Solving equations is one of the most fundamental operations in all of mathematics — from simple linear equations in middle school to complex systems in engineering and science. An equation solver takes mathematical equations as input and finds the values of the unknown variables that make those equations true.
Our equation solver handles two major categories: linear systems with up to 5 equations and 5 unknowns using LU decomposition, and quadratic equations with full discriminant analysis including real and complex roots. Every solution includes step-by-step work and automatic verification by substituting the answers back into the original equations.
Choose the type of equation — linear system or quadratic. For linear systems, enter the coefficients and constants for each equation. For quadratic equations, enter the coefficients a, b, and c. The solver shows the original equations, applies the appropriate method (LU decomposition for systems, quadratic formula for quadratics), and presents the solution with full steps.
Critically, the solver then verifies every solution by substituting the values back into the original equations, confirming that both sides balance. This catches any potential issues and gives you confidence in the results.
Input: x + 2y = 5, 3x + 4y = 11
Output: x = 1, y = 2
Verification: 1 + 2(2) = 5 ✓ and 3(1) + 4(2) = 11 ✓
The solver supports linear systems with up to 5 equations and 5 unknowns. This covers the vast majority of systems encountered in coursework and many practical applications.
If the system is singular (the coefficient matrix has a determinant of zero), the solver will detect this and report that the system either has no solution (inconsistent) or has infinitely many solutions (dependent).
Yes. When the discriminant (b²-4ac) is negative, the solver computes the complex conjugate pair using imaginary numbers. For example, x² + 1 = 0 gives roots x = i and x = -i.
LU decomposition factors the coefficient matrix A into a lower triangular matrix L and an upper triangular matrix U (so A = LU). The system is then solved in two steps: first solving Ly = b (forward substitution), then Ux = y (back substitution). It's more numerically stable and efficient than simple elimination.
No. The solver handles any quadratic equation ax² + bx + c = 0 with any non-zero value of a. It applies the full quadratic formula x = (-b ± √(b²-4ac)) / (2a) regardless of the leading coefficient.
After finding the solution, the solver substitutes each value back into every original equation and confirms that both sides are equal. This catches arithmetic errors, singular system issues, and confirms the solution is correct.
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