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Systems of Equations and Inequalities: Chapter Test

20 multiple-choice questions and 10 free-response questions, drawn from across the chapter and mixed together.

Free response

10 questions in parts, 149 points in total. Work them out on paper. There are no hints here: reveal each question's answer, worked solution, and rubric when you are ready to mark that one.

Free response · work it on paper

Question 1 of 10

1. A grid, one legal move, and the move that walks it back . 9 points. Question 1 of 10.

A matrix is elimination with the letters left out, so every step has to reach the constants as well as the coefficients. Work with the system $2x - 5z = 11$ , $x + 3y + z = -2$ , $4y - z = 7$ , and use the column order $x$ , $y$ , $z$ throughout.
1. Part A.
  Write the augmented matrix of this system.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Apply $R_1 \to R_1 - 2R_2$ to your matrix, report the matrix it produces, and say which rows are different afterwards and which are not.
  
  Carry your own answer forward Apply the operation to the grid you actually wrote down above, whatever it was. What earns credit here is that every entry of the named row moves, the constant among them, and that the other two rows are copied across untouched.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
3. Part C.
  Name the single operation that turns your part B matrix back into your part A matrix, and explain why neither of the two moves can alter which triples solve the system.
  
  Carry your own answer forward Argue from whichever pair of matrices you produced in parts A and B. The credit here is for the account of why a move that can be walked back cannot change the solution set, not for landing on one particular pair of grids.
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
The answer
Part A

$\left[\begin{array}{ccc|c}2 & 0 & -5 & 11 \\ 1 & 3 & 1 & -2 \\ 0 & 4 & -1 & 7\end{array}\right]$ , the two absent variables entering as $0$ so that every column keeps one variable.
- the same nine coefficients and three constants in the same places; what is not the same is a grid whose first or third row has only three entries, because a shorter row shifts the remaining numbers into columns that belong to other variables
Part B

The first row becomes $[\,0 \ \ -6 \ \ -7 \mid 15\,]$ , and rows two and three are exactly as they were.
- $[\,0 \ \ -6 \ \ -7 \mid 15\,]$ only; a first row ending in $11$ has updated three coefficients and left the constant behind, and $[\,0 \ \ -6 \ \ -7 \mid 7\,]$ has taken the constant above as $+2$ rather than $-2$ , computing $11 - 2(2)$
Part C

$R_1 \to R_1 + 2R_2$ turns it back. Any triple satisfying the old pair of equations satisfies the combined one, because it satisfies both pieces; and any triple satisfying the new pair satisfies the old, because row two is still present to be subtracted back out. Nothing is gained or lost in either direction.
Worked solution

Part A

One row per equation, one column per variable in the agreed order, and the constants alone behind the bar. The first equation has no $y$ and the third has no $x$ , so each contributes a $0$ that holds the column open.

$\left[\begin{array}{ccc|c}2 & 0 & -5 & 11 \\ 1 & 3 & 1 & -2 \\ 0 & 4 & -1 & 7\end{array}\right]$

The bar stands for the three equals signs; drop the column behind it and what is left no longer remembers what any equation equals.

Part B

The operation names row one, so every one of its four entries changes, and row two supplies the multiple without being touched itself. The constant runs $11 - 2(-2)$ , and subtracting twice a negative adds.

$[\,2 - 2 \ \ \ 0 - 6 \ \ \ -5 - 2 \mid 11 + 4\,] = [\,0 \ \ \ -6 \ \ \ -7 \mid 15\,]$

So the matrix becomes $\left[\begin{array}{ccc|c}0 & -6 & -7 & 15 \\ 1 & 3 & 1 & -2 \\ 0 & 4 & -1 & 7\end{array}\right]$ , with the last two rows carried over unchanged.

Part C

The move that undoes it. Adding back what was subtracted returns the row entry by entry:

$R_1 \to R_1 + 2R_2 \quad \text{sends} \quad [\,0 \ \ -6 \ \ -7 \mid 15\,] \ \text{back to} \ [\,2 \ \ 0 \ \ -5 \mid 11\,].$

Why the solution set survives. Write the first two equations as $E_1$ and $E_2$ . The step replaces $E_1$ by $E_1 - 2E_2$ and keeps $E_2$ . A triple satisfying $E_1$ and $E_2$ satisfies $E_1 - 2E_2$ as well, so no solution is lost. Going the other way, a triple satisfying $E_1 - 2E_2$ and $E_2$ satisfies $\left(E_1 - 2E_2\right) + 2E_2$ , which is $E_1$ , so no solution is invented either.

Watch out. The reversibility depends on $E_2$ still standing. Scaling a row by $0$ would leave nothing to walk back from, which is why that one move is banned.

In one line

The system encodes as $\left[\begin{array}{ccc|c}2 & 0 & -5 & 11 \\ 1 & 3 & 1 & -2 \\ 0 & 4 & -1 & 7\end{array}\right]$ , with a $0$ in the column of each variable an equation omits. Applying $R_1 \to R_1 - 2R_2$ sends the first row to $[\,0 \ \ -6 \ \ -7 \mid 15\,]$ and leaves the other two exactly as they were. The move $R_1 \to R_1 + 2R_2$ returns the grid to its earlier state, and that is the whole reason the solution set is safe: with row two still standing, the combination can be formed and unformed at will, so every triple that solved the system before solves it after, and no new one appears.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Fixes the column order before recording anything, so each column carries the same variable in every row. . Worth 2 points.

Enters a $0$ wherever an equation omits a variable, and puts the three constants alone in the column behind the bar. . Worth 1 point.

Part B 3 points

Changes all four entries of the row the operation names, the constant behind the bar included. . Worth 2 points.

States that the row supplying the multiple is unaltered, along with the row the operation never mentions. . Worth 1 point.

Part C 3 points

Grounds the claim in the step being undoable and names the operation that undoes it, rather than in the move simply being on a list of allowed ones. . Worth 2 points. needs an explanation, not just an answer

Notes that the row supplying the multiple stays in the system, which is what makes the step reversible. . Worth 1 point.
2. Two equations, and the comparison that has to wait . 14 points. Question 2 of 10.

Neither of these two equations arrives in the standard form $ax + by = c$ , and nothing can be compared until both do: $7y = 4x - 21 \qquad \text{and} \qquad 12x - 21y = 63.$
1. Part A.
  Put both equations in standard form and decide how many pairs satisfy both. Identify which comparison delivered that verdict, and state whether you needed to look at the numbers on the right at all.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 5 points
2. Part B.
  Give the whole solution set at once, as a family carrying a letter rather than as a handful of pairs, and check a general member of it in both of the ORIGINAL equations.
  
  Carry your own answer forward Build the family on the standard form you reached above, whatever it turned out to be. What earns credit is solving for one variable in terms of a letter, and testing with that letter still in place, rather than any particular pair of formulas.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
3. Part C.
  A classmate says the two equations must describe different lines, on the grounds that $12x - 21y = 63$ carries bigger numbers than the other one does. Judge that. Then say what producing a single pair that satisfies both equations would, and would not, have established about the count.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

In standard form they are $4x - 7y = 21$ and $12x - 21y = 63$ . The coefficient ratios $\tfrac{4}{12}$ and $\tfrac{-7}{-21}$ both equal $\tfrac{1}{3}$ , so a single crossing point is ruled out and the constants had to be consulted; $\tfrac{21}{63}$ is $\tfrac{1}{3}$ as well, so there are infinitely many solutions.
Part B

Every pair $(7t,\ 4t - 3)$ as $t$ runs over the real numbers, $(7, 1)$ and $(14, 5)$ among them.
- $\left(x,\ \frac{4x - 21}{7}\right)$ for every real $x$ is the same set written with $x$ as the parameter; what is not the same is any single pair, which names one member and not the set
Part C

Wrong: multiplying an equation through by a nonzero number changes every number in it and none of its solutions, so size says nothing about which points lie on the line. One shared pair would establish only consistency, leaving exactly one and infinitely many both open.
Worked solution

Part A

Gather the variables on one side of the first equation before reading any coefficient off it:

$7y = 4x - 21 \;\Longrightarrow\; 4x - 7y = 21.$

Now the three ratios can be compared in order. The first two agree, which removes the one-solution case and leaves parallel or coincident; the constant ratio then chooses between them:

$\frac{4}{12} = \frac{-7}{-21} = \frac{21}{63} = \frac{1}{3}.$

All three agree, so the second equation is the first multiplied through by $3$ : one line, described twice, and infinitely many solutions.

Part B

Both equations reduce to the one line $4x - 7y = 21$ , so hand a letter to $x$ and let $y$ follow. Choosing $x = 7t$ keeps the division by $7$ clean, and $7t$ still reaches every real number as $t$ does:

$4(7t) - 7y = 21 \;\Longrightarrow\; 7y = 28t - 21 \;\Longrightarrow\; y = 4t - 3.$

Now test the general member in both originals, with the letter still in place:

$7(4t - 3) = 28t - 21 = 4(7t) - 21, \qquad 12(7t) - 21(4t - 3) = 84t - 84t + 63 = 63.$

Every term in $t$ cancels in both, so the family sits inside the solution set for every value of $t$ , not just for the convenient ones.

