How do I solve a system of equations?

How it works

The grid is the augmented matrix [A | b] of the system: each row is one equation, the columns before the last hold the coefficients of x, y, z, and w (in that order, 1 to 4 unknowns), and the last column holds the constant on the right of the equals sign. There can be 1 to 4 equations.

The calculator runs Gauss-Jordan elimination to reduced row echelon form, in exact fractions. Going through the columns from left to right:

1. The pivot is the first row, from the top of the rows not used yet, with a non-zero entry in that column. A column with no such row is skipped.
2. If the pivot row is not the top unused row, the two rows swap.
3. The pivot row is divided by its pivot, so the pivot becomes 1 (left out when it is already 1).
4. Every other row with a non-zero entry in that column has the right multiple of the pivot row subtracted, so its entry becomes 0.

Then:

• No solution when a pivot is in the constant column (a row reads 0 = 1).
• One solution when every unknown's column has a pivot: each unknown equals the constant in its pivot row.
• Infinitely many solutions otherwise. The unknowns without a pivot are free. Each pivot unknown equals the constant of its row minus each free unknown times that row's entry in the free unknown's column.

Exact fractions

Each cell is read as an exact fraction: the fraction nearest to its number with a bottom of at most 1,000,000, when that fraction rounds to the same 64-bit float (so a cell typed as 1/3 is exactly 1/3, as in Python's Fraction(x).limit_denominator(10**6)); otherwise its shortest decimal, exactly (0.1 is 1/10).

How the answer is written

• Solution, one solution: x = 5, y = 3, z = -2: each unknown in order, a fraction in lowest terms (such as 8/7, -9/7) or a whole number.
• Solution, infinitely many: each pivot unknown in order as x = <constant> <terms>, then the free unknowns: x = 1 - z, y = 1 + z, z is any number (with two or three free unknowns: z and w are any numbers, y, z and w are any numbers). The constant comes first and is left out when it is 0 and there are terms; each term is + or - and the coefficient's size in front of the unknown's name, with a coefficient of 1 left out, a whole coefficient written as 2z, and a fraction in brackets as (1/2)z. A first term that is negative starts with -, such as x = -2z. With every term 0, it reads x = 0.
• Solution, no solution: No solution.
• Solution in decimals (one solution only): the same, with each value rounded half up (away from zero) to 10 significant figures from its exact value, trailing zeros dropped, written like a JavaScript number (plain from 0.000001 up to below 10²¹, else e notation).
• Number of solutions: One solution, Infinitely many solutions, or No solution.
• Your equations: each row as an equation, rows separated by ; , written like the terms above with the constant after =: x + y + z = 6; 2y + 5z = -4; 2x + 5y - z = 27. A row with every coefficient 0 reads 0 = <constant>.
• Row operations, separated by ; , rows counted from 1: a swap as R1 ↔ R2; dividing a row as R1 → 1/2 × R1 (the factor that multiplies the row); and subtracting m times row j from row i as Ri → Ri − m × Rj (+ when adding), with m a fraction in lowest terms left out when it is 1. At most 30 are listed, then and N more. With no operation needed, the steps read The matrix is already reduced.

Worked examples by hand

x + y + z = 6, 2y + 5z = −4, 2x + 5y − z = 27. R3 − 2R1 gives 3y − 3z = 15, so y − z = 5. With 2y + 5z = −4: 2(z + 5) + 5z = −4, 7z = −14, z = −2, so y = 3 and x = 6 − 3 + 2 = 5. x = 5, y = 3, z = −2. Check: 2 × 5 + 5 × 3 − (−2) = 27.

2x + 3y = 7, x − y = 1. Divide row 1 by 2: x + (3/2)y = 7/2. Subtract it from row 2: −(5/2)y = −5/2. Divide by −5/2: y = 1. Subtract 3/2 × row 2 from row 1: x = 2. x = 2, y = 1. Steps: R1 → 1/2 × R1; R2 → R2 − R1; R2 → -2/5 × R2; R1 → R1 − 3/2 × R2.

x + 2y = 4, 3x − y = 2. By Cramer's rule, D = 1 × (−1) − 2 × 3 = −7, x = (4 × (−1) − 2 × 2) ÷ −7 = 8/7, y = (1 × 2 − 4 × 3) ÷ −7 = 10/7. In decimals, 1.142857143 and 1.428571429.

x + y = 2, 2x + 2y = 5. Row 2 − 2 × row 1 gives 0 = 1: No solution.

x + 2y − z = 3, 2x + 4y − 2z = 6, x + z = 1. Row 2 is twice row 1, so there are only two independent equations. From x + z = 1, x = 1 − z; then 2y = 3 − x + z = 2 + 2z, so y = 1 + z. x = 1 − z, y = 1 + z, z is any number.

x/2 + y/3 = 1, x/4 − y = 2 (typed as 0.5, 1/3, 1 and 0.25, −1, 2). From the second, x = 8 + 4y; then (8 + 4y)/2 + y/3 = 1, so 7y/3 = −3, y = −9/7, and x = 8 − 36/7 = 20/7.

Four unknowns: x + y + z + w = 10, x − y = −1, y − z = −1, z − w = −1. So y = x + 1, z = x + 2, w = x + 3, and 4x + 6 = 10: x = 1, y = 2, z = 3, w = 4.