numvana

System of 3 Linear Equations Solver

Solve three linear equations in three unknowns (a1x + b1y + c1z = d1, and so on) via Gaussian elimination, including the infinite- and no-solution cases.

x
5
y
3
z
-2

How it works

This solves a1x + b1y + c1z = d1, a2x + b2y + c2z = d2, and a3x + b3y + c3z = d3 by Gaussian elimination with partial pivoting: it repeatedly picks the equation with the largest remaining coefficient in the current column, swaps it into place, and subtracts multiples of it from the equations below to zero out that column — the standard method for keeping the arithmetic numerically stable, taught alongside Cramer's rule as the general-purpose way to solve a linear system.

This site's 2-variable solver can tell 'infinitely many solutions' from 'no solution' just by checking two numbers, because with only two equations that check is both necessary and sufficient. With three equations that shortcut isn't reliable — the coefficient matrix can collapse all the way to a single direction while the three constants still contradict each other. This calculator instead finishes the elimination fully: if it's left with a row that has zero coefficients but a nonzero constant, the equations contradict each other (no solution); otherwise, any leftover redundant row means the system is underdetermined (infinitely many solutions) rather than pinned to one answer.

FAQ

Why isn't this just the 2-equation solver with an extra variable?

Adding a third equation and unknown isn't just 'more of the same' — determining whether a 3-variable system has a unique answer, infinitely many, or none requires checking the rank of the whole system via elimination, not a simple pair of ratio checks the way two equations allow. That's exactly why this is a separate calculator built around full elimination rather than an extension of the 2-variable one.

What does 'infinitely many solutions' mean here?

It means the three equations don't pin down a single point — after removing redundant equations (ones that are combinations of the others), fewer than three independent constraints remain on x, y, and z, so a whole line or plane of points satisfies all three equations at once.

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