Simultaneous equations — substitution, elimination, cross-multiplication: Worked examples

This page is a shooting range. The parent note gave you three methods (substitution, elimination, cross-multiplication). Here we fire every kind of problem the topic can throw at you — clean cases, ugly signs, zero coefficients, the two degenerate cases (parallel lines, identical lines), a real-world word problem, and an exam twist — and solve each one fully.

If you have not yet met the three methods, read the parent first: the topic note and the prerequisites Linear Equations in Two Variables and Graphical Method for Simultaneous Equations.

Before anything, let me define the one quantity that decides which scenario you are in.

Why do we care about $D$ before solving? Because $D$ is the denominator in every formula on this topic. And you can never divide by zero. So the value of $D$ splits every problem into three worlds, shown below.

Figure — Simultaneous equations — substitution, elimination, cross-multiplication

$D \neq = 0$ → lines cross once → one unique solution.
$D = 0$ and the constants don't match up → lines are parallel → no solution.
$D = 0$ and the constants do match up → lines are the same line → infinitely many solutions.

Where the cross-multiplication formulas come from

The next section leans on two formulas. Rather than drop them in, let me derive them once here so you trust them everywhere below.

Start from the general system and eliminate $y$ first. Multiply the top by $b_{2}$ and the bottom by $b_{1}$ : $a_{1} b_{2} x + b_{1} b_{2} y = c_{1} b_{2}, a_{2} b_{1} x + b_{1} b_{2} y = c_{2} b_{1} .$ Why this step? Both now carry the same $y$ -term $b_{1} b_{2} y$ , so subtracting kills $y$ : $(a_{1} b_{2} - a_{2} b_{1}) x = c_{1} b_{2} - c_{2} b_{1} \Rightarrow x = \frac{c _{1} b _{2} - c _{2} b _{1}}{a _{1} b _{2} - a _{2} b _{1}} = \frac{b _{1} c _{2} - b _{2} c _{1}}{D} .$ (The last equality just multiplies top and bottom by $- 1$ to match the standard textbook layout.)

Now eliminate $x$ instead. Multiply the top by $a_{2}$ , the bottom by $a_{1}$ , subtract: $(a_{2} b_{1} - a_{1} b_{2}) y = a_{2} c_{1} - a_{1} c_{2} \Rightarrow y = \frac{a _{2} c _{1} - a _{1} c _{2}}{a _{2} b _{1} - a _{1} b _{2}} = \frac{c _{1} a _{2} - c _{2} a _{1}}{D} .$

The scenario matrix

Every problem on this topic lands in exactly one cell below. The 9 worked examples that follow are labelled with the cell they hit, so together they cover the whole grid.

Cell	Scenario class	What's tricky	Example
A	All positive, clean integer answer	nothing — the warm-up	Ex 1
B	Negative coefficients / mixed signs	sign bookkeeping in cross-mult	Ex 2
C1	A zero $y$ -coefficient (no $y$ in one equation)	equation becomes a vertical line	Ex 3
C2	A zero $x$ -coefficient (no $x$ in one equation)	equation becomes a horizontal line	Ex 4
D	Fractional / non-integer answer	keep fractions, don't round	Ex 5
E	Degenerate: no solution ( $D = 0$ , parallel)	detecting inconsistency	Ex 6
F	Degenerate: infinite solutions ( $D = 0$ , identical)	detecting dependency	Ex 7
G	Word problem (real-world, with units)	translating English → equations	Ex 8
H	Exam twist: equations in $\frac{1}{x}, \frac{1}{y}$ (non-linear disguise)	substitution to make it linear	Ex 9

Example 1 — Cell A: clean, all positive

Forecast: guess before reading. The numbers are friendly — do you expect whole-number answers? Jot a guess.

Isolate $y$ from the second equation. $4 x - y = 5 \Rightarrow y = 4 x - 5$ Why this step? The coefficient of $y$ there is $- 1$ , the easiest thing to isolate — no fractions created. That's the whole art of substitution: pick the variable that's already almost alone.
Substitute into the first equation. $2 x + 3 (4 x - 5) = 13$ Why this step? Every $y$ becomes something made only of $x$ , so the equation now has one unknown.
Expand and collect. $2 x + 12 x - 15 = 13 \Rightarrow 14 x = 28 \Rightarrow x = 2$ Why this step? Distribute the $3$ , gather $x$ -terms, move the constant across.
Back-substitute into $y = 4 x - 5$ : $y = 4 (2) - 5 = 3$ Why this step? We built $y$ as a formula in $x$ in step 1; now feed the found $x$ into it.

