Systems in Three Variables: Free Response

Aim both of the first two combinations at $x$, so that the two rows they leave are about the same pair of unknowns.

Subtract twice equation $(1)$ from equation $(2)$:

$(2x - y + z) - 2(x + y + 2z) = 9 - 18, \qquad -3y - 3z = -9.$

Divide that row through by $-3$, which is allowed because scaling an equation by a nonzero number changes none of its solutions, and it keeps the numbers small:

$y + z = 3.$

Now subtract equation $(1)$ from equation $(3)$:

$(x + 3y - z) - (x + y + 2z) = 10 - 9, \qquad 2y - 3z = 1.$

The two rows carry the same pair of unknowns, which is the whole point of aiming both combinations at $x$. Clear $y$ from the second by subtracting twice the first:

$(2y - 3z) - 2(y + z) = 1 - 6, \qquad -5z = -5.$

The system is now triangular:

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

Each row begins one unknown later than the row above it, which is exactly what makes the climb possible.

The bottom row carries one unknown, so solve it outright:

$-5z = -5 \ \Rightarrow \ z = 1.$

Carry that value into the middle row:

$y + 1 = 3 \ \Rightarrow \ y = 2.$

Carry both into the top row:

$x + 2 + 2(1) = 9 \ \Rightarrow \ x = 5.$

The candidate is $(5, 2, 1)$. Test it in the equations you were given rather than in the rows you built, and above all in equation $(3)$, which was combined last:

$5 + 2 + 2 = 9, \qquad 10 - 2 + 1 = 9, \qquad 5 + 6 - 1 = 10.$

All three hold, so the three planes meet at the single point $(5, 2, 1)$.

Read the staircase rather than the arithmetic. Each row's first surviving unknown is its pivot: $x$ in the top row, $y$ in the middle, $z$ at the bottom. Three rows, three pivots, and the free unknowns are what is left over:

$3 - 3 = 0 \text{ free unknowns}.$

With nothing free there is nothing to choose, so the climb forces one value for each unknown and the solution set is a single point. That is why the check in part B came out exactly, with no letter left standing in it.

Now the other two endings, with the top two rows untouched.

A line. Had the bottom row collapsed to $0 = 0$, it would impose no condition at all and could be struck out. Two rows would survive as genuine constraints, so $3 - 2 = 1$ unknown would be free, and letting it run traces a one-parameter family, which is a line.

Nothing. Had the bottom row collapsed to $0 = k$ for some $k$ that is not zero, no triple could satisfy it. Every combination in the sweep preserves the solution set, so the system you started from would have no solution either, whatever the rows above it said.

The bottom row does not merely finish the problem. It announces which of the shapes you are looking at.

Aim both combinations at $x$. Subtract twice equation $(1)$ from equation $(2)$:

$(2x + 5y - z) - 2(x + 4y + z) = 29 - 44, \qquad -3y - 3z = -15, \qquad y + z = 5.$

Now subtract four times equation $(1)$ from equation $(3)$:

$(4x + 13y + z) - 4(x + 4y + z) = 73 - 88, \qquad -3y - 3z = -15, \qquad y + z = 5.$

The two rows are the same equation, so combining them removes everything at once:

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

A row reading $0 = 0$ is true whatever the unknowns are, so it constrains nothing and can be struck out. Two rows survive, $x + 4y + z = 22$ and $y + z = 5$, so $3 - 2 = 1$ unknown is free and the solution set is a line.

Name the free unknown and solve for the others. Take $z = t$. The surviving second row gives

$y = 5 - t,$

and the first row then gives

$x + 4(5 - t) + t = 22, \qquad x + 20 - 3t = 22, \qquad x = 2 + 3t.$

So every solution has the form

$(x, y, z) = (2 + 3t,\ 5 - t,\ t).$

Check it in all three originals with $t$ left standing, because a family has to hold for every value of the letter and not merely for one:

$(2 + 3t) + 4(5 - t) + t = 22.$

$2(2 + 3t) + 5(5 - t) - t = 29.$

$4(2 + 3t) + 13(5 - t) + t = 73.$

In each one every $t$ cancels and the constant that survives is the right one, so the whole line lies on all three planes.

Test the reported family the way a family has to be tested, with its letter still in place. Substitute it into equation $(1)$:

$(17 - 3s) + 4(s) + (5 + s) = 22 + 2s.$

That equals $22$ only when $s = 0$. Equation $(2)$ tells the same story:

$2(17 - 3s) + 5(s) - (5 + s) = 29 - 2s,$

which equals $29$ only when $s = 0$. So the reported family touches the solution set at exactly one triple and leaves it immediately: at $s = 1$ it offers $(14, 1, 6)$, and

$14 + 4(1) + 6 = 24, \qquad \text{not } 22.$

The check was not wrong, it was insufficient. A report of an infinite solution set claims two things at once: that every triple in the family solves the system, and that no triple outside the family does. Substituting one member tests the first claim at one value of the letter. It says nothing about the other values, and nothing whatever about the second claim. That is why a family is checked with the letter standing: doing so tests the first claim at every value in one line of algebra.

