Comparison Method for Solving Systems of Equations

The comparison method is a classical algebraic procedure for solving a system of linear equations. $$ \begin{cases} a_1x_1 + b_1y_1 = c_1 \\ \\ a_2x_2 + b_2y_2 = c_2 \end{cases} $$

How it works

1. Start by isolating the same variable in both equations, typically y. $$ \begin{cases} y_1 = - \frac{a_1}{b_1} x_1 - \frac{ c_1 }{b_1} \\ \\ y_2 = - \frac{a_2}{b_2} x_2 - \frac{ c_2 }{b_2} \end{cases} $$ This rewrites each equation in an explicit form that makes the functional dependence clear.
2. Set the two expressions for y equal to each other, forming the auxiliary equation y=y. Solving this new equation gives the value of x. $$ \begin{cases} y_1 = - \frac{a_1}{b_1} x_1 - \frac{ c_1 }{b_1} \\ \\ y_2 = - \frac{a_2}{b_2} x_2 - \frac{ c_2 }{b_2} \\ \\ y_1 = y_2 \end{cases} $$ This step collapses the system to a single linear equation in one unknown.
3. Substitute the resulting value of x into either of the original equations to compute y.

Nota. The choice of which variable to isolate is arbitrary. You may solve for x or y depending on which leads to cleaner algebra. In practice, one opts for the expression that minimizes computational effort, in line with the pragmatic reading of Occam’s razor.

The comparison method applies to any first degree linear system with two or more equations and variables, provided the system is consistent and admits a unique solution.

A practical example

Consider the following linear system in the variables x and y.

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

We will solve it using the comparison method.

First, isolate y in both equations.

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

Rewrite the second equation in explicit form.

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

Break down the right-hand side to make the structure explicit.

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

$$ \begin{cases} y=4-2x \\ \\ y=1-\frac{3x}{2} \end{cases} $$

We now have two linear expressions defining y as a function of x.

This allows us to introduce a third equation, y=y, by equating the two expressions.

$$ \begin{cases} y=4-2x \\ \\ y=1-\frac{3x}{2} \\ \\ y=y \end{cases} $$

In the equality y=y, substitute y=4-2x on the left and y=1-3x/2 on the right.

This eliminates y completely and yields an equation solely in x.

$$ \begin{cases} y=4-2x \\ \\ y=1-\frac{3x}{2} \\ \\ 4-2x=1-\frac{3x}{2} \end{cases} $$

Now solve the resulting linear equation.

$$ 4-2x=1-\frac{3x}{2} $$

Move all x-terms to one side.

$$ \frac{3x}{2}-2x=1-4 $$

Simplify.

$$ \frac{3x-4x}{2}=-3 $$

$$ \frac{-x}{2}=-3 $$

Multiply both sides by 2.

$$ \frac{x}{2}=3 $$

$$ x=3 \cdot 2 $$

$$ x=6 $$

The system now becomes

$$ \begin{cases} y=4-2x \\ \\ y=1-\frac{3x}{2} \\ \\ x=6 \end{cases} $$

Substitute x=6 into either equation to find y.

For convenience, use the first equation and disregard the second.

$$ \begin{cases} y=4-2x \\ \\ \require{cancel} \cancel{ y=1-\frac{3x}{2} } \\ \\ x=6 \end{cases} $$

$$ \begin{cases} y=4-2 \cdot 6 \\ \\ x=6 \end{cases} $$

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

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

Therefore, the unique solution of the system is the ordered pair x=6 and y=-8.

Nota. Substituting x=6 into the second equation yields the same result, as expected for a consistent and determined linear system. $$ \begin{cases} \require{cancel} \cancel{ y=4-2x } \\ \\ y=1-\frac{3x}{2} \\ \\ x=6 \end{cases} $$ $$ \begin{cases} y=1-\frac{3 \cdot 6}{2} \\ \\ x=6 \end{cases} $$ $$ \begin{cases} y=1-\frac{18}{2} \\ \\ x=6 \end{cases} $$ $$ \begin{cases} y=1-9 \\ \\ x=6 \end{cases} $$ $$ \begin{cases} y=-8 \\ \\ x=6 \end{cases} $$ The solution is therefore consistently x=6 and y=-8.

And so on.