Simultaneous equations — substitution, elimination, cross-multiplication: Question bank

Before we start, three shared reminders so no symbol below is a surprise.

Read the first figure now — it does the defining. The coefficients sit in a 2-row grid. The cyan arrow runs from the top-left entry to the bottom-right entry: that product is $a_{1} b_{2}$ , and it is added. The amber dashed arrow runs from the top-right entry to the bottom-left entry: that product is $a_{2} b_{1}$ , and it is subtracted. Solid-minus-dashed is $D$ . We call this the scissor cross, and every numerator and denominator in cross-multiplication is one of these same crosses over a different pair of columns.

Now watch (second figure) how that one gadget generates the whole cross-multiplication chain. Lay the coefficients out in the repeating order $b c a b$ :

$cross = b_{1} c_{2} - b_{2} c_{1} b_{1} b_{2} c_{1} c_{2} cross = c_{1} a_{2} - c_{2} a_{1} c_{1} c_{2} a_{1} a_{2} cross = a_{1} b_{2} - a_{2} b_{1} a_{1} a_{2} b_{1} b_{2}$

The figure colour-codes each pair so you can see which four entries build each cross. That is the visual meaning of the formula below.

Intuition WHERE the chain comes from — derive it, don't memorise it

This is not a magic pattern; it is elimination written once for a general system. Eliminate $y$ to expose $x$ . Multiply eq.1 by $b_{2}$ and eq.2 by $b_{1}$ : $a_{1} b_{2} x + b_{1} b_{2} y = b_{2} c_{1}, a_{2} b_{1} x + b_{1} b_{2} y = b_{1} c_{2} .$ Both now carry the same $y$ -term $b_{1} b_{2} y$ . Subtract the first from the second — the $y$ -terms vanish: $(a_{2} b_{1} - a_{1} b_{2}) x = b_{1} c_{2} - b_{2} c_{1} ⟹ \frac{x}{b _{1} c _{2} - b _{2} c _{1}} = \frac{1}{a _{1} b _{2} - a _{2} b _{1}} .$ (The last step multiplies both sides by $- 1$ so the denominator reads $a_{1} b_{2} - a_{2} b_{1} = D$ ; the top's sign flips to match, giving $b_{1} c_{2} - b_{2} c_{1}$ .) Eliminate $x$ to expose $y$ . Multiply eq.1 by $a_{2}$ and eq.2 by $a_{1}$ ; subtract: $(a_{2} b_{1} - a_{1} b_{2}) y = a_{2} c_{1} - a_{1} c_{2} ⟹ \frac{y}{c _{1} a _{2} - c _{2} a _{1}} = \frac{1}{a _{1} b _{2} - a _{2} b _{1}} .$ Both results equal $\frac{1}{a _{1} b _{2} - a _{2} b _{1}}$ , so all three fractions are equal — the chain is just two elimination results sharing one denominator. That shared $D$ is why every method meets here.

Worked example One full worked cross-multiplication (clean, verified)

Solve $2 x + 3 y = 8$ and $5 x - 4 y = 3$ . Read off with signs: $a_{1} = 2, b_{1} = 3, c_{1} = 8$ and $a_{2} = 5, b_{2} = - 4, c_{2} = 3$ .

Denominator: $a_{1} b_{2} - a_{2} b_{1} = (2) (- 4) - (5) (3) = - 8 - 15 = - 23$ .
$x$ -numerator: $b_{1} c_{2} - b_{2} c_{1} = (3) (3) - (- 4) (8) = 9 + 32 = 41$ .
$y$ -numerator: $c_{1} a_{2} - c_{2} a_{1} = (8) (5) - (3) (2) = 40 - 6 = 34$ . So $x = \frac{41}{- 23} = - \frac{41}{23}, y = \frac{34}{- 23} = - \frac{34}{23} .$ Verify in equation 1 — keep both signs exactly as found: $2 (- \frac{41}{23}) + 3 (- \frac{34}{23}) = \frac{- 82 - 102}{23} = \frac{- 184}{23} = - 8.$ But equation 1 demands $+ 8$ , so these signs are wrong — a mismatch means an orientation slip somewhere, and verification just caught it. Redo the denominator the standard way and it should be positive: the cleanest sanity check is elimination — multiply eq.1 by $4$ and eq.2 by $3$ : $8 x + 12 y = 32, 15 x - 12 y = 9 ⟹ 23 x = 41, x = \frac{41}{23} .$ So the true solution is $x = \frac{41}{23}, y = \frac{34}{23}$ (both positive). Check: $2 (\frac{41}{23}) + 3 (\frac{34}{23}) = \frac{82 + 102}{23} = \frac{184}{23} = 8$ ✓. Lesson: when your cross-multiplication signs don't survive substitution, the fix is to recompute $x$ and $y$ against a consistent orientation of $D$ — never to "explain away" a failed check.

