Worked examples — Linear programming — simplex method (intro)

Before we start, one reminder of the vocabulary the parent built, because we will use it in line one of every example:

Recall The five words we reuse everywhere

• Decision variables $x_1,x_2,\dots$ — the numbers you get to choose.
• Slack variable $s_i\ge 0$ — the "unused room" that turns $a\cdot x\le b$ into $a\cdot x+s=b$.
• Basic variable — one we actually solve for; non-basic — one we force to $0$.
• Tableau — the grid of equation coefficients we pivot on.
• Optimal (for maximizing) — reached when every number in the $z$-row is $\ge 0$.

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The scenario matrix

Think of the simplex method as a machine with three moving parts: entering rule, ratio test, stopping rule. Each part has a "normal" behaviour and several "surprising" behaviours. The grid below lists every combination we must be able to survive. There are ten cells (A–J), and the nine worked examples below cover all ten — Example 7 deliberately handles two cells at once (a $\ge$ constraint solved as a minimization), so 9 examples $=$ 10 cells. Each example announces its cell(s) in its heading.

Cell	What is unusual about this case	Covered by
A. Clean run	Nothing — one path straight to the top	Example 1
B. Tie in entering column	Two $z$-row entries equally negative	Example 2
C. Tie in ratio test (degeneracy)	Two rows share the smallest ratio; a basic variable becomes $0$	Example 3
D. Unbounded objective	Entering column has no positive entry; $z\to +\infty$	Example 4
E. Multiple optima	A non-basic $z$-row coefficient equals $0$ at the end; a whole edge is optimal	Example 5
F. Degenerate start / zero RHS	A constraint has $b=0$; corner sits on a wall from the start	Example 6
G. $\ge$ constraint	Wrong-way inequality; needs surplus + artificial variables	Example 7
H. Minimization	We want the smallest $z$; maximize $-z$	Example 7
I. Word problem	Numbers hidden in a story + units; model, then solve	Example 8
J. Exam twist	"How much can a coefficient change before the answer moves?"; sensitivity	Example 9

Cell count reconciled: 10 cells, 9 examples, because Example 7 = Cells G + H together.

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Example 1 — Cell A: the clean run (baseline, with geometry)

Forecast: Before reading on, guess the optimal corner. The steeper "profit per unit" is on $x_1$ ($4$ vs $3$), so bet the winner leans toward large $x_1$. Write your guess down.

Read the figure first. The two constraint lines are drawn — the solid line labelled "$2x_1+x_2=10$" and the dashed line labelled "$x_1+3x_2=15$" (each line is also written on the figure in words, so you never need the colour to tell them apart). The hatched region is the feasible set; its four corners (where lines cross) are marked with dots and labelled with their $z$-values. The big arrow marked "$z$ increases" points into the top corner $(3,4)$ — that is the corner the machine will crawl toward.

1. Add slacks. Why this step? The machine only eats equations, so name the leftover room in each $\le$: $2x_1+x_2+s_1=10,x_1+3x_2+s_2=15,z-4x_1-3x_2=0.$

2. Start at $(0,0)$. Why? Setting $x_1=x_2=0$ makes $s_1=10,s_2=15\ge 0$ — a legal corner (the origin, where the two axes cross, is feasible), $z=0$.

   Basis	$x_1$	$x_2$	$s_1$	$s_2$	RHS
   $s_1$	2	1	1	0	10
   $s_2$	1	3	0	1	15
   $z$	$-4$	$-3$	0	0	0

3. Entering variable. Why? $-4$ is the most negative, so pushing $x_1$ up gains fastest. $x_1$ enters.

4. Ratio test. Why? Divide RHS by positive column entries: $10/2=5$, $15/1=15$. Smallest is $5$ (row 1) — go further and $s_1$ turns negative. $s_1$ leaves.

5. Pivot on the $2$. Why pivot at all? Pivoting makes the entering column a unit vector, so $x_1$ becomes basic (solved for) while $s_1$ drops out — this is exactly the algebra of hopping to the next corner. Row1 $÷2$; Row2 $-$ Row1(new); $z$-row $+4\cdot$Row1(new):

   Basis	$x_1$	$x_2$	$s_1$	$s_2$	RHS
   $x_1$	1	$\frac{1}{2}$	$\frac{1}{2}$	0	5
   $s_2$	0	$\frac{5}{2}$	$-\frac{1}{2}$	1	10
   $z$	0	$-1$	2	0	20

6. Still a $-1$ under $x_2$ → not optimal. $x_2$ enters. Ratios: $5/\frac{1}{2}=10$, $10/\frac{5}{2}=4$. Smallest $4$ → $s_2$ leaves. Why pivot on the $\frac{5}{2}$? Same reason as before: making the $x_2$ column a unit vector turns $x_2$ basic and moves us along an edge to the next, higher corner.