Verify: plug $(2, 3)$ into both originals. $2 (2) + 3 (3) = 4 + 9 = 13$ ✓ and $4 (2) - 3 = 5$ ✓. Both true → genuine intersection point.

Example 2 — Cell B: negative coefficients, cross-multiplication

Forecast: with a $+ 3$ and a $- 4$ floating around, the sign of the answer is easy to get wrong. Predict: will $x$ be positive or negative?

First line up the labels carefully — this is where errors are born: $a_{1} = 2, b_{1} = 3, c_{1} = 8, a_{2} = 5, b_{2} = - 4, c_{2} = 3.$

Compute the determinant $D$ . $D = a_{1} b_{2} - a_{2} b_{1} = (2) (- 4) - (5) (3) = - 8 - 15 = - 23$ Why this step? $D = - 23 \neq = 0$ so we are safely in the "one solution" world — division is allowed.
Compute $x$ using the derived formula $x = \frac{b _{1} c _{2} - b _{2} c _{1}}{D}$ : $x = \frac{( 3 ) ( 3 ) - ( - 4 ) ( 8 )}{- 23} = \frac{9 + 32}{- 23} = \frac{41}{- 23} = - \frac{41}{23}$ Why this step? This is the cross-multiplication formula we earned above; the numerator crosses the $b$ and $c$ columns.

Compute $y$ using $y = \frac{c _{1} a _{2} - c _{2} a _{1}}{D}$ : $y = \frac{( 8 ) ( 5 ) - ( 3 ) ( 2 )}{- 23} = \frac{40 - 6}{- 23} = \frac{34}{- 23} = - \frac{34}{23}$ Why this step? Same pattern, crossing the $c$ and $a$ columns.

Verify: substitute into equation 1: $2 (- \frac{41}{23}) + 3 (- \frac{34}{23}) = \frac{- 82 - 102}{23} = \frac{- 184}{23} = - 8.$ But the equation says it should equal $+ 8$ ! Both signs are flipped. That is the tell-tale sign of a $D$ -sign error. Because $D$ multiplies both numerators, if you accidentally solve with $- D$ you get the exact negative of the true answer.

Corrected root cause: the safest way is to redo directly and trust elimination to referee:

Multiply eqn 1 by $4$ : $8 x + 12 y = 32$ .
Multiply eqn 2 by $3$ : $15 x - 12 y = 9$ .
Add: $23 x = 41 \Rightarrow x = \frac{41}{23}$ (positive).
Then $3 y = 8 - 2 \cdot \frac{41}{23} = \frac{184 - 82}{23} = \frac{102}{23} \Rightarrow y = \frac{34}{23}$ (positive).

So the true answer is $x = \frac{41}{23}, y = \frac{34}{23}$ .

Verify (correct): $2 \cdot \frac{41}{23} + 3 \cdot \frac{34}{23} = \frac{82 + 102}{23} = \frac{184}{23} = 8$ ✓.

Example 3 — Cell C1: a missing $y$ -term

Forecast: the first equation has no $y$ at all (its $b_{1} = 0$ ). What does that do to the picture? Guess before reading.

Read the first equation geometrically. $3 x = 12 \Rightarrow x = 4$ Why this step? With $y$ absent, $x = 4$ is forced regardless of $y$ — this is a vertical line on the graph. See the figure: it pins $x$ instantly.

Substitute $x = 4$ into the second equation. $2 (4) + 5 y = 19 \Rightarrow 8 + 5 y = 19 \Rightarrow 5 y = 11 \Rightarrow y = \frac{11}{5}$ Why this step? The vertical line already fixed $x$ ; the second line then tells us the height $y$ where they cross.

Verify: $3 (4) = 12$ ✓; $2 (4) + 5 \cdot \frac{11}{5} = 8 + 11 = 19$ ✓.

Determinant check: $D = a_{1} b_{2} - a_{2} b_{1} = (3) (5) - (2) (0) = 15 \neq = 0$ , so one solution — consistent with a vertical line meeting a slanted line exactly once.

Example 4 — Cell C2: a missing $x$ -term (the mirror image)

Forecast: now the first variable is missing ( $a_{1} = 0$ ). By symmetry with Ex 3, which coordinate do you think gets pinned first?

