Why Does the Process of Elimination in Solving Systems of Equations Work?

Finding solutions to systems of equations is a fundamental skill in algebra, and one powerful method for doing so is the process of elimination. This article explores why this method works, providing a clear understanding of its underlying principles and practical steps.

Conceptual Understanding

The process of elimination is built on the principle of linear combinations. Linear combinations are formed when equations are added or subtracted to create a new equation that is still valid, without altering the original solution set. This is key to the process's effectiveness.

Equivalence is another fundamental concept. Each step in the elimination process produces equivalent equations. This means that if you perform an operation such as adding or subtracting equations, you are simply transforming the system into a simpler form while maintaining the same solution set.

Isolation of Variables

The process of elimination strategically eliminates variables to simplify the system. By systematically reducing the number of equations and variables, you can isolate one variable at a time. Once a variable is isolated, it can be solved for easily, and then substituted back into the original equations to find the values of the other variables.

Steps in the Process

The steps in the process of elimination are straightforward and structured, making it a reliable method for solving systems of equations:

Aligning Coefficients: You adjust the coefficients of one variable in both equations so that when you add or subtract them, that variable cancels out. Adding or Subtracting: You either add or subtract the equations to eliminate one variable, resulting in a single equation with one variable. Solving for One Variable: Once one variable is eliminated, you can solve the resulting equation for the remaining variable. Back Substitution: After finding the value of one variable, you substitute it back into one of the original equations to find the values of the other variables.

Example

Consider the system of equations:

(2x - 3y 6 quad (1))

(4x - y 5 quad (2))

To eliminate one variable, we can multiply equation (1) by 1 and equation (2) by 3:

(2x - 3y 6 quad (1))

(12x - 3y 15 quad (2))

Now, we add equations (1) and (2) to eliminate the (y) variable:

(2x - 3y 12x - 3y 6 15 )

(14x 21 )

Solving for (x), we get:

(21 14x )

((x ) frac{21}{14} )

((x ) frac{3}{2} )

Next, we substitute (x frac{3}{2}) back into equation (1) to find (y):

(2left(frac{3}{2}right) - 3y 6)

(3 - 3y 6)

((-3y 6 - 3 )

((-3y 3 )

((y -1)

Conclusion

The process of elimination works because it allows us to manipulate equations while maintaining their equivalence, simplifying the system step by step until we can easily solve for the variables. This method is particularly useful for larger systems and provides a clear, systematic approach to finding solutions.