   Basis	$x_1$	$x_2$	$s_1$	$s_2$	RHS
   $x_1$	1	0	$\frac{3}{5}$	$-\frac{1}{5}$	3
   $x_2$	0	1	$-\frac{1}{5}$	$\frac{2}{5}$	4
   $z$	0	0	$\frac{9}{5}$	$\frac{2}{5}$	24

7. All $z$-row entries $\ge 0$ → optimal. Corner $(3,4)$, $z=24$.

Verify: Check constraints: $2(3)+4=10\le 10$ ✓ (tight), $3+3(4)=15\le 15$ ✓ (tight). Both tight, so $(3,4)$ is a genuine corner (two lines crossing). $z=4(3)+3(4)=24$ ✓. Compare corners $(0,0)=0$, $(5,0)=20$, $(0,5)=15$, $(3,4)=24$ — max confirmed, and it is the corner the arrow in the figure pointed at.

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Example 2 — Cell B: a tie for who enters

Forecast: The two variables have identical profit ($5$ each) and the region is symmetric. Guess: the answer sits on the line of symmetry $x_1=x_2$.

1. Initial $z$-row: $-5,-5$. The tie: both are "most negative". Why does the tie not matter? Either choice still increases $z$; simplex converges either way. Apply Bland's rule (stated at the top): break the tie by lowest index, so $x_1$ enters.

2. $x_1$ enters. Ratios $8/1=8$, $8/2=4$ → row 2 leaves ($s_2$). Why pivot on the $2$? To make the $x_1$ column a unit vector so $x_1$ becomes basic — the corner-hop step:

   Basis	$x_1$	$x_2$	$s_1$	$s_2$	RHS
   $s_1$	0	$\frac{3}{2}$	1	$-\frac{1}{2}$	4
   $x_1$	1	$\frac{1}{2}$	0	$\frac{1}{2}$	4
   $z$	0	$-\frac{5}{2}$	0	$\frac{5}{2}$	20

3. $-\frac{5}{2}$ under $x_2$ → $x_2$ enters. Ratios $4/\frac{3}{2}=\frac{8}{3}$, $4/\frac{1}{2}=8$ → row 1 leaves. Why pivot on the $\frac{3}{2}$? Same rule: unit-vector the $x_2$ column, hop to the next corner:

   Basis	$x_1$	$x_2$	$s_1$	$s_2$	RHS
   $x_2$	0	1	$\frac{2}{3}$	$-\frac{1}{3}$	$\frac{8}{3}$
   $x_1$	1	0	$-\frac{1}{3}$	$\frac{2}{3}$	$\frac{8}{3}$
   $z$	0	0	$\frac{5}{3}$	$\frac{5}{3}$	$\frac{80}{3}$

4. All $\ge 0$ → optimal at $(\frac{8}{3},\frac{8}{3})$, $z=\frac{80}{3}\approx 26.67$.

Verify: $x_1=x_2=\frac{8}{3}$ ✓ matches the symmetry forecast. $x_1+2x_2=\frac{8}{3}+\frac{16}{3}=\frac{24}{3}=8$ ✓ tight; $2x_1+x_2=8$ ✓ tight. $z=5\cdot \frac{8}{3}+5\cdot \frac{8}{3}=\frac{80}{3}$ ✓. Had we broken the tie the other way, we'd land on the same corner — the lesson of Cell B.

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Example 3 — Cell C: a ratio-test tie (degeneracy)

Forecast: The three lines $x_1=2$, $x_2=2$, $x_1+x_2=4$ all pass through $(2,2)$. Three walls, one corner: the corner is "over-determined". Guess: the optimum is $(2,2)$ but the machine may take a "stalled" step where the answer doesn't improve.

