Solving 2×2 systems using Cramer's rule: Worked examples

If any symbol below feels unfamiliar, the machine itself is spelled out in the parent note; here we only use it, many times.

The scenario matrix

Every 2×2 system falls into exactly one of these "cells". Read this as a checklist — by the end, each row has a fully worked example tagged to it. (Recall $a, b, c$ are the pieces of $a x + b y = c$ defined just above.)

Cell	What makes it special	Example
A. Clean unique	$D \neq = 0$ , answers are whole numbers	Ex 1
B. Negative & fraction	$D \neq = 0$ but $x, y$ come out negative / fractional	Ex 2
C. Zero in a coefficient	some $a, b, c = 0$ — a slot on the board is blank	Ex 3
D. Zero unknown	one of $x, y$ equals $0$ (so $D_{x}$ or $D_{y} = 0$ , but $D \neq = 0$ )	Ex 4
E. Degenerate — no solution	$D = 0$ , some $D_{x}, D_{y} \neq = 0$ (parallel distinct lines)	Ex 5
F. Degenerate — infinitely many	$D = D_{x} = D_{y} = 0$ (same line)	Ex 6
G. Word problem	real quantities with units	Ex 7
H. Exam twist — literal / symbolic	letters instead of numbers	Ex 8

Related classification lives in Consistency of linear systems; the geometry of "same/parallel/crossing lines" is the same picture used there — and it is exactly the picture in the figure below.

Figure s01 draws the three fates side by side. On the left, two lines of different slope (blue and pink) cross at the single yellow dot — the unique-solution case ( $D \neq = 0$ ). In the middle, two blue/pink lines run parallel and never touch — no solution ( $D = 0$ , some numerator $\neq = 0$ ). On the right, the blue line and the dashed pink line lie exactly on top of each other — infinitely many solutions ( $D = D_{x} = D_{y} = 0$ ). Keep this picture in mind: every example below is really asking "which of these three am I looking at?"

Figure — Solving 2×2 systems using Cramer's rule

Cell A — Clean unique solution

Worked example Ex 1 · whole-number answer

$5 x + 2 y = 16, 3 x - y = 5$ (This is eq.(1) on the left, eq.(2) on the right.)

Forecast: Rewrite each line as $y = slope \cdot x + \dots$ : eq.(1) gives $y = \frac{16 - 5 x}{2}$ , slope $- \frac{5}{2}$ ; eq.(2) gives $y = 3 x - 5$ , slope $3$ . Different slopes ⇒ they cross once ⇒ expect a single tidy answer. Guess: will $D$ be positive or negative?

Compute $D$ . $D = 53 2 - 1 = (5) (- 1) - (2) (3) = - 5 - 6 = - 11.$ Why this step? Always $D$ first: if it were $0$ , Cramer would not apply and we'd stop.
Compute $D_{x}$ — replace column 1 (the $x$ -column) with the constants $[165]$ . $D_{x} = 165 2 - 1 = (16) (- 1) - (2) (5) = - 16 - 10 = - 26.$ Why this step? $x$ owns column 1, so we swap column 1 to isolate $x$ .
Compute $D_{y}$ — replace column 2. $D_{y} = 53165 = (5) (5) - (16) (3) = 25 - 48 = - 23.$ Why this step? $y$ owns column 2, so to isolate $y$ we swap that column (not column 1) for the constants — mirror image of step 2.
Divide. $x = \frac{- 26}{- 11} = \frac{26}{11}, y = \frac{- 23}{- 11} = \frac{23}{11} .$ Why division comes last? Steps 1–3 only measured "spread" numbers; the actual coordinate is a ratio of two spreads. Dividing the swapped-column determinant by $D$ rescales it into the real answer — that final division is where the geometry turns back into numbers.

Verify: Plug into eq.(2): $3 \cdot \frac{26}{11} - \frac{23}{11} = \frac{78 - 23}{11} = \frac{55}{11} = 5.$ ✓ (Forecast $D < 0$ was correct.)

