Enter the coefficients for two linear equations with two variables (x and y).
Equation 1: ax + by = c
Equation 2: dx + ey = f
Solution (x, y)
Enter coefficients and click "Solve System".
Understanding the Elimination Method for Systems of Equations
A system of linear equations is a set of two or more linear equations that share the same variables. For a system with two variables, say x and y, we typically represent it as:
Equation 1: a₁x + b₁y = c₁
Equation 2: a₂x + b₂y = c₂
The goal is to find a pair of values (x, y) that satisfies both equations simultaneously. The elimination method is a powerful algebraic technique to achieve this.
How the Elimination Method Works
The core idea behind the elimination method is to manipulate one or both equations so that the coefficients of one variable (either x or y) are opposites. When the equations are then added together, that variable "eliminates" itself, leaving a single equation with only the other variable.
Align Equations: Ensure both equations are in the standard form (Ax + By = C), with x terms, y terms, and constants aligned.
Choose a Variable to Eliminate: Decide whether to eliminate x or y.
Make Coefficients Opposites: Multiply one or both equations by carefully chosen numbers so that the coefficients of the chosen variable become additive inverses (e.g., 3y and -3y, or -4x and 4x).
Add the Equations: Add the modified equations together. The variable with opposite coefficients will cancel out.
Solve for the Remaining Variable: Solve the resulting single-variable equation.
Substitute Back: Substitute the value found back into one of the original equations to solve for the other variable.
Check the Solution: Substitute both variable values into the *other* original equation to verify that the solution is correct.
Example Calculation
Let's solve the following system using the elimination method:
Equation 1: 2x + 3y = 7
Equation 2: 4x - y = 5
Step 1 & 2: Equations are already aligned. Let's choose to eliminate y.
Step 3: The coefficients of y are 3 and -1. To make them opposites, we can multiply Equation 2 by 3:
Equation 1: 2x + 3y = 7
Equation 2 (multiplied by 3): (4x - y = 5) * 3 => 12x - 3y = 15
Step 4: Add the modified equations:
(2x + 3y) + (12x - 3y) = 7 + 15
14x = 22
Step 5: Solve for x:
x = 22 / 14 = 11 / 7
Step 6: Substitute x = 11/7 into Equation 2 (4x - y = 5):
4 * (11/7) - y = 5
44/7 - y = 5
-y = 5 - 44/7
-y = 35/7 - 44/7
-y = -9/7
y = 9/7
Step 7: Check the solution (x=11/7, y=9/7) in Equation 1 (2x + 3y = 7):
2 * (11/7) + 3 * (9/7) = 22/7 + 27/7 = 49/7 = 7
The solution is correct. The calculator above automates these steps.
Use Cases
Algebraic Problem Solving: Directly solving textbook problems and homework assignments.
Optimization Problems: Certain linear programming or resource allocation problems can be modeled and solved using systems of equations.
Economics: Analyzing market equilibrium where supply and demand equations might form a system.
Engineering: Solving circuit analysis problems (e.g., Kirchhoff's laws) or structural analysis where forces and constraints are represented by linear equations.
Physics: Modeling motion, forces, or other physical phenomena that can be described by linear relationships.
function getInputValue(id) {
var input = document.getElementById(id);
var value = parseFloat(input.value);
return isNaN(value) ? null : value;
}
function displayResult(x, y) {
var solutionElement = document.getElementById('solution');
var errorElement = document.getElementById('error');
errorElement.innerText = "; // Clear previous errors
if (x === null || y === null) {
solutionElement.innerHTML = 'Invalid input. Please enter valid numbers.';
return;
}
// Format the output nicely, avoiding excessive decimals unless necessary
var xFormatted = parseFloat(x.toFixed(6)); // Round to 6 decimal places for display
var yFormatted = parseFloat(y.toFixed(6));
// Further clean up if the number becomes an integer after rounding
if (Math.abs(xFormatted – Math.round(xFormatted)) < 1e-9) { // Check if it's very close to an integer
xFormatted = Math.round(xFormatted);
}
if (Math.abs(yFormatted – Math.round(yFormatted)) < 1e-9) {
yFormatted = Math.round(yFormatted);
}
solutionElement.innerHTML = 'x = ' + xFormatted + ', y = ' + yFormatted;
}
function displayError(message) {
document.getElementById('solution').innerHTML = ''; // Clear any previous solution
document.getElementById('error').innerText = message;
}
function calculateElimination() {
var a1 = getInputValue('a1');
var b1 = getInputValue('b1');
var c1 = getInputValue('c1');
var a2 = getInputValue('a2');
var b2 = getInputValue('b2');
var f2 = getInputValue('f2'); // Corrected to f2 for consistency in input IDs
if (a1 === null || b1 === null || c1 === null || a2 === null || b2 === null || f2 === null) {
displayError('Please fill in all coefficient fields with valid numbers.');
return;
}
// Calculate the determinant of the coefficient matrix
var determinant = a1 * b2 – a2 * b1;
if (determinant === 0) {
// Check for parallel lines (no solution) or coincident lines (infinite solutions)
// If a1/a2 = b1/b2 = c1/f2 (and denominators are non-zero), infinite solutions
// If a1/a2 = b1/b2 != c1/f2, no solution
// Using cross-multiplication to avoid division by zero issues
var check1 = a1 * f2 – a2 * c1; // Corresponds to checking x column determinant
var check2 = b1 * f2 – b2 * c1; // Corresponds to checking y column determinant
if (check1 === 0 && check2 === 0) {
displayError('Infinite solutions (the equations are dependent).');
} else {
displayError('No solution (the equations are inconsistent).');
}
} else {
// Calculate x and y using Cramer's rule (derived from elimination)
var x = (c1 * b2 – c2 * b1) / determinant; // NOTE: The input variable for the second equation's constant was f2, not c2. Need to use f2.
var x_numerator = c1 * b2 – f2 * b1; // Corrected for f2
var x = x_numerator / determinant;
var y_numerator = a1 * f2 – a2 * c1; // Corrected for f2
var y = y_numerator / determinant;
displayResult(x, y);
}
}