Systems of Equations and Inequalities: Chapter Test

Multiply down the main diagonal, then subtract the product up the anti-diagonal. The entry $-4$ carries its sign into that second product, so subtracting $-28$ adds.

$\begin{vmatrix} 6 & -4 \\ 7 & 3 \end{vmatrix} = (6)(3) - (-4)(7) = 18 + 28 = 46$

The symbol $\le$ allows equality, so the points of the line $4x - 3y = -12$ are solutions and the boundary is drawn solid. One point off the line then settles the side, and the origin fails, so the shading goes on the other side.

$4(0) - 3(0) = 0, \qquad 0 \le -12 \text{ is false}$

Both equations are already in standard form, so compare the three ratios in turn. The coefficient ratios agree, which rules out a single crossing point, and then the constants decide between parallel and coincident.

$\frac{10}{-15} = -\frac{2}{3} = \frac{-4}{6}, \qquad \frac{14}{-20} = -\frac{7}{10}$

The constant ratio breaks the pattern, so the lines are parallel and share no point.

One row per equation, one column per variable in the agreed order, and the constant behind the bar. The equation carries no $x$, so its $x$ column holds a $0$ to keep the other coefficients in place.

$4y - 3z = 5 \;\longrightarrow\; [\,0 \ \ 4 \ \ -3 \mid 5\,]$

Climb from the bottom row, one new unknown at a time. The top row subtracts $z$, and $z$ is negative, so that term arrives as $+2$.

$-4z = 8 \Rightarrow z = -2, \qquad 2y + 5(-2) = -6 \Rightarrow y = 2, \qquad x + 3(2) - (-2) = 7 \Rightarrow x = -1$

Expand along the first row. Each entry is multiplied by the two by two determinant left after deleting its own row and column, and the three signs run plus, minus, plus.

$2(0 - 10) - 3(24 - 5) + 1(8 - 0) = -20 - 57 + 8 = -69$

Expanding along the second row instead gives $-4(18 - 2) + 0 - 5(4 - 3) = -69$ as well.

Read the two endings separately. A vanishing row is an equation that restated the others, so it constrains nothing at all; a false row is a demand no triple can meet.

$0 = 0 \ \text{constrains nothing}, \qquad 0 = 9 \ \text{is impossible}$

One row that cannot be satisfied empties the solution set however many other rows vanished, so nothing survives.

One row vanished, so two constraints remain in three unknowns and one variable is free. Solve each surviving row for its own leading variable with $z = t$ in place.

$x = 5 + 2t, \qquad y = -3 - 4t, \qquad z = t$

Substituting back returns $x - 2t = 5$ and $y + 4t = -3$ for every $t$, so the whole family qualifies and nothing outside it does.

The corners are the origin, the two feasible axis points, and the crossing of the slanted boundaries. Subtracting $x + 2y = 24$ from $5x + 2y = 40$ gives $4x = 16$, so that crossing is $(4, 10)$, which satisfies both limits.

$M(0,0) = 0, \quad M(8,0) = 24, \quad M(0,12) = 60, \quad M(4,10) = 62$

The region is bounded with solid boundaries, so the largest of the corner values is the largest value anywhere.

The determinant is the quantity every unknown is divided by, so a nonzero value is exactly the condition for one forced answer. A zero value withdraws that condition and nothing more.

$D = 0 \;\Longrightarrow\; \text{not unique: either none, or infinitely many}$

Which of the two it is depends on the constants, which $D$ never looks at, so the constants have to be compared or the system reduced.

Every entry of the row named by the operation changes, the constant included, and the row supplying the multiple is untouched.

$[\,4 - 4 \ \ \ 5 + 12 \ \ \ -1 - 8 \mid 6 - 28\,] = [\,0 \ \ \ 17 \ \ \ -9 \mid -22\,]$

The $+12$ appears because the entry above is $-3$, and subtracting four times a negative adds.

The coefficient determinant is $25 - k^2$, which vanishes at $k = 5$ and at $k = -5$, so those are the only values where a single point is off the table. Now look at what each one leaves.

