Linear System Solver — 2 Equations, 2 Unknowns
Enter the six coefficients a, b, e, c, d, f and the tool applies Cramer's rule to ax + by = e and cx + dy = f: x = Dx / D, y = Dy / D, where D = ad − bc, Dx = ed − bf and Dy = af − ec. The result is classified into one of three cases: (1) D ≠ 0 → unique solution; (2) D = 0 with Dx = Dy = 0 → infinitely many solutions (the two equations are the same line); (3) D = 0 but Dx or Dy ≠ 0 → no solution (parallel lines, inconsistent). Your input is rendered as a tidy pair of equations so you can sanity-check it — useful for high-school algebra homework, intro linear algebra, and engineering statics problems.
Your system
2x + 3y = 5 x − y = 1
Equation 1
Equation 2
Every coefficient must be a valid number (zero is allowed).
Solution
—
—
x
—
y
—
Coefficient determinant D
—
D_x
—
D_y
—
Cramer's rule: x = Dx / D, y = Dy / D when D ≠ 0.
Formula
D = a · d − b · c coefficient determinant Dx = e · d − b · f determinant with x-column replaced by RHS Dy = a · f − e · c determinant with y-column replaced by RHS x = Dx / D, y = Dy / D when D ≠ 0
- · Cramer's rule is most convenient for 2×2 and 3×3 systems. For larger systems (4×4 and up) Gaussian elimination is asymptotically faster than computing determinants.
- · Geometrically the two equations are lines in the plane. The three cases correspond to: intersecting (unique point), coincident (infinitely many points) and parallel (no intersection).
- · The determinant D = ad − bc is the signed area of the parallelogram spanned by the row vectors (a, b) and (c, d). D = 0 means those rows are parallel (linearly dependent), so the system cannot pin down (x, y) uniquely.
- · Zero coefficients are allowed, including a = 0 or b = 0 — e.g. "3y = 6, x − y = 1" with a = 0 still has D = 0·(−1) − 3·1 = −3 ≠ 0 and the unique solution x = 3, y = 2. The only degenerate case is a = b = c = d = 0, which means there are no equations to speak of.
- · The tool uses a relative tolerance of 1e⁻¹⁰ × max(|coefficient|)² when deciding whether D is zero, so floating-point round-off doesn't mis-classify a genuine unique solution as degenerate.
- · You can run it the other way too — as a consistency check on an experiment. Set e and f to measured values; D = 0 with Dx ≠ 0 means your two observations contradict each other.
- · Sources: Anton & Rorres, Elementary Linear Algebra, 12th ed., §2.1 (Cramer's rule) and §1.2 (existence and uniqueness); Strang, Introduction to Linear Algebra, §3.2.
Frequently asked
I got D = 0 and also Dx = Dy = 0 — does the system have solutions or not?
It has — and there are infinitely many. D = 0 means the two lines have the same slope (parallel or coincident); Dx = Dy = 0 additionally confirms they are the same line (one equation is a scalar multiple of the other, e.g. x + y = 3 and 2x + 2y = 6). The solution set is the entire line, often written parametrically: pick one variable as the parameter t and let x = t, y = (e − a·t)/b (assuming b ≠ 0), for every real t. Textbooks describe it as "a one-dimensional solution subspace". The tool reports "infinitely many solutions" rather than printing a single misleading (x, y).
How is Cramer's rule different from substitution or elimination?
They produce the same numbers, but trade off differently on hand-work and numerical robustness: (1) for 2×2, Cramer is the cleanest pencil-and-paper recipe — three determinants and two divisions, easy to spot-check; (2) for larger systems (4×4 and up), Cramer's O(n!) complexity badly loses to Gaussian elimination's O(n³); (3) numerically, Cramer's rule is unstable for ill-conditioned systems (D close to 0) because you divide by a tiny number — Gaussian elimination with partial pivoting is far better behaved. In practice numerical libraries like LAPACK always use LU or QR decompositions, never Cramer. But in teaching Cramer remains the standard recipe because it ties directly to the central idea that "determinant = 0 ⇔ no unique solution".
My coefficients are very small decimals — will the tool get confused?
No — the tool uses a relative threshold for "D is zero": about 1e⁻¹⁰ × max(|coefficient|)². So even if every coefficient is on the order of 0.001, the threshold drops to ~1e⁻¹⁶, still well above the IEEE 754 double-precision floor (~1e⁻¹⁶ relative). Likewise for coefficients up to ~1e⁹ the threshold scales up to ~1e⁸, so a "true" D of 1e⁻³ that happens to surface as round-off won't be mistaken for a genuine zero. That said, if your system is genuinely close to ill-conditioned (near-parallel lines), x and y can still have significant relative error — that's a limitation of Cramer's rule itself, not a bug.
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