System of Equations Calculator (2×2 & 3×3, Free)

A system of equations calculator finds the values that make every equation true at the same time. For 2x + 3y = 8 and x − y = −1, multiply the second equation by 2 and subtract it from the first: 5y = 10, so y = 2. Then x − 2 = −1 gives x = 1. The solution is (1, 2), where the two lines cross.

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How to use the system of equations calculator

1. Choose 2 × 2 (two equations, two unknowns) or 3 × 3 (three equations, three unknowns).
2. Type each equation, such as 2x + 3y = 8 or y = 2x − 1, or switch to Coefficients and fill in the grid.
3. Pick the method you want to see: elimination, substitution, Cramer’s rule or the matrix (Gauss-Jordan) method. Every method gives the same answer; only the steps change.
4. Read the solution, the type of system and the determinant. For 2 × 2 systems the graph shows both lines and the point where they cross. Open Show steps for the full working and the check.

What is a system of equations?

A system of equations is two or more equations that share the same unknowns. A solution is a set of values that satisfies all of them at once. For x + y = 10 and x − y = 4, the solution is x = 7 and y = 3, because 7 + 3 = 10 and 7 − 3 = 4.

Graphically, each linear equation in two variables is a line, and the solution is the point the lines share. With three variables, each equation is a plane, and the solution is the point where all three planes meet.

One solution, no solution or infinitely many

• One solution (consistent and independent): the lines cross at exactly one point. The determinant D of the coefficients is not 0.
• No solution (inconsistent): the lines are parallel, with the same slope and different intercepts. Solving leads to a false statement such as 0 = −3. Example: x + 2y = 4 and 2x + 4y = 5.
• Infinitely many solutions (dependent): both equations describe the same line, and solving leads to 0 = 0. Example: x − y = 2 and 3x − 3y = 6, whose solutions are x = 2 + t, y = t for any number t.

When D = 0 the system has either no solution or infinitely many, and the calculator tells you which.

Substitution method

Solve one equation for one variable, then put that expression into the other equation.

Example: y = 2x and y = x + 3. The first equation already gives y, so substitute it into the second: 2x = x + 3, so x = 3. Then y = 2(3) = 6. The solution is (3, 6). Check: 6 = 2 × 3 and 6 = 3 + 3.

Substitution is quickest when a variable already has a coefficient of 1 or −1.

Elimination method

Add or subtract the equations so that one variable cancels. Multiply an equation first if the coefficients do not match.

Example: 2x + y = 7 and x − y = 2. The y terms are opposites, so add the equations: 3x = 9, which gives x = 3. Substitute back: 3 − y = 2, so y = 1. The solution is (3, 1).

For 3x + 2y = 7 and 5x − 4y = 19, multiply the first equation by 2 to get 6x + 4y = 14, then add: 11x = 33, so x = 3 and y = −1.

Cramer’s rule

Cramer’s rule uses determinants. For ax + by = e and cx + dy = f:

D = ad − bc, Dx = ed − bf, Dy = af − ec, and then x = Dx/D and y = Dy/D.

For 2x + 3y = 8 and x − y = −1: D = 2(−1) − 3(1) = −5, Dx = 8(−1) − 3(−1) = −5 and Dy = 2(−1) − 8(1) = −10. So x = −5/−5 = 1 and y = −10/−5 = 2. Cramer’s rule needs D ≠ 0. The Cramer’s rule calculator goes deeper into determinants.

Matrix method (Gauss-Jordan elimination)

Write the system as an augmented matrix, one row per equation, then use row operations (swap rows, multiply a row by a nonzero number, add a multiple of one row to another) until the left part is the identity matrix. The last column is then the solution.

For 2x + 3y = 8 and x − y = −1, the augmented matrix [[2, 3, 8], [1, −1, −1]] reduces to [[1, 0, 1], [0, 1, 2]], so x = 1 and y = 2. The same method scales to larger systems; the matrix calculator handles bigger matrices.

Solving a 3 × 3 system

Use elimination twice to reduce three equations to two, then to one. Example:

x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2.

1. Subtract the first equation from the third: y − 2z = −4.
2. Subtract 2 times the first equation from the second: −3y − z = −9, or 3y + z = 9.
3. Eliminate y: (3y + z) − 3(y − 2z) = 9 − 3(−4), so 7z = 21 and z = 3.
4. Back-substitute: y − 2(3) = −4 gives y = 2, and x + 2 + 3 = 6 gives x = 1.

The solution is (1, 2, 3).

Which method should you use?

A word problem

A school sells 20 tickets: adult tickets cost 8 dollars, child tickets cost 5 dollars, and the total is 130 dollars. With a adult tickets and c child tickets, a + c = 20 and 8a + 5c = 130. Multiply the first equation by 5 and subtract: 3a = 30, so a = 10 and c = 10.

Common mistakes with systems of equations

• Multiplying only one side of an equation (or only some terms) before eliminating.
• Sign errors when subtracting equations: subtract every term, including the constant.
• Stopping after one variable. Substitute back to find the other one.
• Not lining up like terms. Rewrite each equation as ax + by = c first.
• Calling 0 = 0 “no solution”. A true statement like 0 = 0 means infinitely many solutions; a false one like 0 = 5 means none.

Checking homework by hand

Solve the system on paper first, then enter it here and compare. Always check the answer by substituting it into every original equation; the calculator’s last step does exactly this. If your answer is different, choose the same method in the calculator and find the first step that differs.

Systems of equations practice quiz: 10 questions

Test yourself with 10 multiple-choice questions. Choose an answer and press Submit to see if you are right and why. Your score appears at the end.

Frequently asked questions

What is a system of equations?

It is a set of equations with the same unknowns that must all be true together. For example, x + y = 10 and x − y = 4 form a system whose solution is x = 7, y = 3.

What is the solution to a system of equations?

The solution is the set of values that satisfies every equation. For y = 2x and y = x + 3, the solution is (3, 6), since 6 = 2 × 3 and 6 = 3 + 3. On a graph it is the point where the lines cross.

How do you solve a system of equations?

Use substitution, elimination, a graph, Cramer's rule or matrices. For 2x + y = 7 and x − y = 2, add the equations to get 3x = 9, so x = 3; substituting gives y = 1. Check (3, 1) in both equations.

How do you solve a system of equations on a graphing calculator?

Solve each equation for y and enter them as Y1 and Y2. For x + y = 5 and x − y = 1, enter Y1 = −X + 5 and Y2 = X − 1, press GRAPH, then use CALC (2nd, TRACE) and choose "intersect". On a TI-84, press ENTER three times to select both curves and a guess. The intersection is (3, 2).

How do you solve systems of equations with a calculator matrix?

Enter the augmented matrix and use the rref function. For 2x + 3y = 8 and x − y = −1, the matrix [[2, 3, 8], [1, −1, −1]] reduces to [[1, 0, 1], [0, 1, 2]], so x = 1 and y = 2.

Is simultaneous equations a calculator paper topic?

"Simultaneous equations" is the UK name for systems of equations, and they can appear on both calculator and non-calculator GCSE papers. Learn to solve them by hand with elimination and substitution, and use a calculator to check.

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