Cell B — Negative and fractional answers

Worked example Ex 2 · watch every sign

$- 2 x + 4 y = 2, 3 x + y = - 5$ (eq.(1) top, eq.(2) bottom.)

Forecast: The constants pull downward/leftward — expect $x$ negative. Don't let the leading minus scare you; it's just $a_{1} = - 2$ (the $a$ of eq.(1)).

$D$ . $- 2 3 41 = (- 2) (1) - (4) (3) = - 2 - 12 = - 14.$ Why: main-diagonal $(- 2) (1)$ minus anti-diagonal $(4) (3)$ — the order matters, in that order every time.
$D_{x}$ (col 1 → $[2 - 5]$ ): $2 - 5 41 = (2) (1) - (4) (- 5) = 2 + 20 = 22.$ Why: $(4) (- 5) = - 20$ , and subtracting $- 20$ adds $20$ — the classic sign trap.
$D_{y}$ (col 2 → constants): $- 2 3 2 - 5 = (- 2) (- 5) - (2) (3) = 10 - 6 = 4.$ Why this step? $y$ lives in column 2, so we swap column 2 (leaving column 1 as the original $x$ -coefficients) to build $y$ 's numerator.
Divide. $x = \frac{22}{- 14} = - \frac{11}{7}, y = \frac{4}{- 14} = - \frac{2}{7} .$ Why division last? The negative $D = - 14$ flips the sign of both positive numerators, which is how the answers turn out negative — you cannot know the sign of $x, y$ until this last ratio is taken.

Verify: eq.(1): $- 2 (- \frac{11}{7}) + 4 (- \frac{2}{7}) = \frac{22 - 8}{7} = \frac{14}{7} = 2.$ ✓

Cell C — Zero coefficients (blank slots on the board)

This cell has three flavours of blank slot: an $a = 0$ (a variable missing), a $b = 0$ (the other variable missing), and a $c = 0$ (a zero constant). We show all three so no arrangement of zeros is ever new.

Worked example Ex 3a ·

$a = 0$ : a variable missing from an equation $0 \cdot x + 3 y = 9, 2 x - y = 1$ (eq.(1) is $3 y = 9$ ; eq.(2) is $2 x - y = 1$ .) A zero coefficient is totally allowed — it just puts a $0$ into the determinant.

Forecast: From eq.(1) alone, $y = 3$ . So expect $y = 3$ and Cramer had better agree.

$D$ . $02 3 - 1 = (0) (- 1) - (3) (2) = 0 - 6 = - 6.$ Why: the $0$ zaps the main-diagonal term, leaving only the anti-diagonal. Still $D \neq = 0$ , so unique solution exists.
$D_{x}$ (col 1 → $[91]$ ): $91 3 - 1 = (9) (- 1) - (3) (1) = - 9 - 3 = - 12.$ Why this step? $x$ owns column 1, so we swap that column for the constants to build $x$ 's numerator.
$D_{y}$ : $0291 = (0) (1) - (9) (2) = - 18.$ Why this step? We keep column 1 (the original $x$ -column, here $[02]$ ) and swap only column 2 for the constants — that is what makes this determinant the numerator for $y$ , not $x$ .
Divide. $x = \frac{- 12}{- 6} = 2, y = \frac{- 18}{- 6} = 3.$ Why division last? Only after dividing by $D = - 6$ does $D_{y} = - 18$ collapse to $y = 3$ — matching the forecast we made straight from eq.(1). The determinants alone weren't the answer; the ratio is.

Verify: $y = 3$ matches the forecast; eq.(2): $2 (2) - 3 = 1.$ ✓

Worked example Ex 3b ·

$b = 0$ and a $c = 0$ together $2 x + 0 \cdot y = 6, x + 4 y = 0$ (eq.(1) is $2 x = 6$ , so here $b_{1} = 0$ ; eq.(2) has constant $c_{2} = 0$ .) Two kinds of zero at once.