Changing a constant moves no coefficient, so every combination you already carried out produces the same left sides and only the right sides need recomputing. The row from equation $(2)$ is untouched:

$y + z = 5.$

The row from equation $(3)$ is worked out again with the new constant:

$(4x + 13y + z) - 4(x + 4y + z) = 70 - 88, \qquad -3y - 3z = -18, \qquad y + z = 6.$

The same expression cannot be both $5$ and $6$. Subtracting one row from the other says so outright:

$0 = 1.$

A false row is satisfied by no triple, and every step of the sweep preserved the solution set, so the system now has no solution at all.

What changed geometrically is worth naming, because it is easy to overstate. The coefficient triples are still $1, 4, 1$ and $2, 5, -1$ and $4, 13, 1$, and no one of them is a multiple of another, so no two planes are parallel and each pair still meets in a line. What moved is the third plane, sliding without tilting until it no longer contains the line the first two share. One value of that constant, the $73$ it started with, put it back on that line; every other value takes the solution set from a whole line to nothing at all.

Work backwards from the ending you want. A sweep announces no solution by reaching a row that reads $0 =$ a number that is not zero, and that happens when some combination wipes out every unknown at once while the constants refuse to agree. So build the left sides to be dependent and the constants not to be.

Start with any two equations that are not proportional:

$(1)\ x + 2y - z = 3, \qquad (2)\ 2x - y + 4z = 5.$

Adding their left sides gives $3x + y + 3z$. If a third equation used that left side with the matching constant $3 + 5 = 8$, it would follow from the two above and add nothing. Choose any other constant, say $14$:

$(3)\ 3x + y + 3z = 14.$

No equation is a multiple of another. Scaling equation $(1)$ scales its coefficient triple $1, 2, -1$, and neither $2, -1, 4$ nor $3, 1, 3$ is such a scaling: doubling gives $2, 4, -2$ and tripling gives $3, 6, -3$. Nor is $3, 1, 3$ a multiple of $2, -1, 4$, since matching the first entries needs the factor $\tfrac{3}{2}$, which would force a second entry of $-\tfrac{3}{2}$ rather than $1$. Every pair has been tested, not just the pair that was combined.

There is no solution. Clear $x$ from both of the others with equation $(1)$:

$(2) - 2(1): \ -5y + 6z = -1, \qquad (3) - 3(1): \ -5y + 6z = 5.$

The same expression cannot be both $-1$ and $5$. Subtracting one row from the other says so with nothing left in it:

$0 = 6.$

Every step of a sweep preserves the solution set, so the system started with none. The test passes this system and its conclusion fails on it, so the claim as stated is false.

Suppose one equation is $k$ times another, both sides included, with $k$ not zero. The case $k = 0$ does not arise, since it would turn an equation into $0 = 0$, which is not a linear equation in three unknowns at all.

Divide the multiple through by $k$, which changes none of its solutions. The scaling reached the constant as well as the coefficients, so the two equations are now identical, right side and left. Subtracting one from the other removes every unknown and leaves

$0 = 0.$

That is the only ending the stated multiple can produce. A false row $0 = c$ with $c$ not zero is not available here: an equation whose constant disagreed would not have been a multiple in the first place.

A row reading $0 = 0$ constrains nothing and can be struck out. At most two of the three equations survive as genuine constraints, so at most two unknowns are pinned down, and the solution set is then a line, or a plane, or empty if what survives is itself contradictory. It is never a single point.

A stronger statement, which is not this one. Proportional LEFT sides are a wider condition than a multiple, because they leave the constants free to disagree, and the wider statement has to be argued separately rather than read off the narrow one. Suppose two equations carry proportional left sides. Scaling one makes the left sides match, and subtracting leaves

$0 = c_2 - c_1,$

where $c_1$ and $c_2$ are the two constants after that scaling. Now both endings are live. Agreeing constants give the $0 = 0$ above and the count that follows it. Disagreeing constants give a row no triple satisfies, so the system has no solution at all. Either way the count is not one.

So a system with exactly one solution can contain neither a multiple of one of its own equations nor two equations with proportional left sides. That is the half of the claim that holds. What part A shows is that the arrow runs only this way: ruling out multiples does not rule out either ending above, because a row can be wiped out by a combination of the OTHER TWO equations rather than by a single one of them.