1. Slacks: $x_1+s_1=2$, $x_2+s_2=2$, $x_1+x_2+s_3=4$. Initial tableau (start at $(0,0)$):

   Basis	$x_1$	$x_2$	$s_1$	$s_2$	$s_3$	RHS
   $s_1$	1	0	1	0	0	2
   $s_2$	0	1	0	1	0	2
   $s_3$	1	1	0	0	1	4
   $z$	$-1$	$-1$	0	0	0	0

2. $x_1$ enters ($z$-row tie $-1,-1$ → Bland's rule picks lowest index $x_1$). Ratios: $2/1=2$ (row 1), $x_2$-row has $0$ in $x_1$ column (skip), $4/1=4$ (row 3). Smallest $2$ → $s_1$ leaves. Why pivot on row-1 $x_1$-entry ($=1$)? To make $x_1$ basic and hop to the corner $(2,0)$:

   Basis	$x_1$	$x_2$	$s_1$	$s_2$	$s_3$	RHS
   $x_1$	1	0	1	0	0	2
   $s_2$	0	1	0	1	0	2
   $s_3$	0	1	$-1$	0	1	2
   $z$	0	$-1$	1	0	0	2

   Now $z=2$ at corner $(2,0)$ (where $x_2=0$ meets $x_1=2$).

3. $-1$ under $x_2$ → $x_2$ enters. Ratios: $2/1=2$ (row 2, $s_2$), and row 3 now gives $2/1=2$ too ($s_3$) — a tie. Why it matters: whichever we pick, the other basic variable will land at exactly $0$ — a degenerate basic feasible solution. Bland's rule breaks it: choose the leaving variable of lowest index, $s_2$ (index below $s_3$).

   Read the figure: the wall labelled "$x_1=2$", the wall labelled "$x_2=2$" and the line labelled "$x_1+x_2=4$" all pierce the single marked point $(2,2)$ (each line is drawn with a different dash pattern and captioned in words). Three walls, one corner — that visual over-crowding is exactly what "degenerate" means, and it is why two rows tied in the ratio test.

4. $s_2$ leaves (pivot on row-2 $x_2$-entry to make $x_2$ basic). We reach $(2,2)$ with $z=4$, but $s_3=0$ sits in the basis at value 0. Why this is the whole point of Cell C: the corner $(2,2)$ can be described by two different bases; the simplex might swap between them with zero improvement (a stall). Bland's lowest-index rule is exactly what stops that swap from looping forever.

5. Check the $z$-row after that pivot: all $\ge 0$ → optimal, $z=4$.

Verify: $(2,2)$: $x_1=2$ ✓ tight, $x_2=2$ ✓ tight, $x_1+x_2=4$ ✓ tight — three tight constraints at one corner is exactly what "degenerate" means. $z=2+2=4$ ✓.

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Example 4 — Cell D: unbounded objective (no maximum)

Forecast: Look at the constraints — both let $x_1$ grow enormously as long as $x_2$ trails behind. Guess: this region is an open wedge and $z$ can be pushed to infinity.

Read the figure first. The two constraint lines (one solid, one dashed, each captioned in words) bound a hatched region that does not close up — it flares open to the upper-right forever. The big arrow marked "$z$ increases" rides that open channel: as you slide along it, both $x_1$ and $x_2$ grow without limit, so $z=x_1+x_2$ has no ceiling. A closed polygon would trap the arrow at a corner; this open wedge never does.

1. Slacks: $-x_1+x_2+s_1=1$, $x_1-2x_2+s_2=2$. Initial $z$-row $-1,-1$.

2. $x_1$ enters ($z$-row tie $-1,-1$ → Bland's rule picks $x_1$). Ratio test needs positive column entries: column $x_1$ is $(-1,1)$. Only row 2's $1$ is positive → ratio $2/1=2$. $s_2$ leaves. Why pivot on that $1$? To unit-vector the $x_1$ column and make $x_1$ basic — the usual corner-hop:

   Basis	$x_1$	$x_2$	$s_1$	$s_2$	RHS
   $s_1$	0	$-1$	1	1	3
   $x_1$	1	$-2$	0	1	2
   $z$	0	$-3$	0	1	2

3. $-3$ under $x_2$ → $x_2$ enters. Column $x_2$ is $(-1,-2)$ — both negative! Why this is the signal: the ratio test only accepts positive entries, and there are none. That means we can raise $x_2$ forever without any constraint stopping us, and $z$ rises without bound.