Part C

The size of the numbers. Scaling both sides of an equation by a nonzero constant is one of the moves that never touches a solution set: a pair satisfying $4x - 7y = 21$ satisfies $12x - 21y = 63$ , and dividing by $3$ walks it back. So the second equation is the first with every number tripled, and the line is the same line.

$3\left(4x - 7y\right) = 3(21) \;\Longleftrightarrow\; 12x - 21y = 63.$

What one pair would settle. A pair satisfying both equations proves that the intersection is not empty, which rules out the parallel case and nothing else. Both surviving cases contain shared pairs: crossing lines have one, coincident lines have infinitely many.

Watch out. Producing several shared pairs would settle it, since two distinct shared points force the lines to coincide. One never does.

In one line

In standard form the equations are $4x - 7y = 21$ and $12x - 21y = 63$ , whose three ratios $\tfrac{4}{12}$ , $\tfrac{-7}{-21}$ and $\tfrac{21}{63}$ all equal $\tfrac{1}{3}$ , so the constants did have to be consulted and the system has infinitely many solutions. Its solution set is every pair $(7t,\ 4t - 3)$ for real $t$ , and substituting that family with the letter still in place returns $63$ and $28t - 21$ in the two original equations for every $t$ . Bigger numbers are no evidence of a different line, since tripling an equation changes every number in it and none of its solutions; and a single shared pair would have shown only that the system is consistent, leaving exactly one and infinitely many both open.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Rearranges into standard form before comparing anything, with the variables on one side and the constant alone on the other. . Worth 2 points.

Compares the coefficient ratios first and states that their agreement leaves two cases open, so the constant ratio is what decides. . Worth 2 points.

Reports the count, and says that the constants were consulted rather than leaving that unstated. . Worth 1 point.

Part B 5 points

Hands a letter to one variable and solves for the other in terms of it, giving a family rather than a list of sample pairs. . Worth 2 points.

Substitutes the family into both original equations with the letter still present, so the check covers every member at once. . Worth 2 points.

States the set as every pair of that shape, over all real values of the letter. . Worth 1 point.

Part C 4 points

Argues from scaling an equation leaving its solutions untouched, rather than from the two equations merely looking alike. . Worth 2 points. needs an explanation, not just an answer

States that one shared pair establishes consistency only, and names the two cases it leaves open. . Worth 2 points. needs an explanation, not just an answer
3. Where two lines meet, and whether anyone may stand there . 15 points. Question 3 of 10.

A library is ordering two kinds of shelving unit. A tall unit costs $3$ hundred dollars and takes $1$ metre of wall; a wide unit costs $2$ hundred dollars and takes $4$ metres of wall. The budget allows at most $36$ hundred dollars of spending and the room offers at most $32$ metres of wall. Let $x$ be the number of tall units and $y$ the number of wide units, so the region of allowable orders is cut out by $3x + 2y \le 36$ , $x + 4y \le 32$ , $x \ge 0$ and $y \ge 0$ .
1. Part A.
  Locate the corner of this region that touches neither axis. Show the pair of equations you had to solve to get there, and say what its two coordinates mean for the order.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Each of the two slanted boundary lines meets an axis at a point that is NOT a corner of this region. Find both of those points and show, for each one, the constraint it breaks.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  List every corner of the region, and say what a point has to satisfy before a crossing of two boundary lines counts as one.
  
  Carry your own answer forward Assemble the list from whatever you found in parts A and B, even if those were not the expected points. The credit here is for the account of what a crossing must satisfy before it counts, not for reproducing one particular list.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer
Part A

$(8, 6)$ : eight tall units and six wide units, an order that spends the whole budget and fills the whole wall.
- $x = 8$ with $y = 6$ is the same point; $(6, 8)$ is not, since the first coordinate counts tall units and the second counts wide ones
Part B

$(0, 18)$ breaks the wall limit, since $0 + 4(18) = 72$ exceeds $32$ , and $(32, 0)$ breaks the budget, since $3(32) = 96$ exceeds $36$ .

Part C

The corners are $(0, 0)$ , $(12, 0)$ , $(8, 6)$ and $(0, 8)$ . A crossing counts as a corner only when it satisfies every constraint in the system, not merely the two whose lines meet there. The two points from part B each satisfy their own pair and fail the remaining limit, which is exactly why they sit outside the region.
Worked solution

Part A

A corner off both axes is where the two slanted boundaries meet, so solve them as equations. Doubling the first and subtracting the second clears $y$ :

$2(3x + 2y) - (x + 4y) = 2(36) - 32 \;\Longrightarrow\; 5x = 40 \;\Longrightarrow\; x = 8,$

and then $8 + 4y = 32$ gives $y = 6$ . Check both limits at that point: the cost is $3(8) + 2(6) = 36$ hundred dollars and the wall used is $8 + 4(6) = 32$ metres, each exactly at its ceiling.

Part B

Set one variable to zero in each boundary equation, then test the point that comes out against the constraint it did not come from.

$3(0) + 2y = 36 \Rightarrow (0, 18), \qquad x + 4(0) = 32 \Rightarrow (32, 0).$

At $(0, 18)$ the wall used is $0 + 4(18) = 72$ metres against a ceiling of $32$ . At $(32, 0)$ the cost is $3(32) = 96$ hundred dollars against a ceiling of $36$ . Each point satisfies the one boundary it was built from and fails the other, so neither is an allowable order at all, let alone a corner.

Part C

Six pairs of boundary lines cross in total, and four of the crossings survive the test:

$(0, 0), \qquad (12, 0), \qquad (8, 6), \qquad (0, 8).$

Here $(12, 0)$ comes from $3x = 36$ and uses $12$ metres of wall, comfortably inside $32$ ; $(0, 8)$ comes from $4y = 32$ and costs $16$ hundred dollars, comfortably inside $36$ .

The test itself. A boundary line records where one constraint is met exactly, so a crossing is guaranteed to satisfy the two constraints that produced it and is guaranteed nothing about the rest. The region is the overlap of all four half-planes, so membership is decided by all four, and a crossing joins the corner list only after every constraint has been checked against it.

In one line

The two slanted boundaries cross at $(8, 6)$ , an order of eight tall units and six wide ones that spends all $36$ hundred dollars and fills all $32$ metres. The other two crossings with the axes are $(0, 18)$ , which needs $72$ metres of wall, and $(32, 0)$ , which costs $96$ hundred dollars, so neither is an allowable order. The region's corners are therefore $(0, 0)$ , $(12, 0)$ , $(8, 6)$ and $(0, 8)$ . A crossing of two boundary lines is guaranteed to satisfy only the two constraints that produced it; it becomes a corner of the region exactly when it satisfies all of them.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Solves the pair of boundary EQUATIONS, with the inequality symbols replaced by equals signs, rather than working with the inequalities. . Worth 2 points.

Carries the elimination through to both coordinates and checks the point against both limits. . Worth 2 points.

Says which count each coordinate is, so the pair reads as an order rather than as two loose numbers. . Worth 1 point.

Part B 5 points

Locates both axis intercepts by setting one variable to zero in each boundary equation. . Worth 2 points.

Substitutes each point into the OTHER constraint and shows the arithmetic that breaks it, rather than asserting that it fails. . Worth 2 points.

Says what the failure means for the library, that neither point is an order the room or the budget would allow. . Worth 1 point.

Part C 5 points

Gives a complete corner list, including the three on the axes that neither earlier part produced. . Worth 2 points.

States the test as satisfying EVERY constraint of the system, and explains that a crossing only ever guarantees the two constraints that built it. . Worth 3 points. needs an explanation, not just an answer
4. Down the staircase, up the staircase, and one claim about the route . 16 points. Question 4 of 10.

A forward sweep is a fixed routine and not a series of improvisations: aim every first-stage combination at one and the same unknown, drop a second unknown from the pair that survives, and the staircase left behind is read upward. Here is the system: $(1)\ 3x + y - z = -2, \qquad (2)\ x - 2y + 4z = -8, \qquad (3)\ 2x + 3y + z = 12.$ The middle equation is the convenient one to lead with, since its $x$ carries no coefficient to divide by.
1. Part A.
  Take $x$ out of equations $(1)$ and $(3)$ using equation $(2)$ , and then take $y$ out of the surviving pair. Set down the staircase this leaves, dividing through any row whose entries carry a factor in common.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Work your staircase upward from its bottom row, give the triple in the order $x$ , $y$ , $z$ , and substitute it into each of the three equations as the stem first wrote them.
  
  Carry your own answer forward Climb the staircase you actually reached above, and report honestly what it delivers. What earns credit is working the rows upward, carrying every value found so far into the row above it, and checking the result against the three equations as the stem first wrote them.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  A classmate says the sweep could just as well have cleared $y$ first instead of $x$ , and that doing so would have produced a different answer. Judge the two halves of that separately, and say what the shape of the solution set actually depends on.
  
