🔡System of Equations Calculator

Solve a system of 2 or 3 linear equations using Cramer's rule. Handles unique solutions, no solution (inconsistent), and infinitely many solutions (dependent). Shows the determinant, step-by-step Cramer's rule, and verification by substitution.

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Solution

x = 3, y = 2

Solution TypeUnique solution
x3
y2
z (3-variable systems only)
Equation 12x +3y = 12
Equation 21x -1y = 1
Equation 3 (3-var)
Verification — Eq. 112 = 12 ✓
Verification — Eq. 21 = 1 ✓
Verification — Eq. 3
System Determinant (D)-5
D_x (Cramer's rule numerator for x)-15
D_y (Cramer's rule numerator for y)-10
D_z (Cramer's rule numerator for z)
Cramer's Rule — xx = D_x / D = -15 / -5 = 3
Cramer's Rule — yy = D_y / D = -10 / -5 = 2
Cramer's Rule — z

Solution Values

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System of Equations Calculator: Solve 2×2 and 3×3 Linear Systems

A system of linear equations has a unique solution when the determinant D ≠ 0. For a 2×2 system a₁x + b₁y = c₁ and a₂x + b₂y = c₂, Cramer's rule gives: x = (c₁b₂ − c₂b₁) / (a₁b₂ − a₂b₁) and y = (a₁c₂ − a₂c₁) / (a₁b₂ − a₂b₁).

Key formula: D = a₁b₂ − a₂b₁ — if D = 0, the system has no solution or infinite solutions

ConditionGeometric meaningSolutions
D ≠ 0Lines intersect at one pointUnique solution
D = 0, D_x ≠ 0 or D_y ≠ 0Lines are parallelNo solution
D = 0, D_x = 0, D_y = 0Lines are identicalInfinite solutions

This calculator uses Cramer's rule — a determinant-based formula that elegantly expresses the solution of Ax = b as ratios of determinants. It works for systems with a unique solution and automatically detects when no unique solution exists.

How Cramer's Rule Works

For a 2×2 system, D = a₁b₂ − a₂b₁ is the determinant of the coefficient matrix. To find x, replace the x-column with the constants and compute D_x = c₁b₂ − c₂b₁. Then x = D_x/D. For y, replace the y-column: D_y = a₁c₂ − a₂c₁, and y = D_y/D. For 3×3, the same principle applies with 3×3 determinants computed by cofactor expansion.

Geometric Interpretation

Each equation in a 2×2 system represents a line in the xy-plane. A unique solution is the intersection point of two non-parallel lines. A 3×3 system represents three planes in 3D space; a unique solution is the point where all three planes meet.

Frequently Asked Questions

How do you solve a system of 2 equations with 2 unknowns?

Three methods work for 2×2 systems: (1) Substitution — solve one equation for x in terms of y, substitute into the other. (2) Elimination — multiply equations to make coefficients match, then add/subtract to eliminate one variable. (3) Cramer's rule — compute D = a₁b₂ − a₂b₁; x = (c₁b₂ − c₂b₁)/D; y = (a₁c₂ − a₂c₁)/D. Example: 2x+3y=12, x−y=1. D = 2(−1)−1(3) = −5. D_x = 12(−1)−1(3) = −15. D_y = 2(1)−12(1) = −10. x = −15/−5 = 3, y = −10/−5 = 2.

What does it mean when a system of equations has no solution?

A system has no solution (inconsistent) when the equations represent parallel lines (2D) or parallel planes (3D) — they never intersect. Algebraically, this happens when D = 0 but the numerator determinants D_x or D_y are non-zero. Example: x + y = 3 and x + y = 5 — the same left-hand side but different right-hand sides, so no point satisfies both simultaneously. The lines are parallel (slope = −1 for both) but have different y-intercepts.

What does infinite solutions mean in a system of equations?

A system has infinitely many solutions (dependent) when the equations represent the same line (2D) or intersecting planes sharing a line (3D). All three determinants D, D_x, D_y equal zero. Example: x + y = 3 and 2x + 2y = 6 — the second equation is just twice the first, so they represent the same line. Any point on that line satisfies both equations: (0,3), (1,2), (2,1), etc. The solution set is typically expressed as {(t, 3−t) : t ∈ ℝ} using a parameter t.

How do you solve a 3×3 system of equations?

For 3×3 systems, the most common methods are: Gaussian elimination (row reduce the augmented matrix to row echelon form, then back-substitute) and Cramer's rule (compute 4 determinants: D, D_x, D_y, D_z; then x=D_x/D, y=D_y/D, z=D_z/D). Example: x+y+z=6, 2x−y+z=3, x+2y−z=2. D = det([[1,1,1],[2,−1,1],[1,2,−1]]) = 1(1−2)−1(−2−1)+1(4+1) = −1+3+5 = 7. D_x = det([[6,1,1],[3,−1,1],[2,2,−1]]) = 7, so x=1. Similarly y=2, z=3.