Forecast: From eq.(1) alone, $x = 3$ . Then eq.(2) gives $3 + 4 y = 0 \Rightarrow y = - \frac{3}{4}$ . Expect $x = 3, y = - \frac{3}{4}$ .

$D$ . $2104 = (2) (4) - (0) (1) = 8 - 0 = 8.$ Why: the $b_{1} = 0$ kills the anti-diagonal term this time; $D = 8 \neq = 0$ , unique solution.
$D_{x}$ (col 1 → $[60]$ ): $6004 = (6) (4) - (0) (0) = 24.$ Why this step? $x$ owns column 1, so we swap that column for the constants (the zero constant $c_{2} = 0$ simply sits in the bottom slot).
$D_{y}$ (col 2 → constants): $2160 = (2) (0) - (6) (1) = - 6.$ Why this step? $y$ owns column 2, so we keep column 1 as the original $x$ -coefficients $[21]$ and swap only column 2 — the zero constant now lands in the top slot.
Divide. $x = \frac{24}{8} = 3, y = \frac{- 6}{8} = - \frac{3}{4} .$ Why division last? The bare determinants $24$ and $- 6$ are not the answer; only dividing by $D = 8$ rescales them into $x = 3, y = - \frac{3}{4}$ , exactly the forecast.

Verify: eq.(1): $2 (3) = 6$ ✓; eq.(2): $3 + 4 (- \frac{3}{4}) = 3 - 3 = 0$ ✓.

Cell D — A zero unknown (answer itself is zero)

Worked example Ex 4 ·

$x = 0$ , but $D \neq = 0$ $4 x + 5 y = 15, 2 x - 3 y = - 9$ (eq.(1) top, eq.(2) bottom.)

Forecast: These constants might make one unknown vanish. Note the trap from the parent note: " $D_{x} = 0$ " does not mean $D = 0$ — it can genuinely mean $x = 0$ .

$D$ . $42 5 - 3 = (4) (- 3) - (5) (2) = - 12 - 10 = - 22.$ Why: $D \neq = 0$ , so there is a unique answer — whatever comes out is trustworthy.
$D_{x}$ (col 1 → constants): $15 - 9 5 - 3 = (15) (- 3) - (5) (- 9) = - 45 + 45 = 0.$ Why this is fine: $D_{x} = 0$ while $D = - 22 \neq = 0$ gives $x = 0/ (- 22) = 0$ . A perfectly valid zero answer, not a breakdown.
$D_{y}$ : $42 15 - 9 = (4) (- 9) - (15) (2) = - 36 - 30 = - 66.$ Why this step? Column 2 (the $y$ -column) is swapped for the constants while column 1 stays as the original $x$ -coefficients — that placement is exactly what makes this the numerator for $y$ .
Divide. $x = \frac{0}{- 22} = 0, y = \frac{- 66}{- 22} = 3.$ Why division last? This is where the difference between " $D_{x} = 0$ " and " $x = 0$ " becomes visible: only by dividing $0$ by the nonzero $D$ do we legally obtain $x = 0$ . Without the final division, $D_{x} = 0$ alone would be meaningless.

Verify: eq.(1): $4 (0) + 5 (3) = 15.$ ✓ Compare with the consistency view — a unique point $(0, 3)$ sits on both lines.

Cell E — Degenerate: no solution (parallel distinct lines)

Worked example Ex 5 ·

$D = 0$ but $D_{x} \neq = 0$ $2 x - 6 y = 4, - x + 3 y = 5$ (eq.(1) top, eq.(2) bottom.)

Forecast: Slopes: eq.(1) is $y = \frac{x - 2}{3}$ , eq.(2) is $y = \frac{x + 5}{3}$ — same slope $\frac{1}{3}$ , different intercepts. Parallel and distinct ⇒ they never meet ⇒ no solution. Watch $D$ hit $0$ .

