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Two-variable linear system solver

Do your two linear equations have one solution, none, or infinitely many?

Enter six coefficients for ax + by = c and dx + ey = f. Find the solution type and, when unique, x and y.

Your two equations

Use ax + by = c and dx + ey = f, with constants on the right. Keep your intended coefficient and unit bases consistent. Enter fractions as decimals, not a/b.

How the estimate works

D = ae − bd. For D ≠ 0, x = (ce − bf)/D and y = (af − cd)/D. For D = 0, consistency is checked separately. If a, b, d and e are all zero, only c = f = 0 gives infinitely many solutions. Otherwise, both ce − bf and af − cd must be zero for infinitely many solutions; if not, there is no solution. Classification needs all six valid inputs; D needs only a, b, d and e.

ax + by = c; dx + ey = f
D = ae − bd
D ≠ 0: x = (ce − bf) ÷ D; y = (af − cd) ÷ D

Only two declared real linear equations. Zero rows such as 0 = 0 and contradictions such as 0 = 1 are included. For infinitely many solutions, no parameter formula or partially fixed coordinate is shown. No larger systems, nonlinear equations, inequalities, complex or symbolic output, graph, fitting, unit conversion, measurement validation, uncertainty estimate or physical or safety guarantee.

OpenStax: determinants and Cramer’s rule ↗
OpenStax: linear systems and solution types ↗

Check the filled example

Our chosen example: 2x − y = 7 and x + 3y = 0. D = 7, x = 3, y = −1. Substitution gives 2×3 − (−1) = 7 and 3 + 3×(−1) = 0. These are mathematical example inputs, not standard values or a forecast.

Accepted input values

First equation: a
2
First equation: b
-1
First equation: c
7
Second equation: d
1
Second equation: e
3
Second equation: f
0

Plan the next step

Use the standard calculator for separate arithmetic. Check that both entered equations describe your intended problem.

Open the standard calculator →