  Carry your own answer forward Argue from whatever your own sweep produced, even if it was not the expected staircase. The credit here is for the account of why a route cannot change the answer and of what the shape does turn on, not for one particular triple.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
The answer
Part A

$x - 2y + 4z = -8$ , then $y - z = 4$ , then $-6z = -6$ .
- $z = 1$ in place of $-6z = -6$ is the same row already divided through, and a middle row left as $7y - 7z = 28$ is the same staircase undivided, since clearing $y$ from $7y - 13z = 22$ with that row gives the same bottom row $-6z = -6$ ; what is not the same is a bottom row of $-6z = 50$ , which subtracted the left sides of those two rows but added their constants
Part B

$(x, y, z) = (-2, 5, 1)$ , and the three original left-hand sides come out as $-2$ , $-8$ and $12$ , matching their right-hand sides.
- $x = -2$ , $y = 5$ , $z = 1$ is the same answer written out; $(1, 5, -2)$ is not, since it lists the values in the order the climb produced them rather than in the order $x$ , $y$ , $z$
Part C

The first half is right and the second wrong. Any unknown may be cleared first. But each step replaces one equation by itself plus a multiple of another, which neither loses nor invents a solution, so the answer cannot move. The shape turns on how many rows survive as constraints, a property of the equations, not of the route.
Worked solution

Part A

Subtract multiples of equation $(2)$ from the other two, so that $x$ leaves both at once:

$(1) - 3(2): \ 7y - 13z = 22, \qquad (3) - 2(2): \ 7y - 7z = 28 \;\Longrightarrow\; y - z = 4.$

Now clear $y$ from the first of those using the simplified second one:

$\left(7y - 13z\right) - 7\left(y - z\right) = 22 - 28 \;\Longrightarrow\; -6z = -6.$

Stacked up, the system is triangular, one unknown dropping away at each step:

$\begin{array}{rrrcr} x & -\, 2y & +\, 4z & = & -8 \\ & y & -\, z & = & 4 \\ & & -6z & = & -6 \end{array}$

Part B

The bottom row gives $z$ outright, and each row above brings in exactly one unknown that is not yet known:

$-6z = -6 \Rightarrow z = 1, \qquad y - 1 = 4 \Rightarrow y = 5, \qquad x - 10 + 4 = -8 \Rightarrow x = -2.$

Now test the triple everywhere, above all in equation $(1)$ , which the sweep combined and then set aside:

$3(-2) + 5 - 1 = -2 \ \checkmark, \qquad -2 - 10 + 4 = -8 \ \checkmark, \qquad -4 + 15 + 1 = 12 \ \checkmark.$

Part C

The first half. Nothing privileges $x$ . Clearing $y$ first would use equation $(1)$ , whose $y$ has coefficient $1$ , to remove $y$ from the other two, and the sweep would run to a staircase in a different order of unknowns. It is a route, not a rule.

The second half. Each combination replaces one equation by itself plus a multiple of another while leaving the supplier standing, and that step is reversible, so the system after it has exactly the solutions the system before it had:

$E_i \to E_i + kE_j \ \text{with} \ E_j \ \text{kept} \quad \Longrightarrow \quad \text{same solution set}.$

A chain of such steps therefore ends on the same set whichever chain it was. Any two honest sweeps of this system land on $(-2, 5, 1)$ .

What the shape does depend on. Three rows survive here as genuine constraints, so there are three pivots and $3 - 3 = 0$ free unknowns, which is what makes the answer a single point. A different route would still find three, because whether an equation restates the others is a fact about the equations.

In one line

Clearing $x$ with equation $(2)$ leaves $7y - 13z = 22$ and $y - z = 4$ , and clearing $y$ leaves $-6z = -6$ , so the triangular system is $x - 2y + 4z = -8$ , then $y - z = 4$ , then $-6z = -6$ . Climbing gives $z = 1$ , $y = 5$ and $x = -2$ , and the triple $(-2, 5, 1)$ returns $-2$ , $-8$ and $12$ in the three original equations. The classmate is half right: any unknown may be cleared first, but the answer cannot move, because every step of a sweep replaces one equation by itself plus a multiple of another while keeping the supplier, which neither loses nor invents a solution. What fixes the shape is the number of rows that still constrain, three here, so no unknown is free and the solution set is the single point.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Aims both first-stage combinations at the SAME unknown, so the two rows that survive share a pair of unknowns. . Worth 2 points.

Carries each combination across every term, the constants included, with the signs of the negative coefficients handled. . Worth 2 points.

Reports a stacked system in which each row begins one unknown later than the row above it. . Worth 1 point.

Part B 5 points

Starts at the bottom row and substitutes every value already found into each row on the way up. . Worth 2 points.

Reaches all three values and reports them as an ordered triple in the order $x$ , $y$ , $z$ . . Worth 2 points.

Tests the triple in all three ORIGINAL equations, including the one the sweep set aside. . Worth 1 point.

Part C 6 points

Accepts that any unknown may be cleared first, and identifies which equation would lead a sweep aimed at a different unknown. . Worth 1 point.

Rejects the second half by appealing to each step preserving the solution set, rather than by re-running the sweep and comparing answers. . Worth 3 points. needs an explanation, not just an answer

Ties the shape to the number of rows that still constrain, and counts the free unknowns as three minus that number. . Worth 2 points. needs an explanation, not just an answer
5. One number computed first, and how much of the answer it is entitled to give . 15 points. Question 5 of 10.

Two systems are put side by side. System I is $9x - 4y = 56, \qquad 7x + 5y = 3.$ System II is $x + 3y - 2z = 5, \qquad 2x + 6y - 4z = 9, \qquad 4x - y + z = 3.$ In each case the coefficient determinant is computed before anything else, because it decides how much further there is to go.
1. Part A.
  Evaluate the coefficient determinant of System I, and then solve the system by forming the two column-swapped determinants and dividing. Check the pair you get in both equations.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Evaluate System II's coefficient determinant by a first-row cofactor expansion, writing out all three minors together with the sign each one carries. Then say what that value settles and what, if anything, it leaves open.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Decide which of the two outcomes System II actually has. Then change exactly one of its constants so that it has the other one instead, and say why no change to any constant could ever give it exactly one solution.
  
  Carry your own answer forward Continue from whatever value you produced in part B, and say honestly what it does and does not permit. The credit here is for choosing between the two outcomes with the constants, for producing a repair that reaches the other one, and for the argument about what the determinant is built from.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
The answer
Part A

The coefficient determinant is $73$ , the swapped determinants are $292$ and $-365$ , and the solution is $(x, y) = (4, -5)$ , which returns $56$ and $3$ in the two equations.
- $x = 4$ with $y = -5$ is the same pair; $\left(\tfrac{73}{292},\ \tfrac{73}{-365}\right)$ is not, since it divides the coefficient determinant by the swapped one rather than the other way round
Part B

The coefficient determinant is $0$ . That settles that the solution is not a single triple, and it leaves open which of the other two outcomes holds: no solution, or infinitely many.

Part C

System II has no solution: its second equation has twice the first's left side while $9$ is not twice $5$ , so subtracting leaves $0 = -1$ . Replacing that $9$ by $10$ vanishes the row instead, leaving two constraints in three unknowns and a line of solutions. No constant can buy one solution: the determinant reads the coefficients alone.
Worked solution

Part A

Take the main diagonal first, then subtract the anti-diagonal, keeping the signs of the entries:

$D = \begin{vmatrix} 9 & -4 \\ 7 & 5 \end{vmatrix} = 45 - (-28) = 73.$

Since $D$ is not zero, one solution exists and the division below is legal. Replace the first column with the constants for the $x$ numerator, and the second column for the $y$ numerator:

$D_x = \begin{vmatrix} 56 & -4 \\ 3 & 5 \end{vmatrix} = 280 + 12 = 292, \qquad D_y = \begin{vmatrix} 9 & 56 \\ 7 & 3 \end{vmatrix} = 27 - 392 = -365.$

Dividing gives $x = \tfrac{292}{73} = 4$ and $y = \tfrac{-365}{73} = -5$ . The check holds: $9(4) - 4(-5) = 56$ and $7(4) + 5(-5) = 3$ .

Part B

Expand along the top row, with the signs running plus, minus, plus, and the middle term genuinely subtracted:

$1\begin{vmatrix} 6 & -4 \\ -1 & 1 \end{vmatrix} - 3\begin{vmatrix} 2 & -4 \\ 4 & 1 \end{vmatrix} + (-2)\begin{vmatrix} 2 & 6 \\ 4 & -1 \end{vmatrix} = 1(2) - 3(18) - 2(-26) = 2 - 54 + 52 = 0.$

The determinant is built from the nine coefficients and never looks at the constants, so a value of $0$ can only report that the answer is not one triple. Choosing between an empty set and an infinite one needs the constants, which this number has not read.

Part C

Which outcome. Compare the first two equations. Their left sides are locked in a ratio, so the constants decide:

$(2) - 2(1): \ 0 = 9 - 10 = -1.$

A false row empties the solution set, whatever the third equation does, so System II has no solution.

