System of Linear Equations Calculator

1. Pick the system size (2×2 or 3×3).
2. Choose a method (Auto is usually best).
3. Enter coefficients in the form a·x + b·y (= + d·z) = c.
4. Click Calculate to get the classification (unique / none / infinite) and the solution when it exists.
5. Optional: toggle Show step-by-step and Prefer exact fractions to see clean algebra.

Tip: If you’re unsure about signs, rewrite your equation so all variable terms are on the left and the constant is on the right.

• 2×2 systems: we classify using the determinant and/or elimination consistency.
• 3×3 systems: we use Gaussian elimination to reach row-echelon form and classify.
• Solution types: unique (one solution), none (inconsistent), or infinite (dependent).

Tip: If you get a row like 0x + 0y + 0z = 5, the system has no solution.

Example 1 — 2×2 unique solution

Solve: 2x + y = 5 and x − y = 1

1. Add the equations to eliminate y: (2x + y) + (x − y) = 5 + 1
2. So 3x = 6 → x = 2
3. Plug into x − y = 1: 2 − y = 1 → y = 1

Example 2 — 2×2 no solution

Solve: 2x + 4y = 6 and x + 2y = 4

1. Multiply the second equation by 2: 2x + 4y = 8
2. Now compare: first says 2x + 4y = 6 but the scaled second says 2x + 4y = 8
3. This contradiction means the system is inconsistent → no solution.

Example 3 — 3×3 unique solution

Solve: x + y + z = 6, 2x − y + z = 3, and x + 2y − z = 3

1. Start with equation (1): x = 6 − y − z
2. Substitute into (2): 2(6 − y − z) − y + z = 3 → 3y + z = 9
3. Substitute into (3): (6 − y − z) + 2y − z = 3 → y − 2z = −3
4. Solve the 2×2 system: from y − 2z = −3 → y = 2z − 3. Plug into 3y + z = 9: 3(2z − 3) + z = 9 → 7z = 18 → z = 18/7.
5. Then y = 2(18/7) − 3 = 15/7.
6. Finally x = 6 − y − z = 6 − 15/7 − 18/7 = 9/7.

Final answer: x = 9/7, y = 15/7, z = 18/7.