If some left side is built from the other two, the count is never one. Suppose the left side of equation $(3)$ equals $m$ times the left side of $(1)$ plus $n$ times the left side of $(2)$. Subtracting those same multiples from equation $(3)$ removes every unknown at once and leaves

$0 = c_3 - mc_1 - nc_2.$

If that number is not zero, the row is false and the system has no solution. If it is zero, the row reads $0 = 0$ and constrains nothing, so at most two constraints survive in three unknowns, at least one unknown is free, and the solutions form a line or a plane, or nothing at all if the two survivors contradict each other. No ending here is a single point.

If no left side is built from the other two, the count is exactly one. Run the sweep, and suppose some row collapsed, every unknown vanishing from it. That row was built by subtracting multiples of the rows above it from one of the original equations, so its left side is that original left side minus a combination of the others. Its vanishing therefore says precisely that one left side WAS a combination of the other two, which is what we assumed away. So no row collapses; all three rows survive carrying a leading unknown, and the climb forces one value for each. (If the unknown you aimed at is missing from a row, aim at one that is present. The argument does not depend on the order.)

Where the constants act. They appear nowhere in the paragraph above: whether a row collapses is settled by the left sides alone. Once a row has collapsed, the constants decide which ending it is: agreeing constants leave $0 = 0$, a redundant equation, and provided no other row is false the survivors leave an unknown free and the solutions are infinite; disagreeing constants leave $0 = k$, and that empties the solution set whatever else survives. That is the division the test blurred, because a multiple is only the smallest of the ways one left side can be built from the others.

Let $x$, $y$ and $z$ be the tons of the $20$, $35$ and $50$ dollar grades. Each is a weight, so none of them can be negative, though none has to be a whole number either.

The first line counts tons:

$x + y + z = 30.$

The second charges each ton at its own grade's price:

$20x + 35y + 50z = 900.$

The third charges the same tonnage again with $5$ dollars per ton added, and since the delivery charge does not depend on the grade, every price rises by the same $5$:

$25x + 40y + 55z = 1050.$

Three unknowns and three recorded conditions. Whether three recorded conditions are three genuinely different conditions is another question, and it is the one the sweep answers.

Clear $x$ first, using the line that counts tons:

$(20x + 35y + 50z) - 20(x + y + z) = 900 - 600, \qquad 15y + 30z = 300, \qquad y + 2z = 20.$

$(25x + 40y + 55z) - 25(x + y + z) = 1050 - 750, \qquad 15y + 30z = 300, \qquad y + 2z = 20.$

The two rows are the same equation, so the sweep ends at

$0 = 0.$

That is no accident. Adding $5$ dollars per ton to every grade makes the third line the second line plus $5$ times the first, so it was built out of the two above it before the customer ever read it.

Two constraints survive in three unknowns, so one unknown is free and the answer is a line. Let $z = t$ be the tons of the most expensive grade. Then

$y = 20 - 2t, \qquad x = 30 - (20 - 2t) - t = 10 + t.$

So every delivery the invoice allows has the form

$(x, y, z) = (10 + t,\ 20 - 2t,\ t) \text{ tons}.$

Check it against all three lines with $t$ standing: $(10 + t) + (20 - 2t) + t = 30$ tons; then $20(10 + t) + 35(20 - 2t) + 50t = 900$ dollars; then $25(10 + t) + 40(20 - 2t) + 55t = 1050$ dollars.

Now bring in what the situation itself demands. No tonnage can be negative, so $10 + t \ge 0$ and $20 - 2t \ge 0$ and $t \ge 0$ hold together, which leaves

$0 \le t \le 10.$

Gravel is weighed rather than counted, so every value in that range is a genuine possibility, from $(10, 20, 0)$ at one end to $(20, 0, 10)$ at the other. The invoice cannot tell them apart.

Why a computed total cannot help. Suppose the yard offers a new line whose left side is $m$ times the tonnage line's left side plus $n$ times the gravel cost's left side, with its constant worked out the same way from $30$ and $900$. Subtracting those same multiples from the new equation removes every unknown and leaves

$0 = 0,$

exactly as the delivery line did. The row vanishes, no new pivot appears, and the free unknown stays free. The delivery total was of this kind, with $m = 5$ and $n = 1$, so it was never going to help. A tax at a fixed percentage of the gravel cost would be of this kind too.

What would help. The new figure's left side has to be one that cannot be built from the two the customer already holds. Any measurement that treats the grades differently in a new way qualifies: the tonnage of one grade on its own, or the cost of the two cheaper grades together.