The repair. Replace the $9$ by $10$ . The same combination then gives $0 = 0$ , the second equation is revealed as the first restated, and what survives is $x + 3y - 2z = 5$ together with $4x - y + z = 3$ : two genuine constraints in three unknowns, so one unknown is free and the solutions form a line.

Why one solution is out of reach. The coefficient determinant reads the nine coefficients and nothing else. Moving a constant leaves all nine where they are, so $D$ stays $0$ and the answer stays not-unique. Only a change to a coefficient could restore a single triple.

In one line

System I has coefficient determinant $73$ , so it has one solution, and the swapped determinants $292$ and $-365$ give $(x, y) = (4, -5)$ , which returns $56$ and $3$ in the two equations. System II has coefficient determinant $0$ , which settles only that its answer is not a single triple. The constants decide the rest: the second equation carries twice the first's left side while $9$ is not twice $5$ , so the combination leaves $0 = -1$ and there is no solution. Replacing that $9$ by $10$ vanishes the row instead and leaves two constraints in three unknowns, a whole line of solutions. No constant at all can restore a single solution, because the determinant reads the nine coefficients and never the constants, so it stays $0$ throughout.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Evaluates the coefficient determinant as main diagonal minus anti-diagonal, with the sign of every entry carried into its product. . Worth 2 points.

Builds each numerator by replacing that unknown's own column with the constant column, and divides the swapped determinant by the coefficient one. . Worth 2 points.

Reports the pair and confirms it in both equations rather than in the one it was read from. . Worth 1 point.

Part B 4 points

Forms each minor by deleting the row and column of its own entry, and attaches the alternating signs so the middle term is subtracted. . Worth 2 points.

Reaches the value and states it as ruling out a single solution, without naming either of the remaining outcomes as settled. . Worth 2 points.

Part C 6 points

Chooses between the two outcomes by comparing the constants of the proportional pair, and shows the row the comparison produces. . Worth 2 points.

Names a single replacement constant and says what the reduced system then leaves standing, including how many unknowns are free. . Worth 2 points.

Argues that the determinant is built from the coefficients alone, so no constant can move it off zero. . Worth 2 points. needs an explanation, not just an answer
6. A region given by its corners, and an objective to maximize on it . 15 points. Question 6 of 10.

A region of the plane is the quadrilateral with corners $(0, 0)$ , $(9, 0)$ , $(6, 4)$ and $(0, 7)$ . All four of its sides are solid. Two of them lie along the axes; the other two are the segment from $(9, 0)$ to $(6, 4)$ and the segment from $(6, 4)$ to $(0, 7)$ . The objective to be maximized on it is $T = 8x + 6y$ .
1. Part A.
  Write the system of inequalities whose solution set is exactly this region, one inequality per side, and say how you fixed the direction of the symbol on each slanted side.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Evaluate $T = 8x + 6y$ at every corner, and report the largest value together with every point of the region that attains it.
  
  Carry your own answer forward Use the corners as the stem gives them, and evaluate the objective at each. If your part A system was not the expected one, that does not affect this part: the corners were supplied rather than derived, and the credit here is for evaluating at all four and for describing the whole set of best points.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  Describe how the lines along which $T$ is constant meet the boundary of this region, say what that forces about the full set of points where $T$ is largest, and decide whether the corner point principle has been contradicted here.
  
  Carry your own answer forward Argue from your own part B values and your own part A inequalities, whatever they were. What earns credit is tying whatever you found there to the slopes involved, and reading the corner point principle correctly, rather than naming one particular pair of corners.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer
Part A

$x \ge 0$ , $y \ge 0$ , $4x + 3y \le 36$ and $x + 2y \le 14$ . A point inside the region, such as $(1, 1)$ , makes both slanted inequalities true, which fixes both directions.
- $8x + 6y \le 72$ in place of $4x + 3y \le 36$ is the same half-plane written with every number doubled; what is not the same is any strict symbol, since the corners themselves belong to the region
Part B

The corner values are $0$ , $72$ , $72$ and $42$ . The largest is $72$ , attained at $(9, 0)$ , at $(6, 4)$ , and at every point of the segment joining them.
- naming the maximizing set as the whole side $4x + 3y = 36$ between those two corners is the same set; reporting $(9, 0)$ alone, or $(6, 4)$ alone, is not, since each leaves out points where $T$ is just as large
Part C

The objective is twice the left side of $4x + 3y \le 36$ , so the lines $T = c$ are parallel to that side and the last one to touch the region meets it all along that side. The principle is not contradicted: it promises at least one corner among the best points, not exactly one point, and both ends of the side are corners.
Worked solution

Part A

Each slanted side supplies one boundary equation, found from the two corners it joins. From $(9, 0)$ to $(6, 4)$ the slope is $-\tfrac{4}{3}$ , and from $(6, 4)$ to $(0, 7)$ it is $-\tfrac{1}{2}$ :

$4x + 3y = 36, \qquad x + 2y = 14.$

Confirm each against both of the corners it was built from, since an equation fitted to one point is not yet a line through two:

$4(9) + 3(0) = 36, \quad 4(6) + 3(4) = 36; \qquad 6 + 2(4) = 14, \quad 0 + 2(7) = 14.$

The symbol on each is then fixed by a point known to be inside, not by the look of the picture. At $(1, 1)$ the left sides are $7$ and $3$ , both below their constants, so both symbols point that way. The axes supply $x \ge 0$ and $y \ge 0$ , and every boundary is solid, so every symbol allows equality.

Part B

Evaluate at the four corners in turn:

$T(0,0) = 0, \qquad T(9,0) = 72, \qquad T(6,4) = 48 + 24 = 72, \qquad T(0,7) = 42.$

Two corners tie at $72$ . Because the region is convex, the whole segment between them lies inside it, and $T$ is $72$ at every point of that segment: a point $t$ of the way along is $\left(9 - 3t,\ 4t\right)$ , and $8(9 - 3t) + 6(4t) = 72 - 24t + 24t = 72$ for every $t$ from $0$ to $1$ .

Part C

Why the tie happens. For each value of $c$ , the set where $T = c$ is the line $8x + 6y = c$ , and all such lines share the slope $-\tfrac{4}{3}$ . That is exactly the slope of the side joining $(9, 0)$ to $(6, 4)$ , since the side lies on $4x + 3y = 36$ and $T = 2\left(4x + 3y\right)$ . So the objective's lines and that side are parallel.

$T = 8x + 6y = 2\left(4x + 3y\right) \le 2(36) = 72.$

Sliding the line outward, the last one to meet the region is $T = 72$ , and it does not graze a point: it lies along the whole side.

Whether the principle breaks. It does not. The corner point principle says that wherever an optimum exists and the region has corners, at least one corner attains it. Here two do, which satisfies that statement comfortably. What would contradict the principle is a maximum attained only at an interior point or only in the middle of a side, and the tie here still reaches both ends of the side.

Watch out. Evaluating at the corners is still the right procedure when an edge ties. What changes is only the report: the answer is the whole edge, and naming one endpoint understates it.

In one line

The region is the solution set of $x \ge 0$ , $y \ge 0$ , $4x + 3y \le 36$ and $x + 2y \le 14$ , with each slanted symbol fixed by an interior point such as $(1, 1)$ and every symbol allowing equality because every side is solid. The objective $T = 8x + 6y$ takes the values $0$ , $72$ , $72$ and $42$ at the four corners, so its largest value is $72$ , attained at $(9, 0)$ , at $(6, 4)$ , and at every point of the side joining them. The tie is forced by $T$ being twice the left side of $4x + 3y \le 36$ , which makes the lines $T = c$ parallel to that very side, so the last one to touch the region lies along it. The corner point principle is untouched: it promises at least one corner among the best points, and here two of them are.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Builds each slanted boundary's equation from the two corners it joins, and confirms it against both of them. . Worth 2 points.

Fixes the direction of each symbol by substituting a point known to be inside the region, rather than reading it off the arrangement of the picture. . Worth 2 points.

Uses symbols that allow equality throughout, since every side is solid, and includes the two axis constraints. . Worth 1 point.

Part B 5 points

Evaluates the objective at all four corners rather than at a chosen few. . Worth 2 points.

Reports the largest value and identifies more than one point attaining it, extending the answer to the whole side between the tied corners. . Worth 3 points.

Part C 5 points

Identifies the objective as a multiple of one constraint's left side, so its level lines share that side's slope. . Worth 2 points. needs an explanation, not just an answer

States the corner point principle as promising at least one corner, not exactly one point, and applies that reading to the tie. . Worth 2 points. needs an explanation, not just an answer

Notes that both ends of the tied side are corners, so the promise is met rather than merely unbroken. . Worth 1 point.
7. Three equations, one matrix, and what its rows report . 17 points. Question 7 of 10.