$D$ . $2 - 1 - 6 3 = (2) (3) - (- 6) (- 1) = 6 - 6 = 0.$ Why: rows point the same direction ( $(- 1, 3) = - \frac{1}{2} (2, - 6)$ ), so the parallelogram they span has zero area — that's what $D = 0$ is. This is the middle panel of Figure s01: two parallel lines that never cross.
Since $D = 0$ , don't stop — test $D_{x}$ . $45 - 6 3 = (4) (3) - (- 6) (5) = 12 + 30 = 42 \neq = 0.$ Why this step: $D = 0$ with any nonzero numerator means the ratio $D_{x} / D$ is $\frac{42}{0}$ — undefined. No point satisfies both. (Here the division itself is the diagnosis: it refuses to happen, which is exactly "no solution".)
Conclude: $D = 0$ and $D_{x} \neq = 0$ ⇒ no solution.

Verify: Multiply eq.(2) by $- 2$ : $2 x - 6 y = - 10$ , but eq.(1) says $2 x - 6 y = 4$ . Since $- 10 \neq = 4$ , the equations contradict — confirmed inconsistent. ✓

Cell F — Degenerate: infinitely many (same line)

Worked example Ex 6 ·

$D = D_{x} = D_{y} = 0$ $3 x + 6 y = 9, x + 2 y = 3$ (eq.(1) top, eq.(2) bottom.)

Forecast: eq.(1) is exactly $3 \times$ eq.(2). Same line drawn twice ⇒ every point on it works ⇒ infinitely many solutions. Expect all three determinants to be $0$ .

$D$ . $3162 = (3) (2) - (6) (1) = 6 - 6 = 0.$
$D_{x}$ . $9362 = (9) (2) - (6) (3) = 18 - 18 = 0.$
$D_{y}$ . $3193 = (3) (3) - (9) (1) = 9 - 9 = 0.$ Why this step? We still swap only column 2 for $y$ 's numerator; checking it (and not just $D_{x}$ ) matters because both numerators must vanish for the lines to truly coincide. This is the right-hand panel of Figure s01: one line drawn twice.
Describe the solution set. From $x + 2 y = 3$ : $x = 3 - 2 y$ , with $y$ free. Solutions: $(3 - 2 t, t)$ for any real $t$ . Why no division here? Every ratio would be $0/0$ — the machine's final division is undefined, which is precisely its way of announcing "not one answer but infinitely many". So instead of dividing we describe the whole line.

Verify: Put $t = 1$ : $(1, 1)$ . eq.(1): $3 (1) + 6 (1) = 9$ ✓; eq.(2): $1 + 2 = 3$ ✓. Put $t = 0$ : $(3, 0)$ also works — infinitely many, as forecast.

Cell G — A word problem (with units)

Worked example Ex 7 · tickets and money

A cinema sells adult tickets at ₹120 and child tickets at ₹80. One evening it sold 50 tickets for a total of ₹5200. How many of each? Let $x =$ adult tickets, $y =$ child tickets. $x + y = 50 (count, eq.(1)), 120 x + 80 y = 5200 (rupees, eq.(2))$

Forecast: If all 50 were children we'd collect ₹4000; we collected ₹1200 more. Each adult adds ₹40 over a child, so ≈ $1200/40 = 30$ adults. Guess $x = 30, y = 20$ ; let Cramer confirm.

$D$ . $1120180 = (1) (80) - (1) (120) = 80 - 120 = - 40.$ Why: $D \neq = 0$ ⇒ a unique ticket split exists (physically it must).
$D_{x}$ (col 1 → $[505200]$ ): $505200180 = (50) (80) - (1) (5200) = 4000 - 5200 = - 1200.$
$D_{y}$ : $1120505200 = (1) (5200) - (50) (120) = 5200 - 6000 = - 800.$ Why this step? $y$ counts child tickets and its coefficients sit in column 2, so we swap column 2 (keeping the original $x$ -column) to build $y$ 's numerator.
Divide. $x = \frac{- 1200}{- 40} = 30$ adults, $y = \frac{- 800}{- 40} = 20$ children. Why division last? The determinants carry the wrong scale (they are "spread" areas, not counts); dividing by $D = - 40$ converts them into actual ticket numbers with the right units. That final ratio is the only step that produces a countable answer.

