System of equations calculator, solved row by row.

A system of equations calculator finds the values of the unknowns that make every equation true at once. Type the equations one per line and it solves them by row reduction, showing every step. It handles two, three or more unknowns, tells you when there is no solution or infinitely many, and can also solve a straight line together with a curve, a parabola with a line, and two equations such as a circle with an ellipse. You can choose row reduction, substitution, elimination or Cramer's rule for the steps.

How to use the system of equations calculator

1. Type one equation per line in the box, like 2x + y = 5 and x − y = 1. A semicolon also separates equations. Any letters work.
2. Read the "Read as" line to see the equations and the unknowns it found.
3. Read the answer cards. Then open the steps to see the augmented matrix and each row operation.
4. Under Method choose how the steps are written. The answer is the same for every method.
5. For two unknowns, the graph draws each equation, lines and curves, and white dots mark the solutions.
6. Use Copy answer or Copy link to keep the result.

A two equation example

Solve 2x + y = 5 and x − y = 1. Add the equations: the y terms cancel, leaving 3x = 6, so x = 2. Put x = 2 in the second equation: 2 − y = 1, so y = 1. Check in the first: 2 times 2 plus 1 is 5. With the default method, the calculator gets the same result by row reduction: it writes the rows as numbers, subtracts a multiple of one row from the other, and divides to get 1s on the diagonal.

A three equation example

Solve x + y + z = 6, 2x − y + z = 3 and x + 2y − z = 2. The answer is x = 1, y = 2, z = 3. Check each: 1 + 2 + 3 = 6, 2 − 2 + 3 = 3, and 1 + 4 − 3 = 2. In the steps you will see rows such as R2 = R2 − 2·R1, which removes x from the second equation, then a division to make each pivot 1.

The three possible outcomes

When there are infinitely many solutions, one or more unknowns are free. The calculator calls the free value t (or s, r) and writes the others in terms of it. Every choice of t gives a solution. For x + y + z = 3 and x − y + 2z = 1, z = t is free, and then y = t/2 + 1 and x = 2 − 3t/2. Try t = 2: z = 2, y = 2, x = −1, and both equations hold.

A line and a curve

For y = x + 1 together with x^2 + y^2 = 25, the calculator puts the line into the circle. That gives x^2 + (x + 1)^2 = 25, which simplifies to x^2 + x − 12 = 0, with roots x = 3 and x = −4. The matching y values are 4 and −3, so the circle and line cross at (3, 4) and (−4, −3). If the roots are not neat, the answers are given exactly when possible and as decimals next to them. Only real solutions are listed.

Four ways to solve the same system

Take 3x + 4y = 10 and 5x − 2y = 4. The answer is x = 18/13 and y = 19/13, about 1.3846 and 1.4615, whichever method you pick.

• Row reduction writes the numbers as a matrix and cleans it up row by row. It works for any size.
• Substitution solves one equation for one letter and puts that into the other. Solving 5x − 2y = 4 for y gives y = (5x − 4)/2, and putting that in the first equation gives 3x + 2(5x − 4) = 10, so 13x = 18. The calculator picks the equation and letter that look easiest.
• Elimination multiplies the equations so one letter has matching numbers, then subtracts. Multiply the first equation by 5 and the second by 3 and both have 15x. Subtracting leaves 26y = 38, so y = 19/13, and then x follows. The calculator shows its own multipliers.
• Cramer's rule uses determinants. D is the determinant of the coefficients, and each unknown is the determinant with its column replaced by the right-hand numbers, divided by D. Here D = 3·(−2) − 4·5 = −26, Dx = 10·(−2) − 4·4 = −36 and Dy = 3·4 − 10·5 = −38, so x = −36/−26 = 18/13 and y = −38/−26 = 19/13. If D is 0, the rule cannot be used, and the calculator says so and shows row reduction.

Elimination, substitution and matrices are three views of the same idea. For a small system by hand, substitution is often quickest. A correct answer satisfies every original equation, and the calculator checks that for you.

Two curves: when substitution works

If one equation has a letter only to the first power, that letter can be written in terms of the other and put into the second equation. For y = x^2 and y = 2x + 3 this gives x^2 = 2x + 3, so x = 3 or x = −1, and the points are (3, 9) and (−1, 1). For x*y = 12 and x + y = 7, y = 7 − x gives x(7 − x) = 12, so x = 3 or 4 and the points are (3, 4) and (4, 3). When both equations only contain x^2 and y^2, like x^2 + y^2 = 25 and x^2 − y^2 = 7, the calculator treats x^2 and y^2 as the unknowns: x^2 = 16 and y^2 = 9, so x = ±4 and y = ±3, and all four pairings are solutions. Only real solutions are listed, and the page tells you when complex ones were left out.

What it cannot do

It does not solve two curves where no letter is to the first power and the squares are mixed with other terms, such as x^2 + y^2 = 5 with x*y = 2. It does not solve systems of inequalities. Coefficients must be plain numbers in a linear system, not other letters. The limit is 8 unknowns and 10 equations.

For a single equation, use the algebra calculator. For matrix maths on its own, try the matrix calculator, and for lines see the slope calculator.