Three equations arrive with left sides that are visibly related to one another: $(1)\ x + 2y - 3z = 4, \qquad (2)\ 2x + 4y - 6z = 8, \qquad (3)\ 3x + 6y - 9z = 15.$ Use the column order $x$ , $y$ , $z$ throughout.
1. Part A.
  Write the augmented matrix and reduce it, recording each step in $R_i \to R_i + cR_j$ or $R_i \leftrightarrow R_j$ notation at the moment you take it. Report the grid you finish on.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Translate each of the two lower rows of your reduced matrix back into an equation, say what each one on its own would report, and then give the system's solution set.
  
  Carry your own answer forward Read back whichever rows your own reduction in part A produced, and report honestly what they say. The credit here is for translating each row into the equation it stands for, for saying what that equation demands on its own, and for letting those readings settle the solution set.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 5 points
3. Part C.
  A classmate stops at the zero row and reports: 'a row vanished, so the system is dependent and has infinitely many solutions, one for each of two free variables'. Say precisely what the zero row does establish and what it does not. Then give the one change to a single constant of the ORIGINAL system that would make the report correct.
  
  Carry your own answer forward Judge the report against whatever your own reduction produced in parts A and B. The credit here is for separating what a vanishing row licenses from what it does not, and for a repair that reaches the reported outcome, not for one particular constant.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 7 points
The answer
Part A

$\left[\begin{array}{ccc|c}1 & 2 & -3 & 4 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0\end{array}\right]$ , after $R_2 \to R_2 - 2R_1$ , then $R_3 \to R_3 - 3R_1$ , then the swap $R_2 \leftrightarrow R_3$ .
- the same three rows with the zero row second and the row ending in $3$ third is the same reduction before the tidying swap; what is not the same is a grid in which the row ending in $3$ has been discarded, since discarding it throws away the only row that decides the answer
Part B

One row reads $0 = 3$ , which no triple can satisfy, and the other reads $0 = 0$ , which every triple satisfies and which therefore reports nothing. The solution set is empty: a row that cannot be satisfied empties it however many other rows vanished.

Part C

The zero row establishes only that equation $(2)$ restated equation $(1)$ , so it constrains nothing. It never settles a count, because another row can still be false, and here one is. Replacing the $15$ in equation $(3)$ by $12$ vanishes that row as well, leaving one constraint, two free variables and a plane.
Worked solution

Part A

Encode the three equations first, one row each, with the constants alone behind the bar:

$\left[\begin{array}{ccc|c}1 & 2 & -3 & 4 \\ 2 & 4 & -6 & 8 \\ 3 & 6 & -9 & 15\end{array}\right]$

Both lower rows have left sides that are multiples of the top row's, so a single pass clears them:

$R_2 \to R_2 - 2R_1: \ [\,0 \ \ 0 \ \ 0 \mid 8 - 8\,], \qquad R_3 \to R_3 - 3R_1: \ [\,0 \ \ 0 \ \ 0 \mid 15 - 12\,].$

That leaves $[\,0 \ \ 0 \ \ 0 \mid 0\,]$ and $[\,0 \ \ 0 \ \ 0 \mid 3\,]$ . Row-echelon form sinks any all-zero row to the bottom, so swap the two:

$\left[\begin{array}{ccc|c}1 & 2 & -3 & 4 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0\end{array}\right]$

Part B

Read each row as the equation it stands for, with the coefficient entries multiplying the unknowns:

$[\,0 \ \ 0 \ \ 0 \mid 3\,] \ \text{reads} \ 0x + 0y + 0z = 3, \qquad [\,0 \ \ 0 \ \ 0 \mid 0\,] \ \text{reads} \ 0 = 0.$

The first is false for every triple, so no triple satisfies the whole system. The second is true for every triple, so it excludes nothing and adds nothing. Since a solution has to satisfy every equation at once, the false row alone settles it and the set is empty.

Part C

What the zero row shows. A row of zeros across the whole width, constant included, is the report that one equation was already implied by the others. It removes a constraint from the count and nothing more.

What it does not show. It says nothing about the rows it did not come from. Counting free variables is only legitimate once no row is false, and here the reduction also produced $0 = 3$ . So the classmate's inference runs in the right shape but on an unchecked premise, and the premise fails.

The repair. Equation $(3)$ has three times the left side of equation $(1)$ , so its row vanishes exactly when its constant is three times $4$ :

$R_3 \to R_3 - 3R_1: \ 0 = 12 - 12 = 0 \quad \text{when the constant is} \ 12.$

With $15$ replaced by $12$ , all three equations describe the single plane $x + 2y - 3z = 4$ . One constraint survives in three unknowns, so $3 - 1 = 2$ variables are free and the solutions are $(s,\ t,\ \tfrac{s + 2t - 4}{3})$ for all real $s$ and $t$ , a whole plane. That, and only that, is the report the classmate wrote.

In one line

The augmented matrix reduces by $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$ , and a swap sinks the all-zero row, giving $\left[\begin{array}{ccc|c}1 & 2 & -3 & 4 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0\end{array}\right]$ . The two lower rows read $0 = 3$ and $0 = 0$ : the first is satisfied by nothing and the second by everything, so the first decides and the solution set is empty. The classmate's zero row establishes only that equation $(2)$ restated equation $(1)$ ; it never settles a count, because a false row elsewhere overrides it, and here one is. Replacing the $15$ in equation $(3)$ by $12$ vanishes that row too, leaving the single plane $x + 2y - 3z = 4$ with two free variables, which is the only version of this system the report describes.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Encodes all three equations with their constants behind the bar before any operation is applied. . Worth 1 point.

Applies each operation across the whole row, so the two lower rows differ from one another only in the column behind the bar. . Worth 2 points.

Names every operation used, including the swap that sinks the all-zero row to the bottom. . Worth 2 points.

Part B 5 points

Translates each row back into an equation, with the constant behind the bar on the right-hand side. . Worth 2 points.

Distinguishes a row nothing satisfies from a row everything satisfies, and says what each one contributes on its own. . Worth 2 points.

Reports the solution set, giving the false row priority over the vanished one. . Worth 1 point.

Part C 7 points

States what a fully zero row does establish, that one equation was already implied, and stops there. . Worth 2 points.

Locates the fault in counting free variables before checking that no row is false, naming the row that is. . Worth 3 points. needs an explanation, not just an answer

Produces a single replacement constant and shows the resulting count of surviving constraints and free variables. . Worth 2 points.
8. Choosing an additive dose . 15 points. Question 8 of 10.

A water plant doses two additives into each batch. One litre of additive X removes $2$ units of iron and $3$ units of manganese; one litre of additive Y removes $3$ units of iron and $1$ unit of manganese. Every batch must lose at least $42$ units of iron and at least $28$ units of manganese, so with $x$ litres of X and $y$ litres of Y the requirements are $2x + 3y \ge 42$ , $3x + y \ge 28$ , $x \ge 0$ and $y \ge 0$ . Additive X costs $4$ dollars a litre and additive Y costs $5$ . The region of allowable doses has exactly three corners: $(0, 28)$ , $(6, 10)$ and $(21, 0)$ .
1. Part A.
  Evaluate the cost $C = 4x + 5y$ at each of the three corners and report the cheapest allowable dose with its litres and its cost. Then show that no allowable dose at all costs less, by combining the two requirements into a lower bound on $C$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Decide whether this plant also has a most expensive allowable dose. Settle it from the requirements themselves rather than from the three corner values, and say which property of the region your argument leans on.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
3. Part C.
  A classmate says the corner point principle cannot be used on this region at all, because the region is unbounded. Say what the principle does and does not require, and say what boundedness would have added here.
  
  Carry your own answer forward Argue from your own findings in parts A and B, whatever they were. The credit here is for stating what the principle actually requires and for saying what boundedness supplies, not for a particular cheapest dose.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer
Part A

The corner costs are $140$ , $74$ and $84$ dollars, and $4x + 5y = \tfrac{11}{7}(2x + 3y) + \tfrac{2}{7}(3x + y) \ge 74$ , so the cheapest dose is $6$ litres of X with $10$ of Y, at $74$ dollars.
- $74$ dollars at $(6, 10)$ is the same answer, and any positive weights on the two requirements that combine to $4x + 5y$ give the same bound; $74$ on its own is not, because a cost with no dose attached does not tell the plant what to pour
Part B

There is none. Take any allowable dose and add a litre of X: both requirements still hold, since more of each impurity is removed, and the cost rises by $4$ dollars. So no dose can carry the largest cost. The argument turns on the region running on without limit in that direction.

Part C

The principle requires an optimum to exist and the region to have corners; it does not require boundedness. The cheapest dose does exist, and a corner attains it. Boundedness with solid boundaries, on a region that is not empty, would have guaranteed a dearest dose too, and that guarantee is what this region withholds.
Worked solution

Part A

Evaluate the cost at each corner in turn, keeping the litres attached to the number:

$C(0, 28) = 140, \qquad C(6, 10) = 24 + 50 = 74, \qquad C(21, 0) = 84.$

The smallest of the three is $74$ dollars, at $6$ litres of X and $10$ litres of Y, and that dose meets both requirements exactly: $2(6) + 3(10) = 42$ units of iron and $3(6) + 10 = 28$ units of manganese.

