How this calculation works
Each entered equation becomes one row of an augmented matrix. Swapping rows, dividing a row by a nonzero pivot and subtracting a multiple of one row from another preserve the solution set. The calculator shows those exact operations, then distinguishes a unique solution from free variables or a contradictory row.
Worked example
This example uses illustrative inputs. Your current result is shown above.
For x + y = 5 and 2x − y = 1, subtract twice the first equation from the second: −3y = −9, so y = 3. Substituting in x + y = 5 gives x = 2. Checks: 2 + 3 = 5 and 2×2 − 3 = 1. The three-variable default has x = 1, y = 2 and z = 3.
Assumptions and limits
- Provide exactly two equations in two variables or three equations in three variables. Every row includes a coefficient for each variable followed by the right-hand value.
- Finite decimal inputs are interpreted through their JavaScript numeric values, with about 15–17 significant digits. Rescale an equation if its input numbers exceed this range.
- Exact rational elimination determines rank and contradictions without discarding small nonzero pivots. Fractions are retained when short; long values are displayed to 12 significant digits.
- A unique solution is additionally checked with 80 significant digits. Relative substitution error must not exceed 10⁻¹², using the sum of the absolute equation terms as denominator.
- An exact answer to the entered decimals cannot recover precision already lost in rounded measurements. Near-dependent systems can be very sensitive to that rounding.
Questions people ask
How do I enter 2x − y + 3z = 7?
Choose three variables and enter 2 -1 3 7 on one line. Enter the other two equations on separate lines, using 0 for a missing coefficient. The final number in each row is the right-hand value.
What creates infinitely many solutions?
Some equations repeat information already supplied by others. For x + y = 2 and 2x + 2y = 4, one equation is redundant: y can be any value and x = 2 − y. The reduced rows identify the free variables.
What creates no solution?
The equations contradict each other. For x + y = 2 and 2x + 2y = 5, elimination produces 0 = 1. No values of x and y can satisfy both original equations.
Can small coefficients be mistaken for zero?
The solver uses exact fractions formed from the entered finite decimal numbers. It does not discard small nonzero pivots using a rank tolerance. It also substitutes unique solutions numerically at 80 significant digits and checks relative error against 10⁻¹².