SOLVETUTORMATH SOLVER

Instrument MI-01-197 · Mathematics

Elimination Method Calculator

Two equations, two unknowns: scale one until a variable's coefficients match its twin, subtract to cancel that letter outright, then back-substitute for the other.

Instrument MI-01-197
Sheet 1 OF 1
Rev A
Verified
Type 05 — Algebra SER. 2026-01197

x

2.20000000

x = (c₁b₂ − c₂b₁) ⁄ (a₁b₂ − a₂b₁)

1.20000000 y
The working Every figure verified twice
  1. x = (8·-1 − 1·3) ⁄ (2·-1 − 1·3) = 2.20000000
  2. y = (2·1 − 1·8) ⁄ (2·-1 − 1·3) = 1.20000000
Worksheet log
  1. No entries yet — change an input to log a scenario.

How this instrument works

Elimination solves a pair of linear equations by turning two true statements into a third one with a variable removed. Multiply an equation through by a constant and it stays just as true — 2x + 3y = 8 says exactly what 4x + 6y = 16 says — so choosing a multiplier that makes one variable's coefficient identical in both equations lets a straight subtraction cancel that letter completely, leaving one equation in one unknown to solve outright.

The technique is older than the name usually attached to it. Arrays of coefficients reduced by this same row-by-row cancellation appear in the Chinese text The Nine Chapters on the Mathematical Art, compiled roughly two thousand years ago and worked with counting rods rather than pen and ink. 'Gaussian elimination' only entered common use in the nineteenth century, borrowed for Carl Friedrich Gauss's unrelated work on planetary orbits, and the name stuck to the method by association rather than by priority.

Elimination can fail in two distinct ways worth telling apart. Subtract and land on 0 = 0, and the two equations were secretly the same line scaled differently — every point on it satisfies both, so infinitely many pairs solve the system rather than one. Subtract and land on something like 0 = 5, and the lines run parallel: no multiplier ever brings a false statement into agreement, and the system has no solution at all.

x=c1b2c2b1a1b2a2b1x = \dfrac{c_1 b_2 - c_2 b_1}{a_1 b_2 - a_2 b_1}y=a1c2a2c1a1b2a2b1y = \dfrac{a_1 c_2 - a_2 c_1}{a_1 b_2 - a_2 b_1}
a₁, b₁, c₁ — first equation's coefficients and constant; a₂, b₂, c₂ — second equation's own trio. x, y — the single point solving both equations, reached here by cancelling one letter at a time rather than by reading off a determinant.
  • Enter the first equation's numbers into a₁, b₁, and c₁ — the coefficients and constant in a₁x + b₁y = c₁.
  • Enter the second equation's numbers into a₂, b₂, and c₂ — the coefficients and constant in a₂x + b₂y = c₂.
  • The sheet scales one equation so a variable's coefficient matches the other, then subtracts them — that working is shown step by step.
  • Read x and y as the single point where the two lines cross; a warning appears instead if the lines turn out parallel.

Worked example — eliminating x from 2x + 3y = 8 and x − y = 1

Take the pair 2x + 3y = 8 and x − y = 1, so a₁ = 2, b₁ = 3, c₁ = 8, a₂ = 1, b₂ = −1, c₂ = 1. The second equation's x-coefficient is 1, exactly half the first equation's 2, so multiplying the second equation through by 2 gives 2x − 2y = 2 without changing what it actually means.

Subtracting that scaled equation from the first cancels x outright: (2x + 3y) − (2x − 2y) = 8 − 2, which simplifies to 5y = 6, so y = 6 ⁄ 5 = 1.2. Back-substituting into the untouched x − y = 1 gives x − 1.2 = 1, so x = 2.2 — matching exactly what the sheet reports for these six coefficients.

Questions

What does it mean to 'eliminate' a variable?

It means combining two true equations so one unknown drops out entirely. Scaling an equation by a nonzero constant leaves it just as true, so picking a scale that makes one variable's coefficient match the other equation's — in size, or in size with an opposite sign — lets a plain addition or subtraction erase that letter, leaving one equation in one unknown to solve directly.

How do I know whether to add or subtract the two equations?

Compare the signs of the matching coefficient once scaling is done. If the two coefficients come out equal, subtracting cancels them; if they come out equal in size but opposite in sign, adding does the same job. The two choices are really one move in disguise, since subtracting a negative number is identical to adding its positive.

How is elimination different from Cramer's rule on the same system?

Both land on identical numbers for a₁x + b₁y = c₁ and a₂x + b₂y = c₂, but Cramer's rule gets there by building two determinants and dividing one by the other, while elimination gets there by scaling and subtracting the original equations directly, with no determinant involved. Elimination is the technique taught first; determinants formalize the same bookkeeping once matrices enter the picture.

What does it mean if subtracting the equations leaves 0 = 0?

It means the two equations describe the same line in disguise — one is just a scaled copy of the other — so every point on it satisfies both, and infinitely many (x, y) pairs solve the system instead of one. No further scaling can break the tie, because the two lines were never actually distinct to begin with.

What does it mean if subtracting leaves something like 0 = 5?

That contradiction means the two lines run parallel and never meet, so no multiplier turns a false statement into a true one and the system has no solution at all. This sheet flags exactly that case whenever a₁b₂ − a₂b₁ works out to zero, since that shared quantity vanishing is precisely what parallel lines produce.

Does elimination always require scaling both equations first?

No — only when neither variable already shares a common coefficient between the two equations. If a₁ already equals a₂, or b₁ already equals b₂ or its negative, one equation can be added to or subtracted from the other immediately, skipping the scaling step; matching coefficients by luck is common in textbook problems built to have tidy answers.

References