2×2 Linear System Solver
Solve two linear equations a₁x+b₁y=c₁, a₂x+b₂y=c₂; output intersection (x,y) with steps.
Input Data
Results
At a glance:Cramer's rule: x=(c1·b2−b1·c2)/D, y=(a1·c2−c1·a2)/D, D=a1·b2−a2·b1. D=0 with Dx,Dy not both zero ⇒ no solution; all zero ⇒ infinitely many.
Formula
D = a1·b2 − a2·b1.
x = (c1·b2 − b1·c2)/D, y = (a1·c2 − c1·a2)/D.
Elimination: scale then subtract to remove a variable.
Substitution: express one variable, plug into the other.
$$D=a_{1}b_{2}-a_{2}b_{1}$$$$x=\frac{c_{1}b_{2}-c_{2}b_{1}}{D}$$$$y=\frac{a_{1}c_{2}-a_{2}c_{1}}{D}$$How to Use
- Pick a method (elimination / substitution).
- Enter the six coefficients.
- The tool returns (x,y), D and the steps.
Outcome by determinant
| D | Dx, Dy | Conclusion |
|---|---|---|
| D ≠ 0 | — | unique (intersect) |
| D = 0 | not both 0 | no solution (parallel) |
| D = 0 | both 0 | infinite (coincident) |
Dx = c1·b2 − c2·b1, Dy = a1·c2 − c1·a2.
Case Studies
x+y=5, 2x−y=1
D=−3; x=2, y=3; intersection (2,3).
FAQ
What if D = 0?
The lines are parallel or coincident: no unique solution unless all determinants are zero (infinitely many).
Elimination or substitution?
Use elimination when coefficients line up; substitution when one equation easily gives a single variable. Same result.
Related Tools
References
Content review: Calculatorism Science Team. Cramer's rule plus elimination and substitution steps have been verified. Results are for reference only.