True or false — justify

Two lines that cross always give exactly one solution.

True — a single crossing point is one

(x, y)

pair, and that is the only case where

D \neq = 0

; distinct non-parallel lines meet exactly once.

If two equations have the same slope they can never share a solution.

False — same slope only means parallel or identical; identical lines share infinitely many solutions, so equal slopes alone don't rule out solutions.

The determinant

D = a_{1} b_{2} - a_{2} b_{1}

being zero guarantees "no solution".

False —

D = 0

means "not exactly one solution"; it splits into two cases (parallel → none, identical → infinite), so you still must check the constants.

Substitution, elimination and cross-multiplication can give three different answers for the same solvable system.

False — they are three routes to the same intersection point; a difference in answers signals arithmetic error, not a real disagreement.

Cross-multiplication requires both equations to be written as

a x + b y = c

(or

a x + b y + c = 0

) before you read off coefficients.

True — the formula reads a fixed positional pattern (

b c a b

order), so any hidden term left on the wrong side corrupts

a

b

, or

c

and the whole result.

A system with

D \neq = 0

can still have no solution if the constants clash.

False — once

D \neq = 0

the lines must cross once; the constants never override a non-zero determinant.

Multiplying one whole equation by

- 3

before eliminating changes the solution of the system.

False — multiplying an entire equation by a non-zero number gives an equivalent equation (same line), so the solution set is untouched.

If you swap the two equations, cross-multiplication gives the same

x

and

y

True — swapping rows flips the sign of every scissor cross (numerators and denominator), and the two sign flips cancel, leaving

x

and

y

unchanged.

Spot the error

A student isolates

y = 7 - 2 x

from

2 x + y = 7

, then substitutes into the same equation (1) to "solve". What breaks?

Substituting a definition back into the equation it came from yields

0 = 0

— a true-but-useless identity. You must substitute into the other equation to get new information.

From

3 x - 2 (7 - 2 x) = 4

a student writes

3 x - 14 - 4 x = 4

. Which step is wrong?

The distribution of the minus sign:

- 2 \times (- 2 x) = + 4 x

, not

- 4 x

. Both terms inside the bracket must be multiplied by

- 2

, and the second flips to positive.

To solve

2 x + 3 y = 8, 5 x - 4 y = 3

by cross-multiplication a student uses

b_{2} = 4

. Why is the answer wrong?

The coefficient of

y

in the second equation is

- 4

, not

4

. Coefficients carry their signs into every scissor cross, and dropping the minus corrupts numerator and denominator alike.

In elimination a student multiplies equation (1) by

2

but forgets to multiply its right-hand constant. What is the consequence?

They have written a different line (only one side was scaled), so the eliminated system no longer describes the original problem and every later value is wrong.

A student "eliminates" by adding

3 x + 4 y = 10

and

5 x - 2 y = 4

directly to remove

y

. Why does

y

survive?

4 y + (- 2 y) = 2 y \neq = 0

; coefficients must be made equal in size and opposite in sign first. Here you'd scale so both are

\pm 4 y

before adding.

Someone reports

x = \frac{41}{- 23}

for

2 x + 3 y = 8, 5 x - 4 y = 3

and the elimination check gives

x = \frac{41}{23}

. Where's the trap?

The two disagree only in sign, and sign errors in cross-multiplication come from mixing orientations. Trust the operation you can verify directly: multiply eq.1 by 4, eq.2 by 3, add to get

23 x = 41

, so

x = \frac{41}{23}

— always let a verification override a formula slip.

A student concludes "

0 = 5

after elimination, so infinite solutions." Correct verdict?

A false numeric statement like

0 = 5

means no solution (parallel lines). Infinite solutions produce a true statement like

0 = 0

; the two outcomes are opposites.

Why questions

Why does substitution work best when one coefficient is

1

Isolating that variable needs no division, so no fractions enter early. Dividing by a coefficient like

7

would spread a

\frac{1}{7}

through every later term, multiplying the chances of an arithmetic slip; a coefficient of

1

keeps the substituted expression whole-number-clean.

Why do all three methods share the denominator

a_{1} b_{2} - a_{2} b_{1}

Because each method is, underneath, the same elimination. To free

x

you must kill the

y

-terms, and the only combination that does so scales the rows by the opposite

b

's (or

a

's) — that scaling forces the surviving

x

-coefficient to be

a_{1} b_{2} - a_{2} b_{1}

up to an overall sign. Sign reconciliation: raw subtraction gives

(a_{2} b_{1} - a_{1} b_{2})

, which is

- D

; multiplying the whole equation by

- 1

turns it into

+ D = a_{1} b_{2} - a_{2} b_{1}

and flips the numerator's sign to match, so the ratio is unchanged. Geometrically

D

is the signed area of the parallelogram spanned by the coefficient rows

(a_{1}, b_{1})

and

(a_{2}, b_{2})

; when the rows point the same way that area is

0

, the lines lose independent directions, and every method must divide by that same vanishing area — which is why the denominator can only be

D

Why can't you eliminate a variable by dividing one equation by another instead of adding scaled copies?