4. Conclusion: the LP is UNBOUNDED. There is no finite maximum.

Verify: Set $x_2=t$ and follow the basis: $x_1=2+2t$, $s_1=3+t$ (all stay $\ge 0$ for every $t\ge 0$). Then $z=x_1+x_2=(2+2t)+t=2+3t\to \infty$ as $t\to \infty$ ✓. The forecast (open wedge) holds.

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Example 5 — Cell E: multiple optima (a whole optimal edge)

Forecast: The objective $z=2x_1+2x_2=2(x_1+x_2)$ is maximized by making $x_1+x_2$ as big as possible — but $x_1+x_2\le 4$ caps it at $4$. Guess: every point on the segment $x_1+x_2=4$ inside the region gives $z=8$. Not one corner — a whole edge.

Read the figure first. The solid line captioned "$x_1+x_2=4$" is parallel to the dashed "objective level lines" (lines of constant $z$). Because they are parallel, the objective doesn't touch the region at a single corner — it lies flat along the whole bold edge from $(3,1)$ to $(0,4)$. Every point on that bold edge scores the same $z=8$; that is what "multiple optima" looks like.

1. Slacks: $x_1+x_2+s_1=4$, $x_1+s_2=3$. $z$-row $-2,-2$.

2. $x_1$ enters ($z$-row tie $-2,-2$ → Bland's rule picks $x_1$). Ratios $4/1=4$, $3/1=3$ → $s_2$ leaves. Why pivot on the $x_1$-entry $=1$ in row 2? Unit-vector the column, make $x_1$ basic, hop to corner $(3,0)$:

   Basis	$x_1$	$x_2$	$s_1$	$s_2$	RHS
   $s_1$	0	1	1	$-1$	1
   $x_1$	1	0	0	1	3
   $z$	0	$-2$	0	2	6

3. $-2$ under $x_2$ → $x_2$ enters. Ratio $1/1=1$ → $s_1$ leaves. Why pivot on the $x_2$-entry $=1$ in row 1? Same rule; hop to corner $(3,1)$:

   Basis	$x_1$	$x_2$	$s_1$	$s_2$	RHS
   $x_2$	0	1	1	$-1$	1
   $x_1$	1	0	0	1	3
   $z$	0	0	2	0	8

4. All $z$-row $\ge 0$ → optimal, $z=8$ at $(3,1)$. But look at $s_2$: its $z$-row coefficient is $0$ and $s_2$ is non-basic. Why this signals multiple optima: a $0$ under a non-basic variable means we could bring it into the basis and $z$ would not change. Doing so slides us to the other corner $(0,4)$ — same $z=8$, exactly the bold edge in the figure.

Verify: Corner $(3,1)$: $z=2(3)+2(1)=8$ ✓. Corner $(0,4)$: $z=2(0)+2(4)=8$ ✓. Midpoint $(1.5,2.5)$ on the edge: $z=2(1.5)+2(2.5)=8$ ✓ — the entire segment is optimal, exactly as forecast.

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Example 6 — Cell F: degenerate start ($b=0$ constraint)

Forecast: The first constraint $x_1\le x_2$ with $b=0$ passes through the origin. So the starting corner $(0,0)$ already sits on that wall — the very first tableau is degenerate. Guess: the first pivot may look "stuck" (no visible improvement) before things move.

1. Slacks: $x_1-x_2+s_1=0$, $x_1+x_2+s_2=4$. Start $(0,0)$: $s_1=0$, $s_2=4$. Note $s_1=0$ already — degenerate start (the origin already lies on the wall $x_1-x_2=0$).

2. $x_1$ enters ($-3$ most negative). Ratios: row 1 $0/1=0$, row 2 $4/1=4$. Smallest is $0$ → $s_1$ leaves. Why the zero ratio is fine: it means $x_1$ can increase by $0$ this step — a degenerate pivot that swaps the basis without moving the corner.

3. Why pivot on row-1 $x_1$-entry ($=1$)? To make $x_1$ basic even though we won't physically move — the basis-swap is what lets the next step escape the wall:

   Basis	$x_1$	$x_2$	$s_1$	$s_2$	RHS
   $x_1$	1	$-1$	1	0	0
   $s_2$	0	2	$-1$	1	4
   $z$	0	$-5$	3	0	0

   Still $(0,0)$, $z=0$ — the stall the forecast predicted.