Verify (units too): tickets $30 + 20 = 50$ ✓; money $120 (30) + 80 (20) = 3600 + 1600 = 5200$ ₹ ✓. Matches the forecast of 30 adults.

Cell H — Exam twist: solve with letters

Worked example Ex 8 · symbolic / literal coefficients

Solve for $x, y$ in terms of the constant $k$ (with $k \neq = 3$ ): $x + y = 2, k x + 3 y = 6$ (eq.(1) top, eq.(2) bottom.)

Forecast: Answers will be formulas in $k$ , not numbers. Because we divide by $D$ , whatever makes $D = 0$ will be a "forbidden" value — spot it early.

$D$ . $1 k 13 = (1) (3) - (1) (k) = 3 - k .$ Why: $D = 3 - k$ , so $D = 0$ exactly at $k = 3$ — that's the excluded value the problem warned about (the lines coincide there). Every other $k$ , including $k = 0$ , is perfectly fine.
$D_{x}$ (col 1 → $[26]$ ): $2613 = (2) (3) - (1) (6) = 6 - 6 = 0.$ Why this step? $x$ owns column 1, so we swap that column for the constants to build $x$ 's numerator.
$D_{y}$ (col 2 → constants): $1 k 26 = (1) (6) - (2) (k) = 6 - 2 k .$ Why this step? Column 2 (the $y$ -column, here $[13]$ ) is swapped for the constants while column 1 keeps the $x$ -coefficients $[1 k]$ — the placement that makes this $y$ 's numerator, keeping the $k$ where it belongs.
Divide. $x = \frac{D _{x}}{D} = \frac{0}{3 - k} = 0$ (for any allowed $k$ ), and $y = \frac{D _{y}}{D} = \frac{6 - 2 k}{3 - k} = \frac{2 ( 3 - k )}{3 - k} = 2.$ Why division last? Only after dividing do the $k$ 's cancel: $\frac{2 ( 3 - k )}{3 - k} = 2$ . Before dividing, both the numerator $6 - 2 k$ and $D = 3 - k$ still depend on $k$ ; the ratio is what reveals the surprising $k$ -independence — and why $k = 3$ (making $D = 0$ ) must be excluded, since you cannot divide by $0$ .

Verify: $(x, y) = (0, 2)$ independent of $k$ . eq.(1): $0 + 2 = 2$ ✓; eq.(2): $k (0) + 3 (2) = 6$ ✓ for every $k$ . Nice surprise — the answer doesn't depend on $k$ , yet $k = 3$ still breaks $D$ (there the two equations become the identical line $x + y = 2$ , so infinitely many solutions instead of a unique one).

Recall check

Recall Which cell is each situation?

$D \neq = 0$ , answer whole numbers ::: Cell A — clean unique. $D \neq = 0$ but $D_{x} = 0$ ::: Cell D — the unknown $x$ is genuinely $0$ , still a unique solution. $D = 0$ and $D_{x} \neq = 0$ ::: Cell E — no solution, parallel distinct lines. $D = D_{x} = D_{y} = 0$ ::: Cell F — infinitely many, same line. A coefficient is $0$ (e.g. $0 \cdot x + 3 y = 9$ ) ::: Cell C — allowed; just a $0$ enters the determinant, $D$ can still be nonzero. Coefficients are letters ::: Cell H — answers are formulas; the value making $D = 0$ is excluded.

Connections

Solving 2×2 systems using Cramer's rule — the parent machine these examples exercise.
Determinant of a 2×2 matrix — the criss-cross number computed in every step.
Consistency of linear systems — Cells E and F are the "inconsistent" and "dependent" classes.
Matrix form of linear equations Ax=b — every example above is one $A x = c$ .
Cramer's rule for 3×3 systems — same case-checklist, one more column to swap.