Comparing three corner values does not by itself produce a cheapest dose, because the corner principle assumes an optimum exists and this region runs on without limit. So bound the cost from the requirements themselves, using weights that are positive so that both inequalities may be applied:

$4x + 5y = \tfrac{11}{7}\left(2x + 3y\right) + \tfrac{2}{7}\left(3x + y\right) \ge \tfrac{11}{7}(42) + \tfrac{2}{7}(28) = 74.$

No allowable dose costs less than $74$ dollars, and $(6, 10)$ costs exactly that, so a cheapest dose does exist and it is the one found.

Part B

Suppose some allowable dose $(x, y)$ carried the largest cost. Move one litre to the right, to $(x + 1,\ y)$ , and check the requirements:

$2(x + 1) + 3y = \left(2x + 3y\right) + 2 \ge 42, \qquad 3(x + 1) + y = \left(3x + y\right) + 3 \ge 28,$

and $x + 1 \ge 0$ , $y \ge 0$ still hold. So the new dose is allowable too, and its cost is

$C(x + 1,\ y) = C(x, y) + 4 > C(x, y).$

That contradicts the supposed largest cost, so no dose holds one. What made the step always available is that the requirements are all lower bounds, so nothing caps $x$ from above: the region runs on forever to the right, and the cost climbs as it goes.

Part C

What the principle asks for. It says that wherever an optimum of a linear objective exists on a region with corners, at least one corner attains it. Existence is a hypothesis, not a conclusion, and boundedness appears nowhere in the statement.

What boundedness is for. Boundedness together with solid boundaries is what supplies the existence hypothesis, in both directions at once, so long as the region has a point in it to begin with: a nonempty closed bounded region has a largest and a smallest value for any linear objective. Losing boundedness loses that guarantee and nothing else.

$\text{nonempty, bounded and closed} \;\Longrightarrow\; \text{both extremes exist} \;\Longrightarrow\; \text{a corner attains each}$

This region. Part B showed the cost is unbounded above, so no dearest dose exists and there is nothing for a corner to attain in that direction. The cheapest dose does exist, since part A bounded the cost below by $74$ dollars and a corner reached that bound, so the principle did its work on the half of the question that had an answer.

Watch out. The classmate's rule and its opposite are both wrong. Unboundedness never guarantees that an optimum is missing, and it never guarantees that one is present; it only withdraws the guarantee.

In one line

The cost $C = 4x + 5y$ is $140$ , $74$ and $84$ dollars at $(0, 28)$ , $(6, 10)$ and $(21, 0)$ , and combining the two requirements as $4x + 5y = \tfrac{11}{7}\left(2x + 3y\right) + \tfrac{2}{7}\left(3x + y\right) \ge 74$ shows that no allowable dose costs less, so the cheapest dose is $6$ litres of X with $10$ litres of Y at $74$ dollars a batch. There is no dearest dose: from any allowable dose, one more litre of X keeps both requirements met and adds $4$ dollars, so no dose can hold the largest cost, and what makes that step always available is the region running on without limit to the right. The classmate is wrong about the principle. It requires an optimum to exist and the region to have corners, never boundedness; boundedness with solid boundaries, on a region that is not empty, is merely what would have guaranteed existence in both directions. Here existence holds for the minimum, which the bound above establishes, and fails for the maximum, and the corner that attains the minimum is exactly where the principle says it will be.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Evaluates the cost at all three corners, and settles the cheapest dose by bounding the cost below from the two requirements rather than by the corner comparison alone. . Worth 2 points.

Reports the cheapest dose in litres of each additive as well as in dollars, so the answer names a plan and not only a number. . Worth 2 points.

Part B 6 points

Produces, from an arbitrary allowable dose, another allowable dose that costs strictly more, rather than testing sample points. . Worth 3 points. needs an explanation, not just an answer

Checks that the new dose still meets every requirement, using the requirements as stated rather than the corner list. . Worth 2 points.

Names the region running on without limit in the direction the cost grows as the feature that removes the maximum. . Worth 1 point.

Part C 5 points

States the principle with existence as a hypothesis and boundedness absent from it. . Worth 2 points. needs an explanation, not just an answer

Identifies boundedness with solid boundaries as what guarantees existence, and says which extreme this region therefore loses. . Worth 2 points. needs an explanation, not just an answer

Applies the corrected reading to this region, noting that a corner does attain the cost that exists. . Worth 1 point.
9. One constant left unnamed, and the row that waits for it . 17 points. Question 9 of 10.

A system arrives with one of its constants still unsettled: $(1)\ x + 2y + 5z = 4, \qquad (2)\ 2x + 5y + 8z = 5, \qquad (3)\ 3x + 7y + 13z = p.$ All nine coefficients are fixed; only $p$ is free to be chosen, and it rides through the arithmetic untouched until the very last row.
1. Part A.
  Lead with equation $(1)$ and take $x$ out of the other two, treating $p$ as a number whose value simply has not been said yet, then finish the sweep. Report the bottom row as an equation in $p$ alone.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Give the solution set for every value of $p$ , in whatever form each case turns out to need, and check whatever family you produce in all three ORIGINAL equations.
  
  Carry your own answer forward Use whichever bottom row your part A produced, and split the cases by whatever value makes it vanish. The credit here is for treating a false row and a vanished row differently, for parameterizing rather than naming a sample triple, and for testing with the letter still in place.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 6 points
3. Part C.
  At the value of $p$ that leaves more than one solution, a classmate writes: 'equation $(3)$ was redundant, so it can simply be crossed out; and since a redundant equation turned up at all, the answer would have been infinitely many whatever $p$ was'. Judge the two halves separately.
  
  Carry your own answer forward Judge the claim against whatever you found in parts A and B, even if your value of $p$ was not the expected one. The credit here is for separating what the coefficients settle from what the constants settle, not for a particular number.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 6 points
The answer
Part A

Clearing $x$ leaves $y - 2z = -3$ and $y - 2z = p - 12$ , and clearing $y$ from the second of those leaves the bottom row $0 = p - 9$ .
- $p - 9 = 0$ is the same row written the other way round; a bottom row of $0 = p - 15$ has added the constant $-3$ instead of subtracting it
Part B

For every $p$ other than $9$ the bottom row is false and there is no solution. At $p = 9$ it vanishes and the solutions are the triples $(10 - 9t,\ 2t - 3,\ t)$ , one for each real $t$ .
- any family reaching the same set is equivalent, for instance giving the letter to $y$ instead and reporting $\left(\tfrac{-9y - 7}{2},\ y,\ \tfrac{y + 3}{2}\right)$ ; what is not the same is a single sample triple, which names one member of an infinite set
Part C

The first half is right at that value: equation $(3)$ collapses to $0 = 0$ there, so crossing it out changes no solution. The second half reverses the dependence. The left sides cancel for every $p$ ; only a constant matching that same combination makes the row vanish, and any other $p$ leaves a false row and an empty set.
Worked solution

Part A

Subtract multiples of equation $(1)$ from the other two, letting $p$ ride along untouched:

$(2) - 2(1): \ y - 2z = -3, \qquad (3) - 3(1): \ y - 2z = p - 12.$

The two rows carry the identical left side, so clearing $y$ removes $z$ at the same moment and nothing with an unknown in it survives:

$\left(y - 2z\right) - \left(y - 2z\right) = \left(p - 12\right) - (-3) \;\Longrightarrow\; 0 = p - 9.$

The sweep has run out of unknowns and left a statement about $p$ alone, which is the report the system was always going to give, whatever $p$ turned out to be.

Part B

The bottom row $0 = p - 9$ is false for every $p$ except one, and a false row empties the set, so only $p = 9$ has anything further to report.

At $p = 9$ the bottom row vanishes and two genuine constraints remain: equation $(1)$ and $y - 2z = -3$ . One unknown is therefore free, so hand a letter to $z$ :

$y = 2t - 3, \qquad x = 4 - 2y - 5z = 4 - \left(4t - 6\right) - 5t = 10 - 9t.$

Now test the family, with the letter still in place, in all three originals at $p = 9$ :

$(10 - 9t) + 2(2t - 3) + 5t = 4, \quad 2(10 - 9t) + 5(2t - 3) + 8t = 5, \quad 3(10 - 9t) + 7(2t - 3) + 13t = 9.$

Every term in $t$ cancels in all three, so each holds for every $t$ rather than for a lucky one.

Part C

The first half. At $p = 9$ the combination $(3) - 3(1) - \left[(2) - 2(1)\right]$ sends equation $(3)$ to $0 = 0$ . An equation that reduces to a statement every triple satisfies excludes nothing, so deleting it leaves the solution set exactly where it was, and the family from part B is unchanged.

The second half. The left side of equation $(3)$ is fixed, and it is the sum of the other two left sides:

$\left(x + 2y + 5z\right) + \left(2x + 5y + 8z\right) = 3x + 7y + 13z.$

So the left sides cancel for every $p$ : that part of the collapse was decided by the coefficients alone. What $p$ decides is only the right-hand side, which the same combination sends to $4 + 5 = 9$ . When $p$ matches, the row reads $0 = 0$ ; when it does not, the row reads $0 = p - 9$ with $p - 9$ nonzero, and one false row empties the set however many others vanished.