Read the first equation geometrically. $4 y = 20 \Rightarrow y = 5$ Why this step? With $x$ absent, $y = 5$ holds no matter what $x$ is — this is a horizontal line. It's the exact mirror of Ex 3: there a vertical line fixed $x$ ; here a horizontal line fixes $y$ .
Substitute $y = 5$ into the second equation. $3 x + 2 (5) = 21 \Rightarrow 3 x + 10 = 21 \Rightarrow 3 x = 11 \Rightarrow x = \frac{11}{3}$ Why this step? The horizontal line fixed the height $y$ ; the slanted line then tells us at what $x$ they cross.

Verify: $4 (5) = 20$ ✓; $3 \cdot \frac{11}{3} + 2 (5) = 11 + 10 = 21$ ✓.

Determinant check: $D = a_{1} b_{2} - a_{2} b_{1} = (0) (2) - (3) (4) = - 12 \neq = 0$ , so one solution — a horizontal line and a slanted line cross exactly once.

Recall Zero-coefficient shortcut

Missing $y$ (i.e. $b = 0$ ) → ::: vertical line, $x$ is fixed immediately. Missing $x$ (i.e. $a = 0$ ) → ::: horizontal line, $y$ is fixed immediately.

Example 5 — Cell D: fractional answer, elimination

Forecast: will the answers be whole numbers or fractions? Predict the denominator you expect.

Compute $D$ first. $D = (3) (- 2) - (5) (4) = - 6 - 20 = - 26.$ Why this step? Non-zero → unique solution. Also, $∣ D ∣ = 26$ hints the denominators will involve $13$ after simplifying.
Eliminate $y$ . Multiply eqn 2 by $2$ : $10 x - 4 y = 8$ . Keep eqn 1: $3 x + 4 y = 10$ . Why this step? Now $y$ has coefficients $+ 4$ and $- 4$ — opposite, so adding kills it.
Add the two equations. $13 x = 18 \Rightarrow x = \frac{18}{13}$ Why this step? $+ 4 y + (- 4 y) = 0$ ; the $y$ -term vanishes, leaving one unknown.
Back-substitute into eqn 1: $3 \cdot \frac{18}{13} + 4 y = 10 \Rightarrow \frac{54}{13} + 4 y = 10 \Rightarrow 4 y = \frac{130 - 54}{13} = \frac{76}{13} \Rightarrow y = \frac{19}{13}$ Why this step? Isolate $y$ ; keep everything as fractions to stay exact.

Verify in eqn 2: $5 \cdot \frac{18}{13} - 2 \cdot \frac{19}{13} = \frac{90 - 38}{13} = \frac{52}{13} = 4$ ✓.

Example 6 — Cell E: no solution (parallel lines)

Forecast: notice the second equation looks like the first "doubled". Do you think a solution exists? Guess yes / no.

Compute $D$ . $D = (2) (6) - (4) (3) = 12 - 12 = 0$ Why this step? $D = 0$ means we cannot use any of the divide-by- $D$ formulas. It signals a degenerate case — parallel or identical lines. We must look closer.
Try to eliminate anyway. Multiply eqn 1 by $2$ : $4 x + 6 y = 10$ . Subtract this from eqn 2: $(4 x + 6 y) - (4 x + 6 y) = 20 - 10 \Rightarrow 0 = 10.$ Why this step? Both variables cancel. We're left with a statement about numbers only.
Read the contradiction. $0 = 10$ is false for every $x, y$ . No pair can satisfy both equations.

Verify (geometric): rewrite as slopes. $2 x + 3 y = 5 \Rightarrow y = - \frac{2}{3} x + \frac{5}{3}$ ; $4 x + 6 y = 20 \Rightarrow y = - \frac{2}{3} x + \frac{10}{3}$ . Same slope $- \frac{2}{3}$ , different intercepts → two distinct parallel lines that never meet.

Conclusion: no solution (inconsistent system).

Example 7 — Cell F: infinitely many solutions (identical lines)

Forecast: the second equation is exactly $3 \times$ the first. Same tell as Ex 6 — but is the outcome the same?

Compute $D$ . $D = (1) (- 6) - (3) (- 2) = - 6 + 6 = 0$ Why this step? Again $D = 0$ , degenerate. Parallel or identical — decide by testing the constants.
Eliminate. Multiply eqn 1 by $3$ : $3 x - 6 y = 9$ . Subtract eqn 2: $(3 x - 6 y) - (3 x - 6 y) = 9 - 9 \Rightarrow 0 = 0.$ Why this step? Both variables cancel and the leftover is $0 = 0$ , which is always true.
Read the identity. $0 = 0$ holds for every point on the line $x - 2 y = 3$ . So any $(x, y)$ on that line works.

