Equation Solver

Two linear equations, two unknowns — the classic simultaneous-equations problem, solved instantly with Cramer's rule instead of substitution or elimination by hand.

Inputs

Result

x = 1.9286, y = 2.7143

How the equation solver works

For a₁x + b₁y = c₁ and a₂x + b₂y = c₂, the determinant of the coefficient matrix is a₁b₂ − a₂b₁.

When that determinant is non-zero, x = (c₁b₂ − c₂b₁) ÷ determinant and y = (a₁c₂ − a₂c₁) ÷ determinant.

A determinant of zero means the two lines are parallel (no solution) or identical (infinite solutions) — there's no single (x, y) pair to report either way.

Worked example: 2x + 3y = 12 and 4x − y = 5

  1. Determinant = 2×(−1) − 4×3 = −2 − 12 = −14.
  2. x = (12×(−1) − 5×3) ÷ (−14) = (−12 − 15) ÷ (−14) = 27 ÷ 14 ≈ 1.929.
  3. y = (2×5 − 4×12) ÷ (−14) = (10 − 48) ÷ (−14) = 38 ÷ 14 ≈ 2.714.

Common mistakes to avoid

Assuming 'no unique solution' means a data-entry error

It means the two equations describe parallel or identical lines — a genuine mathematical property of that system, not a bug to fix by re-entering numbers.

Mismatching which equation is '1' and which is '2'

The labelling doesn't matter mathematically, but keep each equation's own a, b, c together — mixing a₁ with c₂ produces a different, wrong system.

Frequently asked questions

What does it mean geometrically when there's no unique solution?

The two equations represent either two parallel lines that never meet (no solution) or the exact same line (every point is a solution).

Can this handle three equations and three unknowns?

No, this tool is built specifically for the 2×2 case; a 3-variable system needs a larger determinant method this simple form can't represent.

How do I check the answer is right?

Substitute the computed x and y back into both original equations — both should balance exactly (allowing for small rounding).

Does the order of the two equations change the answer?

No, swapping which equation is labelled 1 versus 2 gives the identical x and y — the system itself hasn't changed.

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