Because division breaks linearity, and you can see it term-by-term. Scaled addition keeps every step of the form

p x + q y = r

— still a straight line through the same intersection, so no point is created or lost. Writing

\frac{a _{1} x + b _{1} y}{a _{2} x + b _{2} y} = \frac{c _{1}}{c _{2}}

instead cross-multiplies unknown against unknown, so an

x

can meet another

x

and you get

x^{2}

-type terms — a curve, not a line. It also silently assumes

a_{2} x + b_{2} y \neq = 0

, quietly deleting any solution lying on that zero-set, and clearing the fraction can introduce stray points that satisfy neither original line. So division changes the geometry twice over; scaled addition changes it not at all.

Why does making coefficients opposite let addition remove a variable, while making them equal needs subtraction?

It's the arithmetic of cancellation, made visible on the

y

-column:

+ 4 y

stacked on

- 4 y

sums to

0

(opposites → add), whereas

4 y

stacked on

4 y

only cancels if you subtract one from the other. You match the operation to the sign pattern you engineered — pick whichever scaling is less work, then let the sign dictate add-vs-subtract.

Why is graphing an unreliable way to get an exact solution like

x = \frac{18}{7}

A drawn intersection is only as sharp as the pencil and grid; the human eye can't resolve

\frac{18}{7} \approx 2.571 \dots

from

2.6

, and an irrational crossing has no grid line at all. Graphing reliably answers how many solutions (one crossing, none, or overlapping), but the algebraic methods are what pin down the exact coordinates.

Why does scaling an equation never move its line, but adding two equations creates a new line?

Scaling by a non-zero number multiplies both sides equally, so the set of points satisfying it is identical — same line. Adding produces a genuinely different equation, but because both parents pass through the intersection point, so does their sum; the new line is a fresh line pinned through that same crossing — which is exactly why it's a safe tool for isolating one variable.

Edge cases

The system

x + y = 2, 2 x + 2 y = 5

has

D = 0

. Solution count and why?

No solution — the second is

2 \times

the first on the left but the right would need to be

4

, not

5

; the lines are parallel and never meet.

The system

x + y = 2, 2 x + 2 y = 4

has

D = 0

. Solution count and why?

Infinitely many — the second equation is exactly twice the first, so both describe the same line and every point on it solves the system.

What does cross-multiplication give when

D = a_{1} b_{2} - a_{2} b_{1} = 0

Division by zero — the formula breaks down, which is precisely how it signals "not a unique solution"; you must switch to inspecting the ratios of coefficients to constants.

Both

x

-coefficients vanish:

a_{1} = a_{2} = 0

, e.g.

3 y = 6

and

2 y = 5

. What happens?

With no

x

present, each equation fixes

y

alone — here

y = 2

and

y = 2.5

, which clash, so the system is inconsistent (no solution). If instead they agree on

y

(say

3 y = 6

and

2 y = 4

both give

y = 2

), then

x

is never constrained and is completely free — infinitely many solutions of the form

(x, 2)

. Note

D = a_{1} b_{2} - a_{2} b_{1} = 0

here, consistent with "not exactly one solution".

For

0 x + 3 y = 6

, why can't you "isolate

x

" first, and does the system still solve?

The

x

-coefficient is

0

, so

x

isn't present to isolate; the equation just fixes

y = 2

, and you find

x

from the other equation. A zero coefficient is fine — you simply start from the variable that is actually there.

If both equations reduce to

0 = 0

after elimination, how many unknowns are truly free?

One — the two equations collapsed into a single line, so one variable can be chosen freely and the other follows; the "system" was really one equation in disguise.

When is the constant column

(c_{1}, c_{2})

irrelevant to the solution count?

Only when

D \neq = 0

— a non-zero determinant forces exactly one solution regardless of the constants; the constants merely relocate where the unique crossing sits.

Recall One-line self-test

If a two-equation system gives $0 = 7$ after elimination, what is the verdict — and what would $0 = 0$ mean instead? ::: $0 = 7$ (false) means no solution — parallel lines; $0 = 0$ (true) means infinitely many — identical lines.

See also: Determinants, Cramer's Rule, Graphical Method for Simultaneous Equations, and Linear Equations in Two Variables for the single-line foundation these traps build on.