The yard's report. Four tons of the most expensive grade is the equation $z = 4$, whose left side is $0x + 0y + z$. Could that be built from the other two? A combination $m(x + y + z) + n(20x + 35y + 50z)$ carries $y$ with coefficient $m + 35n$, which is zero only when $m = -35n$, and the $x$ coefficient is then $-35n + 20n = -15n$, which is zero only when $n = 0$. Both coefficients vanish only in the combination that is nothing at all, so $z = 4$ is genuinely new. It names the free unknown outright, so it settles the order:

$(x, y, z) = (10 + 4,\ 20 - 8,\ 4) = (14, 12, 4) \text{ tons}.$

Check: $14 + 12 + 4 = 30$ tons; $20(14) + 35(12) + 50(4) = 900$ dollars; and with delivery, $25(14) + 40(12) + 55(4) = 1050$ dollars.

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

Line 1 is sound. Multiplying equation $(1)$ by $-2$ gives $-6x - 2y + 4z = -10$, both sides included, which is equation $(2)$ exactly. So the combination really does give

$2(3x + y - 2z) + (-6x - 2y + 4z) = 2(5) + (-10), \qquad 0 = 0,$

and equation $(2)$ genuinely says nothing new.

Line 2 is the first line that does not follow. It checks the left sides and stops there. Multiplying equation $(1)$ by $3$ gives

$9x + 3y - 6z = 15,$

not $21$. So equation $(3)$ is not a multiple of equation $(1)$: it carries that left side scaled and a constant that does not match. Carrying out the subtraction the line skipped:

$(9x + 3y - 6z) - 3(3x + y - 2z) = 21 - 15, \qquad 0 = 6.$

What that settles. A row reading $0 = 6$ is satisfied by no triple whatever, and every step of a sweep preserves the solution set, so the system has no solution at all. The vanishing row from Line 1 does not soften the verdict: a redundant equation removes a constraint, but it cannot supply a triple to a system that has none. A false row anywhere overrides every vanishing row.

Geometrically there are two planes here rather than three. Equations $(1)$ and $(2)$ are the same sheet, and equation $(3)$ is a parallel sheet that never meets it.

Which constant. For Line 2 to be right, equation $(3)$ has to be $3$ times equation $(1)$ in full, so its constant must be $3(5) = 15$. Changing the $21$ to $15$ does it, and nothing else in the write-up moves.

Could a different single constant have done the job instead? Line 1 needs equation $(2)$ to be $-2$ times equation $(1)$, which pins equation $(2)$'s constant at $-2(5) = -10$, where it already is; changing it breaks Line 1. Changing equation $(1)$'s constant to some $c$ would ask for

$-2c = -10 \quad \text{and} \quad 3c = 21,$

that is $c = 5$ and $c = 7$ at once, which no number satisfies. So the $21$ is the only constant a single change can repair.

The repaired solution set. All three equations now describe the same plane, so exactly one constraint survives:

$3x + y - 2z = 5.$

With $3 - 1 = 2$ unknowns free, the answer is a plane and it takes two letters. Solve for $y$, whose coefficient is $1$ so that no fractions appear, and let the other two run free as $x = s$ and $z = t$:

$y = 5 - 3x + 2z = 5 - 3s + 2t.$

Every solution has the form

$(x, y, z) = (s,\ 5 - 3s + 2t,\ t),$

one triple for each pair of values $(s, t)$.

Every member is a solution. Substitute the family into the surviving equation with the letters left standing:

$3s + (5 - 3s + 2t) - 2t = 5.$

The $s$ terms cancel, the $t$ terms cancel, and $5 = 5$ is what remains, whatever $s$ and $t$ are. The other two equations are multiples of this one, so they hold as well. That is the first claim, established at every pair of values in a single line.

Nothing outside the family is a solution. Take any triple $(x, y, z)$ that solves the system. It satisfies $3x + y - 2z = 5$, and solving that equation for $y$ gives

$y = 5 - 3x + 2z.$

So choosing $s = x$ and $t = z$ produces exactly that triple from the family. Nothing was assumed about the triple beyond its being a solution, so no solution can sit outside the family.

What a sample check is worth. Testing $(0, 5, 0)$ and $(1, 2, 0)$ and finding both work supports the first claim at two pairs of values and says nothing at all about the second. The two directions are what make a report of an infinite set an answer rather than a guess, and only one of them can be reached by sampling. Part A cuts the other way, which is exactly why testing in all three ORIGINAL equations earns its keep: the plane reported there does hold triples satisfying equation $(1)$, but one sample carried into all three would have exposed the report at once, since $(0, 5, 0)$ lies on that plane and, in equation $(3)$ before the repair, gives $9(0) + 3(5) - 6(0) = 15$ rather than the $21$ demanded. Sampling reaches that far. What nothing sampled could have settled is the second claim, that the reported plane held every solution.