4. $-5$ under $x_2$ → $x_2$ enters. Ratios: row 1 has $-1$ (skip, negative), row 2 $4/2=2$ → $s_2$ leaves. Why pivot on the $2$? Unit-vector the $x_2$ column, make $x_2$ basic, finally hop off the wall to a new corner:

   Basis	$x_1$	$x_2$	$s_1$	$s_2$	RHS
   $x_1$	1	0	$\frac{1}{2}$	$\frac{1}{2}$	2
   $x_2$	0	1	$-\frac{1}{2}$	$\frac{1}{2}$	2
   $z$	0	0	$\frac{1}{2}$	$\frac{5}{2}$	10

5. All $\ge 0$ → optimal, $(2,2)$, $z=10$.

Verify: $x_1-x_2=0\le 0$ ✓ tight; $x_1+x_2=4\le 4$ ✓ tight; $z=3(2)+2(2)=10$ ✓. The degenerate first pivot cost us one "empty" step but reached the right answer — matching the forecast.

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Example 7 — Cells G + H: a $\ge$ constraint via the two-phase simplex

Forecast: Minimizing a positive-cost objective pushes both variables down toward $0$; the $\ge 2$ constraint stops them. Guess the cheapest corner is where $x_1+x_2=2$ meets an axis.

1. Turn minimization into maximization (Cell H). Why? The machine maximizes. Min $z$ = $-\max (-z)$, so maximize $w=-2x_1-3x_2$.

2. Convert the $\ge$ properly (Cell G) — surplus + artificial. Why not just multiply by $-1$? That would give a negative RHS, so the origin is infeasible and plain simplex cannot start. Instead, for $x_1+x_2\ge 2$ we subtract a surplus variable $s_1\ge 0$ (the amount by which we exceed $2$) and add an artificial variable $a_1\ge 0$ (a temporary crutch that gives us a legal starting basis): $x_1+x_2-s_1+a_1=2,x_1+2x_2+s_2=6.$ Start with basis $\{a_1,s_2\}$: $a_1=2,s_2=6\ge 0$ — legal.

3. Phase 1 — drive the artificial to zero. Why? $a_1>0$ means we're cheating (the real constraint isn't yet satisfied). Minimize $a_1$, i.e. maximize $w_1=-a_1$. Express $w_1$ in non-basic terms using $a_1=2-x_1-x_2+s_1$, so $w_1=-2+x_1+x_2-s_1$; the $w_1$-row is $-1,-1,+1$ under $x_1,x_2,s_1$.

   Basis	$x_1$	$x_2$	$s_1$	$a_1$	$s_2$	RHS
   $a_1$	1	1	$-1$	1	0	2
   $s_2$	1	2	0	0	1	6
   $w_1$	$-1$	$-1$	1	0	0	$-2$

   $x_1$ enters ($z$-row tie $-1,-1$ → Bland's rule picks $x_1$). Ratios $2/1=2$, $6/1=6$ → $a_1$ leaves. Why pivot on the $a_1$-row $x_1$-entry ($=1$)? To make $x_1$ basic and kick the artificial out:

   Basis	$x_1$	$x_2$	$s_1$	$a_1$	$s_2$	RHS
   $x_1$	1	1	$-1$	1	0	2
   $s_2$	0	1	1	$-1$	1	4
   $w_1$	0	0	0	1	0	0

   $w_1=0$ with $a_1$ non-basic → Phase 1 done, artificial is gone, and $(x_1,x_2)=(2,0)$ is a legal starting corner.

4. Phase 2 — now optimize the real objective $w=-2x_1-3x_2$. Drop the $a_1$ column. Express $w$ in non-basic terms: with $x_1=2-x_2+s_1$, $w=-2(2-x_2+s_1)-3x_2=-4-x_2-2s_1$, so the $w$-row is $-1$ under $x_2$ and $-2$ under $s_1$:

   Basis	$x_1$	$x_2$	$s_1$	$s_2$	RHS
   $x_1$	1	1	$-1$	0	2
   $s_2$	0	1	1	1	4
   $w$	0	$-1$	$-2$	0	$-4$

   $s_1$ enters ($-2$, most negative). Ratio test uses positive entries in the $s_1$ column $(-1,1)$: only row 2's $1$ qualifies → ratio $4/1=4$ → $s_2$ leaves. Why pivot on that $1$? Unit-vector the $s_1$ column, make $s_1$ basic, hop to a new corner:

   Basis	$x_1$	$x_2$	$s_1$	$s_2$	RHS
   $x_1$	1	2	0	1	6
   $s_1$	0	1	1	1	4
   $w$	0	1	0	2	4

   This new corner is $(x_1,s_1)=(6,4)$, i.e. $(x_1,x_2)=(6,0)$, with $w$-row RHS $=4$, meaning $-w=4\Rightarrow w=-4$ — the same $w$ as the previous tableau. So the $s_1$ pivot did not improve $w$: both tableaux record $w=-4$. The stored optimum is $w=-4$ at corner $(2,0)$ (the pre-pivot tableau was already optimal). Hence the true optimum is $w=-4$, i.e. $z=-w=4$, at $(2,0)$.

Verify: Feasibility of $(2,0)$: $2+0=2\ge 2$ ✓ tight, $2+0=2\le 6$ ✓. Cross-check every corner of $\{x_1+x_2\ge 2,x_1+2x_2\le 6,x\ge 0\}$: $(2,0)\to z=4$, $(0,2)\to 6$, $(6,0)\to 12$, $(0,3)\to 9$; the two lines cross at $x_1=-2<0$ (discard). Minimum $z=\min \{4,6,12,9\}=4$ at $(2,0)$ ✓ — matching the two-phase result and the forecast (a corner on an axis).

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Example 8 — Cell I: a real-world word problem (with units)

Forecast: Tables earn more per unit but hog carpentry. Guess a mix, not "all tables".

1. Model it (the hard part of word problems). Why define units first? So the constraints balance. Let $x_1=$ chairs/day, $x_2=$ tables/day (both dimensionless counts).

   \underbrace{2x_1+1x_2\le 8}_{\text{polishing hr}},\quad x_1,x_2\ge0.$$ Objective (dollars/day): $z=20x_1+30x_2$.

2. Slacks $s_1$ (spare carpentry hr), $s_2$ (spare polishing hr). $z$-row $-20,-30$.

3. $x_2$ enters ($-30$ most negative). Ratios $12/3=4$, $8/1=8$ → $s_1$ leaves. Why pivot on the $3$? Unit-vector the $x_2$ column, make $x_2$ basic, hop to the next corner:

   Basis	$x_1$	$x_2$	$s_1$	$s_2$	RHS
   $x_2$	$\frac{1}{3}$	1	$\frac{1}{3}$	0	4
   $s_2$	$\frac{5}{3}$	0	$-\frac{1}{3}$	1	4
   $z$	$-10$	0	10	0	120

4. $-10$ under $x_1$ → $x_1$ enters. Ratios $4/\frac{1}{3}=12$, $4/\frac{5}{3}=\frac{12}{5}$ → $s_2$ leaves. Why pivot on the $\frac{5}{3}$? Same rule: make $x_1$ basic, hop to the final corner:

   Basis	$x_1$	$x_2$	$s_1$	$s_2$	RHS
   $x_2$	0	1	$\frac{2}{5}$	$-\frac{1}{5}$	$\frac{16}{5}$
   $x_1$	1	0	$-\frac{1}{5}$	$\frac{3}{5}$	$\frac{12}{5}$
   $z$	0	0	8	6	144

5. All $\ge 0$ → optimal: $x_1=\frac{12}{5}=2.4$ chairs, $x_2=\frac{16}{5}=3.2$ tables, profit z=\144$/day.

Verify (with units!): Carpentry used $=1\cdot \frac{12}{5}+3\cdot \frac{16}{5}=\frac{12+48}{5}=\frac{60}{5}=12$ hr $\le 12$ ✓ (all used, tight). Polishing $=2\cdot \frac{12}{5}+1\cdot \frac{16}{5}=\frac{24+16}{5}=\frac{40}{5}=8$ hr $\le 8$ ✓ (all used, tight). Both constraints tight ⇒ genuine corner. Profit $z=20\cdot \frac{12}{5}+30\cdot \frac{16}{5}=\frac{240+480}{5}=\frac{720}{5}=144$ ✓. (Sanity: solving the two tight equations $x_1+3x_2=12$, $2x_1+x_2=8$ directly gives the same $(\frac{12}{5},\frac{16}{5})$.)

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Example 9 — Cell J: the exam twist (sensitivity)

Forecast: If chairs