Watch out. The classmate has read a vanished row as a cause when it is a symptom. The cancellation of the left sides happens every time; whether the result is redundancy or contradiction is settled by the constants alone.

In one line

Clearing $x$ with equation $(1)$ leaves $y - 2z = -3$ and $y - 2z = p - 12$ , and clearing $y$ leaves the bottom row $0 = p - 9$ . So for every $p$ other than $9$ the row is false and the system has no solution; at $p = 9$ it vanishes, two constraints survive in three unknowns, and the solutions are the triples $(10 - 9t,\ 2t - 3,\ t)$ for every real $t$ , which return $4$ , $5$ and $9$ in the three original equations with every term in $t$ cancelling. The classmate's first half is right at that value, since a row reading $0 = 0$ excludes nothing. The second half reverses the dependence: the left side of equation $(3)$ is the sum of the other two left sides for every $p$ , so the cancellation always happens, and only the constants decide whether it leaves a redundant row or a false one.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Aims both combinations at $x$ , so the two rows that survive share the same pair of unknowns. . Worth 2 points.

Carries $p$ through the arithmetic as an ordinary number rather than dropping or evaluating it, reaching the constants $-3$ and $p - 12$ and then the bottom row $0 = p - 9$ . . Worth 2 points.

Reports a bottom row with no unknowns left on either side, stated as an equation in $p$ . . Worth 1 point.

Part B 6 points

Splits the values of $p$ at the one that vanishes the bottom row, and reports the empty set for all the others. . Worth 2 points.

Counts the surviving constraints against the three unknowns and hands a letter to the unknown that is left free. . Worth 2 points.

Substitutes the family into all three original equations with the letter still present, so every member is covered at once. . Worth 2 points.

Part C 6 points

Accepts the first half at that value, on the grounds that a row reading $0 = 0$ excludes nothing. . Worth 2 points.

Shows that the LEFT sides cancel for every value of the constant, so the cancellation itself carries no verdict. . Worth 2 points. needs an explanation, not just an answer

Concludes that the constants decide between a vanished row and a false one, and that a false row empties the set whatever else vanished. . Worth 2 points. needs an explanation, not just an answer
10. One coefficient still to be chosen, and what the determinant makes of it . 16 points. Question 10 of 10.

The system $4x - 3y = 7, \qquad kx + 6y = 2$ carries an unspecified number $k$ in front of $x$ in its second equation, and nowhere else. Every other number in it is fixed.
1. Part A.
  Express the coefficient determinant in terms of $k$ . Say for which $k$ a single solution is unavailable, and decide, at any such $k$ , which of the two remaining outcomes actually occurs.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  For the values of $k$ at which the division is legal, write $x$ as a single expression in $k$ , simplified as far as it goes, and say what licenses the simplification.
  
  Carry your own answer forward Divide by whichever coefficient determinant you produced in part A, and exclude whichever value made it vanish. The credit here is for putting the column-swapped determinant on top, for simplifying only where a factor is genuinely nonzero, and for saying which condition licenses that.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
3. Part C.
  Find the value of $k$ for which the solution has $x = 2$ . Then decide whether any value of $k$ at all gives this system infinitely many solutions, and argue that decision rather than testing values.
  
  Carry your own answer forward Solve for $k$ using whichever expression for $x$ you produced in part B, and argue the second question from the ratios of the system as it is stated. The credit here is for reaching a value of $k$ honestly from your own expression, and for an argument that covers every $k$ at once rather than a sample of them.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
The answer
Part A

The coefficient determinant is $3k + 24$ , which is zero only at $k = -8$ . There the two equations demand $4x - 3y = 7$ and $4x - 3y = -1$ at once, so the system has no solution.
- $24 + 3k$ or $3(k + 8)$ is the same expression; a determinant of $24 - 3k$ has taken the anti-diagonal product as $+3k$ rather than $-3k$ , and it puts the failure at the wrong value of $k$
Part B

$x = \dfrac{48}{3k + 24} = \dfrac{16}{k + 8}$ , for every $k$ other than $-8$ . Cancelling the $3$ is licensed because $3$ is nonzero, and the expression exists at all only because $k + 8$ is.
- $\frac{48}{3(k + 8)}$ unsimplified is the same function of $k$ ; what is not the same is $\frac{16}{k} + \frac{16}{8}$ , which splits a single denominator into two
Part C

Setting the part B expression equal to $2$ gives $k + 8 = 8$ , so $k = 0$ . No value of $k$ gives infinitely many: that needs $\tfrac{4}{k} = \tfrac{-3}{6} = \tfrac{7}{2}$ , and the last two ratios carry no $k$ and are already unequal.
Worked solution

Part A

Main diagonal minus anti-diagonal, with the sign of the $-3$ carried into the second product:

$D = \begin{vmatrix} 4 & -3 \\ k & 6 \end{vmatrix} = 24 - (-3)k = 3k + 24.$

This vanishes exactly when $k = -8$ , so for every other $k$ the system has exactly one solution. At $k = -8$ the second equation reads $-8x + 6y = 2$ ; dividing by $-2$ gives

$4x - 3y = -1,$

while the first equation insists the same expression equals $7$ . Nothing can do both, so the outcome there is no solution.

Part B

Replace the first column of the coefficient determinant with the constant column, then divide by the coefficient determinant:

$D_x = \begin{vmatrix} 7 & -3 \\ 2 & 6 \end{vmatrix} = 42 + 6 = 48, \qquad x = \frac{48}{3k + 24} = \frac{48}{3\left(k + 8\right)} = \frac{16}{k + 8}.$

Factoring $3$ out of the denominator is always legal, and cancelling it against the numerator is legal because $3$ is not zero. The expression as a whole is defined only where $k + 8 \ne 0$ , and that is not an extra assumption: it is the condition part A already imposed before any division was allowed.

Watch out. The swapped determinant goes on top. Writing $\frac{D}{D_x}$ instead returns the reciprocal, and here it would report $x = \frac{k + 8}{16}$ , which fails the equations.

Part C

The value of $k$ . Set the expression from part B equal to $2$ :

$\frac{16}{k + 8} = 2 \;\Longrightarrow\; k + 8 = 8 \;\Longrightarrow\; k = 0.$

Check it directly. With $k = 0$ the second equation is $6y = 2$ , so $y = \tfrac{1}{3}$ , and the first gives $4x = 7 + 1 = 8$ , that is $x = 2$ .

Why infinitely many is unreachable. Two equations in two unknowns have infinitely many solutions exactly when one is a rescaling of the other, coefficients and constant alike. Written as ratios, that demands

$\frac{4}{k} = \frac{-3}{6} = \frac{7}{2}.$

The last two of those carry no $k$ : they are $-\tfrac{1}{2}$ and $\tfrac{7}{2}$ , fixed numbers that are not equal. So the chain breaks at a link $k$ cannot reach, whatever value it is given. The only degenerate case available is therefore the one part A found, where the coefficient ratios line up at $k = -8$ and the constants refuse to follow.

Reading it back through the determinant. The determinant vanishes only at $k = -8$ , so that is the only candidate for a non-unique answer at all, and there the system is inconsistent. That is the determinant test working exactly as advertised: it identifies where uniqueness fails and declines to say which failure it is, and the constants supply the rest.

In one line

The coefficient determinant is $3k + 24$ , which vanishes only at $k = -8$ ; there the second equation reduces to $4x - 3y = -1$ against the first equation's $4x - 3y = 7$ , so the system has no solution. For every other $k$ there is exactly one solution, with $x = \frac{48}{3k + 24} = \frac{48}{3\left(k + 8\right)} = \frac{16}{k + 8}$ : cancelling the factor $3$ is licensed by $3$ being nonzero, while $k + 8 \ne 0$ is separately what permitted the division in the first place and is where the expression is defined. Setting that equal to $2$ gives $k = 0$ , and the system $4x - 3y = 7$ with $6y = 2$ does return $x = 2$ . No value of $k$ produces infinitely many solutions: that would need $\tfrac{4}{k} = \tfrac{-3}{6} = \tfrac{7}{2}$ , and the last two ratios carry no $k$ and are already unequal, so the only degenerate case this system has is the inconsistent one.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Evaluates the determinant with the sign of the negative entry carried into the anti-diagonal product, so the expression is a sum rather than a difference in $k$ . . Worth 2 points.

Solves the determinant to zero and states that every other value of $k$ leaves exactly one solution. . Worth 1 point.

Chooses between the two remaining outcomes at that value by comparing the two equations, rather than reading the outcome off the determinant. . Worth 2 points.

Part B 5 points

Builds the numerator by replacing the $x$ column with the constants, and places it over the coefficient determinant rather than under it. . Worth 2 points.

Simplifies the quotient to a single fraction in $k$ , cancelling only a factor that is genuinely nonzero. . Worth 2 points.

States the excluded value and identifies it as the same condition that permitted the division in the first place. . Worth 1 point.

