Solving systems of two linear equations

A single linear equation in $x$ and $y$ is satisfied by every point on its graph, so to fix one pair of values we need a second equation that holds at the same time.

For example, consider the following system with two variables, $x$ and $y$:

Here:

• Each line is a linear equation in the variables $x$ and $y$.
• The system consists of both equations together.
• A solution is a pair of values for $x$ and $y$ that satisfy both equations at once.

Algebraic representation

We often write a system of equations by listing all the equations together, often using a curly brace to show they are part of the same system:

Visual representation

We can also represent a system of two equations in two variables by drawing both lines on the same graph. Each equation corresponds to a straight line. The point where the two lines cross (if they do) is the solution to the system, because it is the point that satisfies both equations.

For the example above, the two lines intersect at a single point. This point is the solution to the system.

Below is an interactive graph showing both equations. The blue and red lines represent the two equations. The point where they intersect, $(3, 1)$, is the solution to the system. This means that $x = 3$ and $y = 1$ makes both equations true.

Have a play around by modifying the lines and watch the system update.

When we have a system of two linear equations with two variables, we can solve it using the substitution method. This method involves the following steps:

1. Solve one equation for one variable.

   We start by choosing one of the equations and rearranging it so that one variable is written in terms of the other. For example, consider the system:

   $$\begin{cases} x + y = 5 \\ 2x - y = 4 \end{cases}$$

   We can solve the first equation for $x$:

   $$x = 5 - y$$

2. Substitute this expression into the other equation.

   Next, we substitute $x = 5 - y$ into the second equation wherever $x$ appears:

   $$\begin{align*} 2x - y &= 4 \\ 2(5 - y) - y &= 4 \end{align*}$$

3. Solve for the remaining variable, then back-substitute.

   Now we solve for $y$:

   $$\begin{align*} 2 \times 5 - 2y - y &= 4 \\ 10 - 2y - y &= 4 \\ 10 - 3y &= 4 \\ 3y &= 6 \\ y &= 2 \end{align*}$$

   Now substitute $y = 2$ back into the expression for $x$:

   $$\begin{align*} x &= 5 - y \\ x &= 5 - 2 \\ x &= 3 \end{align*}$$

So the solution to the system is:

The steps we followed here form the general method.

Solve the system of equations:

using the substitution method.

Solution

Step 1: The first equation is already solved for $y$:

Step 2: Substitute $y$ into the second equation by replacing $y$ with $2x + 1$:

Step 3: Solve for $x$:

Step 4: Substitute $x$ back into the expression for $y$:

So the solution to the system is:

When solving a system of two linear equations with two variables, we can also use the elimination method (also called the addition or subtraction method). The idea is to add or subtract the equations so that one of the variables is eliminated, allowing us to solve for the other variable.

Suppose we have the system:

We use elimination when the coefficients of one variable are equal or opposites (possibly after multiplying one or both equations by a constant). This allows us to combine the equations so that one variable cancels out.

Note:

To add or subtract two equations, we perform the operation on both sides of the equations. For example, adding the equations means adding the left sides together and the right sides together:

Solve the system:

Solution

Since the coefficients of $x$ are already equal and the same sign, we can subtract the second equation from the first to eliminate $x$. Note that it does not matter which equation we subtract from the other, as long as we are consistent on both sides - the solution will be the same.

Here, we chose to subtract the second equation from the first equation so that the right-hand side stays positive, but this is just a preference.

Now substitute $y = 1$ into one of the original equations (for example, the second):

So the solution to the system is: