even. For $s$ to be zero or more, $2l \ge 7$, so $l \ge 4$; for $m$ to be zero or more, $5l \le 36$, so $l \le 7$. The even values left are $l = 4$ and $l = 6$:

$l = 4: \ (s, m, l) = (1, 8, 4), \qquad l = 6: \ (s, m, l) = (5, 3, 6).$

Two schedules, not a whole family of them. That is worth noticing: had the counts been allowed to take any value at all, the two surviving totals would have left a whole line of possibilities. Being whole and not negative cuts that line down to two points.

The finishing total is what separates them. It demands $s + 3m + 2l = 26$, and

$1 + 24 + 8 = 33, \qquad 5 + 9 + 12 = 26.$

The first schedule would need $33$ hours of finishing, so the third condition is not decoration: it is what makes the day's work reconstructible.

Test $(10, -5, -1)$ against each equation in turn, and do not stop at the first one that works.

$10 + (-5) + (-1) = 4,$

so $x + y + z = 4$ holds.

$2(10) + 3(-5) + (-1) = 20 - 15 - 1 = 4,$

so $2x + 3y + z = 4$ holds too.

$3(10) + 3(-5) + 2(-1) = 30 - 15 - 2 = 13,$

against the $9$ that $3x + 3y + 2z = 9$ demands. That equation fails, by $4$.

Two out of three is not a pass mark here. A triple is a solution only when every equation accepts it, so the reported triple is not a solution, and the work that produced it went wrong somewhere.

Test the lines in order, each against what stands above it.

Line 1 is sound. Subtracting equation $(1)$ from equation $(2)$ takes $2x - x = x$, $3y - y = 2y$, $z - z = 0$ and $4 - 4 = 0$, so $x + 2y = 0$ is right.

Line 2 is the first bad line. Equation $(1)$ is multiplied by $2$ before subtracting, and multiplying an equation multiplies BOTH of its sides. The left side was doubled correctly, giving $2x + 2y + 2z$, but the right side was left at $4$ instead of $2(4) = 8$:

$(3x + 3y + 2z) - 2(x + y + z) = 9 - 8, \qquad x + y = 1.$

The line should read $x + y = 1$, not $x + y = 5$.

Everything below Line 2 is then carried out correctly on a wrong equation, which is why the work looks orderly to the end.

Finishing from the repair. With $x + 2y = 0$ and $x + y = 1$, subtracting gives $y = -1$, then $x = 1 - (-1) = 2$, and equation $(1)$ returns

$2 + (-1) + z = 4, \qquad z = 3.$

Check $(2, -1, 3)$ in all three:

$2 - 1 + 3 = 4, \qquad 4 - 3 + 3 = 4, \qquad 6 - 3 + 6 = 9.$

Take the two accepted equations one at a time and ask what in the work guaranteed them.

Why $x + y + z = 4$ was forced. Line 5 does not test that equation, it solves it. Whatever values Lines 3 and 4 had produced for $x$ and $y$, choosing $z$ to satisfy that equation makes it true by construction. It could not have failed.

Why $2x + 3y + z = 4$ was forced. Line 1 combined the equations correctly, and what it produced was

$(2x + 3y + z) - (x + y + z) = 0, \qquad \text{that is} \qquad x + 2y = 0.$

Read it the other way round: the equation $2x + 3y + z = 4$ is exactly $x + y + z = 4$ with $x + 2y = 0$ added on. Lines 3 and 4 solved $x + 2y = 0$ exactly, and $x + y + z = 4$ was forced a moment ago, so their sum was forced too. Substituting confirms it: $10 + 2(-5) = 0$.

What that leaves. Equation $(3)$ is the only one the work touched through the botched combination, so it is the only one nothing guaranteed, and it is the only one able to report the slip. That is the general shape of it: a check is worth something exactly on the equations your own steps did not already force, and since you cannot always see in advance which those are, the safe habit is to substitute into all three.

Every equation here carries $y$ with coefficient $1$ or $-1$, so the pairs add straight away. Equations $(1)$ and $(2)$:

$(x - y + 2z) + (2x + y - z) = 4 + 5, \qquad 3x + z = 9.$

Now a different pair, bringing in the equation not yet used. Equations $(2)$ and $(3)$:

$(2x + y - z) + (4x - y + 3z) = 5 + k, \qquad 6x + 2z = 5 + k.$

The two results are about the same unknowns, and the second has exactly twice the coefficients of the first. Subtract twice the first from the second and both unknowns leave together:

$(6x + 2z) - 2(3x + z) = (5 + k) - 18, \qquad 0 = k - 13.$

No unknown survives, so the whole family's behaviour is decided by one statement about $k$.

The collapsed statement is $0 = k - 13$, and it is either true or false.

When $k = 13$ it reads $0 = 0$. Nothing is contradicted, and equation $(3)$ has added no new information, so the system is really just $(1)$ and $(2)$: two equations carrying three unknowns. They are not multiples of one another, so they leave one unknown free. Writing $x = t$, the first pair gives $z = 9 - 3t$ and then $y = 14 - 5t$, so every triple

$(t, \ 14 - 5t, \ 9 - 3t)$

solves the system, one for each $t$. Test $t = 0$ and $t = 1$ to see that these really are different points on the same line: $(0, 14, 9)$ and $(1, 9, 6)$ both satisfy all three equations when $k = 13$.

When $k \ne 13$ the statement is false, and since every step preserved the solution set, the original system has no solution at all.

Why exactly one solution never happens. Look at the coefficients rather than the constants. Doubling the list $1, -1, 2$ and adding the list $2, 1, -1$ gives $4, -1, 3$, which is the list in $(3)$, for every value of $k$:

$2(1, -1, 2) + (2, 1, -1) = (4, -1, 3).$

So the left side of $(3)$ carries no condition the other two do not already impose. Only its constant is free, and a constant can agree or disagree but cannot pin down a point. That is why the count is always infinite or zero, never one.

Two planes are parallel exactly when one list of coefficients is a multiple of the other, since the coefficients are what fix a plane's tilt. Test the three pairs:

$(1, -1, 2), \qquad (2, 1, -1), \qquad (4, -1, 3).$

No list is a multiple of another: the first two disagree already on the ratio of the first two entries, and any multiple of $(1, -1, 2)$ has its second entry the negative of its first, which $(4, -1, 3)$ does not; and $(4, -1, 3)$ is not a multiple of $(2, 1, -1)$ either, since doubling that list gives $(4, 2, -2)$. So no two planes are parallel, and each pair meets in a line.

Take $k = 3$, one of the values with no solution, and find those three lines by pairing the equations. Equations $(1)$ and $(2)$ give $3x + z = 9$, hence $z = 9 - 3x$ and $y = 14 - 5x$. Equations $(2)$ and $(3)$ give $6x + 2z = 8$, hence $z = 4 - 3x$ and $y = 9 - 5x$. Equations $(1)$ and $(3)$ give $3x + z = -1$, hence $z = -1 - 3x$ and $y = -5x - 6$. Writing $x = t$:

$(t, \ 14 - 5t, \ 9 - 3t), \qquad (t, \ 9 - 5t, \ 4 - 3t), \qquad (t, \ -5t - 6, \ -3t - 1).$

Each line steps the same way, $1$ across in $x$ against $-5$ in $y$ and $-3$ in $z$, so the three are parallel; and at $t = 0$ they sit at $(0, 14, 9)$, $(0, 9, 4)$ and $(0, -6, -1)$, three different points, so the lines are distinct and never meet.

That is the triangular prism: three flat faces, no two of them parallel, meeting in three distinct parallel edges. A point on two of the planes lies on one edge, and no edge lies on the third plane, so nothing is common to all three.

Split on whether any pair of planes is parallel.

Family one: a parallel pair. Two planes are parallel when one list of coefficients is a multiple of the other. That alone is not enough to lose every common point, and this is the trap. If the constants stand in the same ratio as the coefficients, the two equations describe the SAME plane and contradict nothing:

$x + y + z = 1 \qquad \text{and} \qquad 2x + 2y + 2z = 2$

are one plane written twice. It is when the constants break that ratio,

$x + y + z = 1 \qquad \text{and} \qquad 2x + 2y + 2z = 5,$

that the planes are parallel and distinct, and then no point can be on both, whatever the third plane does.

Family two: no parallel pair at all. Every pair then meets in a line, and there are two ways that ends. If no list of coefficients can be built from the other two by scaling and adding, the three planes cut each other at exactly one point. If one list can be built that way, as in this question's system, the three edges run parallel; they either fall together as one shared line, giving infinitely many solutions, or stay apart, giving the prism and no solution.

What to look at, and what not to. The collapsed statement cannot tell these apart: a parallel pair and a prism both end in $0$ equalling something that is not $0$. The coefficients can. Compare the three lists first: a list that is a multiple of another means a parallel pair, and the constants then say whether the planes are distinct or the same one twice; no such multiple, but one list built from the other two, means the prism family.