$k = 5: \ x + y = 4 \ \text{and} \ x + y = 3, \qquad k = -5: \ x - y = 4 \ \text{and} \ x - y = -3$

Each value forces one expression to equal two different numbers, so both are inconsistent and neither produces a line.

Use the first equation to clear $x$ from the other two, which leaves a pair in $y$ and $z$ alone.

$-7y - z = 5, \qquad -3y - 5z = -7 \;\Longrightarrow\; 32y = -32$

So $y = -1$, then $z = -7(-1) - 5 = 2$, then $x = 4 - 2(-1) - 2 = 4$. The triple $(4, -1, 2)$ checks in all three originals, including the one set aside.

The corners are $(0, 18)$, the crossing $(3, 6)$ of the two slanted boundaries, and $(12, 0)$; each satisfies both requirements.

$C(0,18) = 36, \qquad C(3,6) = 27, \qquad C(12,0) = 60$

Comparing corner values cannot by itself produce a smallest value on a region that runs on forever, so bound $C$ from the two requirements instead. Both weights below are positive, so both inequalities may be used:

$5x + 2y = \tfrac{3}{10}\left(2x + 3y\right) + \tfrac{11}{10}\left(4x + y\right) \ge \tfrac{3}{10}(24) + \tfrac{11}{10}(18) = 27$

No allowable point brings $C$ below $27$, and $(3, 6)$ reaches it, so $27$ is the smallest value and a corner attains it. Up and to the right the region runs on without limit and $C$ grows with it, so there is no largest value.

The zero row says $0 = 0$, so it adds nothing, but it also takes nothing away: the first three rows still hold a pivot in the $x$, $y$ and $z$ columns, so no variable is free. Climb from the third row.

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

A zero row is a redundant equation, not a missing pivot, and here every unknown is still pinned down.

Expand along the first row, minding the minus on the middle term.

$D = 1(-6) - 2(7) + 4(5) = -6 - 14 + 20 = 0$

So the answer is not a single triple. Adding the first two equations gives $3x + y + 5z = 8$, which is the third exactly, so the third constrains nothing and two genuine equations remain in three unknowns: infinitely many solutions.

A crossing of two boundary lines counts as a corner only when it satisfies every constraint in the system, not merely the two whose lines meet there. Subtracting $2x + 5y = 60$ from $4x + 5y = 80$ gives $2x = 20$, so $x = 10$, and then $5y = 60 - 20$ gives $y = 8$.

$2(10) + 5(8) = 60 \ \checkmark, \qquad 4(10) + 5(8) = 80 \ \checkmark, \qquad 10 \ge 0, \ 8 \ge 0 \ \checkmark$

Every constraint holds, so $(10, 8)$ is a corner. The two crossings with the axes fail the constraint they did not come from: $(30, 0)$ needs $4(30) = 120$ to be at most $80$, and $(0, 16)$ needs $5(16) = 80$ to be at most $60$, so neither is even in the region.

All three left sides are multiples of $x - 2y + z$, so both lower rows lose their variables at once. Only the constants distinguish them.

$(2) - 3(1): \ 0 = 0, \qquad (3) - 2(1): \ 0 = 11 - 6 = 5$

The second equation restates the first and the third contradicts it. A false row empties the set however many rows vanished, so nothing satisfies all three.

The two left sides are already locked together: multiplying $8x - 12y$ by $-\tfrac{3}{4}$ gives $-6x + 9y$. For one line rather than two, the constant has to travel by the same factor.

$-\tfrac{3}{4}(8x - 12y) = -6x + 9y, \qquad -\tfrac{3}{4}(20) = -15$

Any other $c$ leaves $0 = (\text{nonzero})$ and no solution; no $c$ gives exactly one, since the slopes never differ.

Two constraints in three unknowns leave one variable free, so the answer is a whole line and no single triple can be all of it. Clearing $z$ between the two equations gives $3x + 5y = 13$, which every member of the family has to satisfy. Test the candidate with the letter still in place.

$(5t + 1) + 3(2 - 3t) - 4(-t - 1) = 5t + 1 + 6 - 9t + 4t + 4 = 11$

The second equation checks the same way, the terms in $t$ canceling for every $t$, which is what a family must do and what one sample triple never shows.