Elimination Method Calculator
Enter the coefficients of your system of linear equations and this calculator solves for all variables using the elimination method, showing every step of the working. Switch between a 2-equation system (two unknowns x and y) and a 3-equation system (three unknowns x, y and z). Special cases such as no solution and infinitely many solutions are detected and explained.
Formula
Worked example
Solve 2x + y = 8 and 3x - 2y = 1. Multiply Eq 1 by 2: 4x + 2y = 16. Add to Eq 2: 4x + 2y + 3x - 2y = 16 + 1, so 7x = 17, x = 17/7. Substitute back: 2(17/7) + y = 8, y = 8 - 34/7 = 22/7.
What is the elimination method?
The elimination method (also called the addition method) is a technique for solving systems of linear equations. The core idea is to combine the equations in a way that cancels out one variable, leaving a simpler equation with just one unknown. Once that value is found it is substituted back to find the remaining variable or variables. For a 2x2 system, this means one round of elimination followed by back-substitution. For a 3x3 system, the process is repeated - first eliminating one variable from two pairs of equations to produce a 2x2 system, then solving that in turn. The method is systematic, always terminable, and straightforward to check by substituting the solution back into the original equations.
How to use this calculator
Choose your system size: two equations with two unknowns (x and y) or three equations with three unknowns (x, y and z). Enter the coefficient of each variable in each equation and the right-hand side constant. For example, if your first equation is 2x + 3y = 12, enter 2 in the "coefficient of x" field, 3 in the "coefficient of y" field, and 12 in the "right-hand side" field. The calculator solves instantly and shows the type of system, the solution values, and verification that the values satisfy every original equation. Open the steps panel to see the elimination working carried through with your exact numbers.
Step-by-step elimination for a 2x2 system
To eliminate y from the system a1x + b1y = c1 and a2x + b2y = c2, multiply the first equation by b2 and the second by b1. The y-terms then have the same coefficient. Subtracting one scaled equation from the other removes y entirely, leaving (a1*b2 - a2*b1)x = c1*b2 - c2*b1. Dividing both sides by the determinant D = a1*b2 - a2*b1 gives x directly. Substituting x back into either original equation gives y. The method also signals impossible or under-determined systems: if D equals zero, the lines are parallel (no solution) or identical (infinitely many solutions).
Elimination vs. substitution and Cramer's Rule
Substitution isolates one variable first, then inserts it into the other equation. It is often easier when a coefficient is already 1. Elimination is generally preferable when no coefficient is 1 or when working with 3x3 or larger systems, because each step involves only addition, subtraction and scalar multiplication. Cramer's Rule expresses the solution as a ratio of determinants and is algebraically equivalent to elimination, but computing determinants for large systems is computationally expensive. Gaussian elimination (the algorithm underlying this calculator's 3x3 solver) generalises the hand-elimination method to any size system by systematically reducing the matrix to upper-triangular form before back-substitution.
Types of 2x2 linear systems
| Determinant | System type | Geometric interpretation | Number of solutions |
|---|---|---|---|
| Non-zero | Consistent and independent | Two lines intersect at exactly one point | Exactly one |
| Zero (lines parallel) | Inconsistent | Two parallel lines - no intersection | None |
| Zero (lines identical) | Consistent and dependent | Same line described twice | Infinitely many |
A 2-equation, 2-unknown system falls into one of three categories determined by the determinant of the coefficient matrix.
Frequently asked questions
What does it mean when the calculator says "no solution"?
The system is inconsistent: the two equations represent parallel lines that never cross. No values of x and y can satisfy both equations at the same time. This happens when the ratio of the x-coefficients equals the ratio of the y-coefficients but not the ratio of the right-hand sides - in other words, the equations are contradictory.
What does "infinitely many solutions" mean?
The two equations are dependent - one is a scalar multiple of the other, so they describe the same line. Every point on that line is a solution. You effectively have only one independent constraint for two unknowns, so the solution set is a whole line rather than a single point. Adding a second genuinely independent equation would pin down a unique answer.
How does the calculator handle 3-equation systems?
For 3x3 systems the calculator uses Gaussian elimination with partial pivoting. It builds a 3x4 augmented matrix, selects the largest available pivot at each step to minimise rounding error, and reduces the matrix to upper-triangular form. It then back-substitutes to find z, y and x in turn. The method is numerically stable for all well-conditioned systems and correctly flags singular (no unique solution) cases.
Can I enter decimal or fractional coefficients?
Yes. The coefficient fields accept any real number, including decimals. If your system involves fractions such as 1/2 x + 3/4 y = 5, enter the decimal equivalents 0.5 and 0.75. The solver handles them the same way as integers and the solution will be reported as a decimal.
How do I check that the answer is correct?
Substitute the reported x and y (and z for 3x3) back into every original equation and confirm the left-hand side equals the right-hand side. This calculator does this automatically: the "Check" outputs show the computed left-hand side value next to the right-hand side so you can see at a glance that they match.
Why multiply equations by the other equation's coefficient?
The goal of elimination is to make the coefficient of the variable you want to remove identical in both equations. Multiplying Equation 1 by b2 (the y-coefficient from Equation 2) and Equation 2 by b1 (the y-coefficient from Equation 1) ensures both scaled equations have the same y-coefficient, so subtracting them wipes out y entirely.