Elimination Method

The elimination method is a standard technique for solving systems of linear equations. It works by adding or subtracting the equations term by term so that one of the variables is eliminated. $$ \begin{cases} a_1x_1 + b_1y_1 = c_1 \\ \\ a_2x_2 + b_2y_2 = c_2 \end{cases} $$

This process is called the elimination method because combining the equations reduces the number of variables and equations in the system, simplifying the problem to a single equation in one unknown.

How it works

1. Add or subtract the equations term by term to eliminate one of the unknowns. $$ \begin{matrix} a_1x & +b_1y & =c_1 & + \\ a_2x & +b_2y & =c_2 & \\ \hline (a_1+a_2)x & +(b_1+b_2)y & = (c_1+c_2) \end{matrix} $$
2. Solve the resulting equation for the remaining variable (x or y).
3. Substitute that value into one of the original equations to find the other variable.

Note. If adding or subtracting the equations does not eliminate a variable, apply the invariance property of equations by multiplying or dividing both sides of one equation by a suitable nonzero constant so that the coefficients of one variable match (or are opposites).

The elimination method can be applied to any system of linear equations with two or more equations and two or more unknowns.

A worked example

Let’s solve the following system of two linear equations in the unknowns x and y:

$$ \begin{cases} 2x+y=4 \\ \\ 3x+2y=2 \end{cases} $$

Adding or subtracting these equations directly does not eliminate any variable.

Note. Adding the two equations gives \(5x + 3y = 6\): $$ \begin{matrix} 2x & +y & =4 & + \\ 3x & +2y & =2 & \\ \hline 5x & +3y & = 6 \end{matrix} $$ Subtracting them instead gives \(x - y = 2\): $$ \begin{matrix} 2x & +y & =4 & - \\ 3x & +2y & =2 & \\ \hline x & -y & = 2 \end{matrix} $$

Notice that the coefficient of \(y\) in the first equation is half that in the second equation.

$$ \begin{cases} 2x+y=4 \\ \\ 3x+2y=2 \end{cases} $$

Using the invariance property, multiply both sides of the first equation by 2 so the coefficients of \(y\) match.

$$ \begin{cases} 2 \cdot (2x+y)= 2 \cdot 4 \\ 3x+2y=2 \end{cases} $$

$$ \begin{cases} 4x+2y= 8 \\ 3x+2y=2 \end{cases} $$

Now the coefficient of \(y\) is the same in both equations, making it easy to eliminate.

Note. The choice of which variable to eliminate is arbitrary. Typically, you choose the one that can be removed with the fewest arithmetic steps.

Now subtract the second equation from the first to eliminate \(y\).

$$ \begin{matrix} 4x & +2y & =8 & - \\ 3x & +2y & =2 & \\ \hline x & & = 6 \end{matrix} $$

After this step, the system reduces to a single linear equation in one unknown, since \(y\) has been eliminated.

This immediately gives the value of \(x\).

$$ x=6 $$

Note. In this example, no additional algebraic manipulation is required to isolate \(x\). In other cases, some algebraic simplification might be needed, but the process is still straightforward because only one variable remains.

Once we know that \(x=6\), we substitute this value into one of the original equations to determine \(y\).

$$ \begin{cases} 2x+y=4 \\ \\ 3x+2y=2 \\ \\ x=6 \end{cases} $$

For instance, substituting \(x=6\) into the first equation and discarding the second yields:

$$ \begin{cases} 2(6)+y=4 \\ \\ \require{cancel} \cancel{3x+2y=2} \\ \\ x=6 \end{cases} $$

$$ \begin{cases} 12+y=4 \\ \\ x=6 \end{cases} $$

Note. It doesn’t matter which equation you use for substitution. You could use either one and obtain the same result. The best choice is usually the equation that leads to the simplest algebraic work.

Solving for \(y\) gives y = -8.

$$ \begin{cases} y=4-12 \\ \\ x=6 \end{cases} $$

$$ \begin{cases} y=-8 \\ \\ x=6 \end{cases} $$

Therefore, the solution to the system of linear equations is \(x=6\) and \(y=-8\).