Solve a system of two linear equations using the elimination method.
Equation 1: ax + by = c
Equation 2: dx + ey = f
Understanding the Elimination Method for Systems of Equations
Solving systems of linear equations is a fundamental skill in algebra, with applications ranging from economics and engineering to everyday problem-solving. The elimination method, also known as the addition method, is a powerful technique for finding the unique solution (if one exists) to a system of two or more linear equations.
What is a System of Linear Equations?
A system of linear equations is a collection of two or more linear equations involving the same set of variables. For a system of two variables, say x and y, the general form is:
a1*x + b1*y = c1
a2*x + b2*y = c2
The solution to the system is an ordered pair (x, y) that satisfies both equations simultaneously. Graphically, this solution represents the point where the lines represented by the equations intersect.
The Elimination Method Explained
The core idea behind the elimination method is to manipulate one or both equations (by multiplying them by a non-zero constant) so that the coefficients of one of the variables are opposites. When the equations are then added together, this variable is "eliminated," leaving a single equation with only one variable, which can be easily solved.
Here's a step-by-step breakdown:
Standard Form: Ensure both equations are in the standard form Ax + By = C.
Align Variables: Align the equations vertically so that the x terms, y terms, and constants are in the same columns.
Multiply to Match/Oppose Coefficients: Multiply one or both equations by suitable constants so that the coefficients of either the x variable or the y variable become additive inverses (e.g., 3 and -3, or 5 and -5).
Add the Equations: Add the two modified equations together. One of the variables should cancel out.
Solve for the Remaining Variable: Solve the resulting single-variable equation for the variable that remains.
Substitute and Solve: Substitute the value found in step 5 back into either of the original equations to solve for the other variable.
Check the Solution: Substitute the values of both variables into both original equations to verify that the solution satisfies both.
Example Calculation
Let's solve the system:
2x + 3y = 7
4x - 2y = 10
To eliminate y, we can multiply the first equation by 2 and the second equation by 3:
(2x + 3y = 7) * 2 => 4x + 6y = 14
(4x - 2y = 10) * 3 => 12x - 6y = 30
Now, add the two new equations:
(4x + 6y) + (12x - 6y) = 14 + 30
16x = 44
Solve for x:
x = 44 / 16 = 11 / 4 = 2.75
Substitute x = 2.75 into the first original equation (2x + 3y = 7):
2(2.75) + 3y = 7
5.5 + 3y = 7
3y = 7 - 5.5
3y = 1.5
y = 1.5 / 3 = 0.5
The solution is (x, y) = (2.75, 0.5).
Use Cases
Resource Allocation: Determining optimal distribution of limited resources based on multiple constraints.
Mixture Problems: Calculating the amounts of different components needed to achieve a desired mixture concentration.
Rate Problems: Solving problems involving speeds, distances, and times, or work rates.
Economic Modeling: Analyzing supply and demand equilibrium points.
Physics and Engineering: Analyzing circuits, forces, and motion where multiple conditions must be met.
function calculateElimination() {
var a1 = parseFloat(document.getElementById("a1").value);
var b1 = parseFloat(document.getElementById("b1").value);
var c1 = parseFloat(document.getElementById("c1").value);
var a2 = parseFloat(document.getElementById("a2").value);
var b2 = parseFloat(document.getElementById("b2").value);
var c2 = parseFloat(document.getElementById("c2").value);
var resultDiv = document.getElementById("result");
resultDiv.innerHTML = "; // Clear previous results
// Check if inputs are valid numbers
if (isNaN(a1) || isNaN(b1) || isNaN(c1) || isNaN(a2) || isNaN(b2) || isNaN(c2)) {
resultDiv.innerHTML = "Please enter valid numbers for all coefficients and constants.";
return;
}
// Calculate the determinant of the coefficient matrix
// | a1 b1 |
// | a2 b2 |
var determinant = (a1 * b2) – (a2 * b1);
if (determinant === 0) {
// Check for infinite solutions or no solution
// If (a1*c2 – a2*c1) == 0 AND (b1*c2 – b2*c1) == 0, then infinite solutions
// Otherwise, no solution
var det_x_numerator_check = (c1 * b2) – (c2 * b1);
var det_y_numerator_check = (a1 * c2) – (a2 * c1);
if (det_x_numerator_check === 0 && det_y_numerator_check === 0) {
resultDiv.innerHTML = "The system has infinitely many solutions (dependent equations).";
} else {
resultDiv.innerHTML = "The system has no solution (parallel lines).";
}
} else {
// Calculate x and y using Cramer's Rule or equivalent elimination logic
// x = | c1 b1 | / determinant
// | c2 b2 |
var x = ((c1 * b2) – (c2 * b1)) / determinant;
// y = | a1 c1 | / determinant
// | a2 c2 |
var y = ((a1 * c2) – (a2 * c1)) / determinant;
// Check for very small numbers close to zero due to floating point precision
var tolerance = 1e-9;
if (Math.abs(x) < tolerance) x = 0;
if (Math.abs(y) < tolerance) y = 0;
resultDiv.innerHTML = "Solution:x = " + x.toFixed(4) + "y = " + y.toFixed(4) + "";
}
}