Verify: pick two points on $x - 2 y = 3$ : $(3, 0)$ and $(1, - 1)$ . Check both in eqn 2. $3 (3) - 6 (0) = 9$ ✓ and $3 (1) - 6 (- 1) = 3 + 6 = 9$ ✓. Both satisfy — because it's literally the same line.

Conclusion: infinitely many solutions; the solution set is the whole line $x - 2 y = 3$ .

Recall The two degenerate cases, side by side

$D = 0$ and elimination gives $0 = k$ with $k \neq = 0$ ::: no solution (parallel) $D = 0$ and elimination gives $0 = 0$ ::: infinite solutions (identical line)

Example 8 — Cell G: word problem with units

Forecast: guess roughly — are there more adults or more children? (The average price ₹122 sits closer to ₹150, so lean which way?)

Name the unknowns and translate. Let $a$ = number of adult tickets, $c$ = number of child tickets. $a + c = 200 (tickets), 150 a + 80 c = 24400 (rupees)$ Why this step? One equation counts tickets, the other counts money. Matching units on each side is how you know the translation is right.
Substitute. From the first, $c = 200 - a$ . Put into the money equation: $150 a + 80 (200 - a) = 24400$ Why this step? Reduce to one unknown $a$ — this is substitution, chosen because $c$ is already isolated.
Expand and solve. $150 a + 16000 - 80 a = 24400 \Rightarrow 70 a = 8400 \Rightarrow a = 120$ Why this step? Collect $a$ -terms, move the constant across, divide by $70$ .
Find $c$ . $c = 200 - a = 200 - 120 = 80.$ Why this step? Feed the found $a$ back into the isolated expression for $c$ .

Answer: $a = 120$ adults, $c = 80$ children — and indeed $a > c$ , matching the forecast (average ₹122 leans toward the ₹150 side).

Verify (units): tickets $120 + 80 = 200$ ✓; money $150 (120) + 80 (80) = 18000 + 6400 = 24400$ ✓ (₹). Both counts are non-negative whole numbers → physically sensible.

Example 9 — Cell H: exam twist (non-linear disguise)

Forecast: these aren't linear in $x, y$ — the unknowns are downstairs. Can you spot a trick that makes them linear?

Rename the reciprocals. Let $u = \frac{1}{x}, v = \frac{1}{y}$ . Why this step? Then $\frac{2}{x} = 2 u$ , $\frac{3}{y} = 3 v$ , and the system becomes linear in $u, v$ : $2 u + 3 v = 13, 5 u - 4 v = - 2.$
Compute $D$ . $D = (2) (- 4) - (5) (3) = - 8 - 15 = - 23 \neq = 0.$ Why this step? Unique solution exists in $u, v$ ; safe to divide.
Eliminate $v$ . Multiply eqn 1 by $4$ : $8 u + 12 v = 52$ . Multiply eqn 2 by $3$ : $15 u - 12 v = - 6$ . Add: $23 u = 46 \Rightarrow u = 2$ Why this step? Opposite $v$ -coefficients cancel on adding.
Find $v$ . From $2 u + 3 v = 13$ : $4 + 3 v = 13 \Rightarrow 3 v = 9 \Rightarrow v = 3$ . Why this step? Back-substitute the found $u$ .
Undo the rename. $u = \frac{1}{x} = 2 \Rightarrow x = \frac{1}{2}$ ; $v = \frac{1}{y} = 3 \Rightarrow y = \frac{1}{3}$ . Why this step? We solved for the reciprocals; flip them back to the real unknowns.

Verify in the original: $\frac{2}{1/2} + \frac{3}{1/3} = 4 + 9 = 13$ ✓; and $\frac{5}{1/2} - \frac{4}{1/3} = 10 - 12 = - 2$ ✓.

Recall

Recall Which cell decides no-solution vs infinite?

Compute $D = a_{1} b_{2} - a_{2} b_{1}$ . If $D \neq = 0$ → unique. If $D = 0$ , eliminate: leftover $0 = k (k \neq = 0)$ → no solution; leftover $0 = 0$ → infinite. ::: After a cross-multiplication answer fails by exactly a sign flip in both variables, what happened? ::: You divided by $- D$ instead of $D$ ; negate both $x$ and $y$ . In Ex 3, why did $x = 4$ appear before touching the second equation? ::: The first equation had $b_{1} = 0$ (no $y$ ), so it's a vertical line fixing $x$ alone. In Ex 4, what gets fixed first and why? ::: $y = 5$ , because $a_{1} = 0$ (no $x$ ) makes it a horizontal line. In Ex 9, what turned the non-linear system linear? ::: Substituting $u = 1/ x, v = 1/ y$ .