Part C 6 points

Solves for $k$ from the expression for $x$ and confirms the value in the system as originally written. . Worth 2 points.

Argues over all values of $k$ at once, by identifying a comparison in the system that $k$ does not appear in. . Worth 3 points. needs an explanation, not just an answer

Connects the conclusion back to the determinant, noting that it locates where uniqueness fails without naming which failure. . Worth 1 point.

Marking rubrics

Work every question first. Then mark your own paper against these, giving yourself the points for each element you actually wrote down.

1. A grid, one legal move, and the move that walks it back 9 points

Part A 3

Fixes the column order before recording anything, so each column carries the same variable in every row. 2
Enters a $0$ wherever an equation omits a variable, and puts the three constants alone in the column behind the bar. 1

Part B 3

Changes all four entries of the row the operation names, the constant behind the bar included. 2
States that the row supplying the multiple is unaltered, along with the row the operation never mentions. 1

Part C 3

Grounds the claim in the step being undoable and names the operation that undoes it, rather than in the move simply being on a list of allowed ones. 2
Notes that the row supplying the multiple stays in the system, which is what makes the step reversible. 1

2. Two equations, and the comparison that has to wait 14 points

Part A 5

Rearranges into standard form before comparing anything, with the variables on one side and the constant alone on the other. 2
Compares the coefficient ratios first and states that their agreement leaves two cases open, so the constant ratio is what decides. 2
Reports the count, and says that the constants were consulted rather than leaving that unstated. 1

Part B 5

Hands a letter to one variable and solves for the other in terms of it, giving a family rather than a list of sample pairs. 2
Substitutes the family into both original equations with the letter still present, so the check covers every member at once. 2
States the set as every pair of that shape, over all real values of the letter. 1

Part C 4

Argues from scaling an equation leaving its solutions untouched, rather than from the two equations merely looking alike. 2
States that one shared pair establishes consistency only, and names the two cases it leaves open. 2

3. Where two lines meet, and whether anyone may stand there 15 points

Part A 5

Solves the pair of boundary EQUATIONS, with the inequality symbols replaced by equals signs, rather than working with the inequalities. 2
Carries the elimination through to both coordinates and checks the point against both limits. 2
Says which count each coordinate is, so the pair reads as an order rather than as two loose numbers. 1

Part B 5

Locates both axis intercepts by setting one variable to zero in each boundary equation. 2
Substitutes each point into the OTHER constraint and shows the arithmetic that breaks it, rather than asserting that it fails. 2
Says what the failure means for the library, that neither point is an order the room or the budget would allow. 1

Part C 5

Gives a complete corner list, including the three on the axes that neither earlier part produced. 2
States the test as satisfying EVERY constraint of the system, and explains that a crossing only ever guarantees the two constraints that built it. 3

4. Down the staircase, up the staircase, and one claim about the route 16 points

Part A 5

Aims both first-stage combinations at the SAME unknown, so the two rows that survive share a pair of unknowns. 2
Carries each combination across every term, the constants included, with the signs of the negative coefficients handled. 2
Reports a stacked system in which each row begins one unknown later than the row above it. 1

Part B 5

Starts at the bottom row and substitutes every value already found into each row on the way up. 2
Reaches all three values and reports them as an ordered triple in the order $x$ , $y$ , $z$ . 2
Tests the triple in all three ORIGINAL equations, including the one the sweep set aside. 1

Part C 6

Accepts that any unknown may be cleared first, and identifies which equation would lead a sweep aimed at a different unknown. 1
Rejects the second half by appealing to each step preserving the solution set, rather than by re-running the sweep and comparing answers. 3
Ties the shape to the number of rows that still constrain, and counts the free unknowns as three minus that number. 2

5. One number computed first, and how much of the answer it is entitled to give 15 points

Part A 5

Evaluates the coefficient determinant as main diagonal minus anti-diagonal, with the sign of every entry carried into its product. 2
Builds each numerator by replacing that unknown's own column with the constant column, and divides the swapped determinant by the coefficient one. 2
Reports the pair and confirms it in both equations rather than in the one it was read from. 1

Part B 4

Forms each minor by deleting the row and column of its own entry, and attaches the alternating signs so the middle term is subtracted. 2
Reaches the value and states it as ruling out a single solution, without naming either of the remaining outcomes as settled. 2

Part C 6

Chooses between the two outcomes by comparing the constants of the proportional pair, and shows the row the comparison produces. 2
Names a single replacement constant and says what the reduced system then leaves standing, including how many unknowns are free. 2
Argues that the determinant is built from the coefficients alone, so no constant can move it off zero. 2

6. A region given by its corners, and an objective to maximize on it 15 points

Part A 5

Builds each slanted boundary's equation from the two corners it joins, and confirms it against both of them. 2
Fixes the direction of each symbol by substituting a point known to be inside the region, rather than reading it off the arrangement of the picture. 2
Uses symbols that allow equality throughout, since every side is solid, and includes the two axis constraints. 1

Part B 5

Evaluates the objective at all four corners rather than at a chosen few. 2
Reports the largest value and identifies more than one point attaining it, extending the answer to the whole side between the tied corners. 3

Part C 5

Identifies the objective as a multiple of one constraint's left side, so its level lines share that side's slope. 2
States the corner point principle as promising at least one corner, not exactly one point, and applies that reading to the tie. 2
Notes that both ends of the tied side are corners, so the promise is met rather than merely unbroken. 1

7. Three equations, one matrix, and what its rows report 17 points

Part A 5

Encodes all three equations with their constants behind the bar before any operation is applied. 1
Applies each operation across the whole row, so the two lower rows differ from one another only in the column behind the bar. 2
Names every operation used, including the swap that sinks the all-zero row to the bottom. 2

Part B 5

Translates each row back into an equation, with the constant behind the bar on the right-hand side. 2
Distinguishes a row nothing satisfies from a row everything satisfies, and says what each one contributes on its own. 2
Reports the solution set, giving the false row priority over the vanished one. 1

Part C 7

States what a fully zero row does establish, that one equation was already implied, and stops there. 2
Locates the fault in counting free variables before checking that no row is false, naming the row that is. 3
Produces a single replacement constant and shows the resulting count of surviving constraints and free variables. 2

8. Choosing an additive dose 15 points

Part A 4

Evaluates the cost at all three corners, and settles the cheapest dose by bounding the cost below from the two requirements rather than by the corner comparison alone. 2
Reports the cheapest dose in litres of each additive as well as in dollars, so the answer names a plan and not only a number. 2

Part B 6

Produces, from an arbitrary allowable dose, another allowable dose that costs strictly more, rather than testing sample points. 3
Checks that the new dose still meets every requirement, using the requirements as stated rather than the corner list. 2
Names the region running on without limit in the direction the cost grows as the feature that removes the maximum. 1

Part C 5

States the principle with existence as a hypothesis and boundedness absent from it. 2
Identifies boundedness with solid boundaries as what guarantees existence, and says which extreme this region therefore loses. 2
Applies the corrected reading to this region, noting that a corner does attain the cost that exists. 1

9. One constant left unnamed, and the row that waits for it 17 points

Part A 5

Aims both combinations at $x$ , so the two rows that survive share the same pair of unknowns. 2
Carries $p$ through the arithmetic as an ordinary number rather than dropping or evaluating it, reaching the constants $-3$ and $p - 12$ and then the bottom row $0 = p - 9$ . 2
Reports a bottom row with no unknowns left on either side, stated as an equation in $p$ . 1

Part B 6

Splits the values of $p$ at the one that vanishes the bottom row, and reports the empty set for all the others. 2
Counts the surviving constraints against the three unknowns and hands a letter to the unknown that is left free. 2
Substitutes the family into all three original equations with the letter still present, so every member is covered at once. 2

Part C 6

Accepts the first half at that value, on the grounds that a row reading $0 = 0$ excludes nothing. 2
Shows that the LEFT sides cancel for every value of the constant, so the cancellation itself carries no verdict. 2
Concludes that the constants decide between a vanished row and a false one, and that a false row empties the set whatever else vanished. 2

10. One coefficient still to be chosen, and what the determinant makes of it 16 points

Part A 5

Evaluates the determinant with the sign of the negative entry carried into the anti-diagonal product, so the expression is a sum rather than a difference in $k$ . 2
Solves the determinant to zero and states that every other value of $k$ leaves exactly one solution. 1
Chooses between the two remaining outcomes at that value by comparing the two equations, rather than reading the outcome off the determinant. 2

Part B 5

Builds the numerator by replacing the $x$ column with the constants, and places it over the coefficient determinant rather than under it. 2
Simplifies the quotient to a single fraction in $k$ , cancelling only a factor that is genuinely nonzero. 2
States the excluded value and identifies it as the same condition that permitted the division in the first place. 1

Part C 6

Solves for $k$ from the expression for $x$ and confirms the value in the system as originally written. 2
Argues over all values of $k$ at once, by identifying a comparison in the system that $k$ does not appear in. 3
Connects the conclusion back to the determinant, noting that it locates where uniqueness fails without naming which failure. 1

Multiple choice

Free response