A system of equations is a set of two or more algebraic equations that are meant to be solved together. The equations are connected because they share the same variables. A common example is a system with two equations and two variables, $x$ and $y$. That’s what this chapter focuses on.

Identifying a system of equations

Solving a system of equations takes a specific method, so it helps to confirm that the problem really is a system before you start.

There are two main clues:

• The problem includes two different equations.
• Both equations use the same two variables.

Examples of problems involving systems of equations are as follows:

Solve the following system of equations for $x$ and $y$.

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

Here’s another example:

What is the sum of the solutions to the following system of equations? (The solutions refer to the values of $x$ and $y$.)

$$\begin{array}{rl}-3y& =x+4\\ 12& =3x-2y\end{array}$$

Solving a system of equations - substitution method

One way to solve a system is to use one equation to rewrite one variable in terms of the other, and then substitute that expression into the second equation. This is called the substitution method.

In the steps below, we will reference the following example system of equations:

$$\begin{array}{rl}x+y& =7\\ y-3& =3x\end{array}$$

Step 1: Solve the first equation for one of the variables

Start with one equation (here, the first one) and solve for one variable. In this example, solve for $x$:

$$\begin{array}{rl}x+y& =7\\ x& =7-y\end{array}$$

Step 2: Substitute new expression for $x$ in second equation

Now replace $x$ in the second equation with the expression you found. This creates an equation with only one variable.

$$\begin{array}{rl}y-3& =3x\\ y-3& =3\ast (7-y)\end{array}$$

Step 3: Solve for $y$ in the second equation

Solve the new equation for $y$. This gives you one of the two solution values.

$$\begin{array}{rl}y-3& =3\ast (7-y)\\ y-3& =21-3y\\ 4y& =24\\ y& =6\end{array}$$

Step 4: Solve for $x$ in either equation

Now substitute $y=6$ into either original equation to find $x$. Both equations should give the same value for $x$.

First equation:

$$\begin{array}{rl}y-3& =3x\\ 6-3& =3x\\ x& =1\end{array}$$

Second equation:

$$\begin{array}{rl}x+y& =7\\ x+6& =7\\ x& =1\end{array}$$

Our final solutions to the system of equations are:

$$\begin{gathered} x=1 \\ y=6 \end{gathered}$$

Solving a system of equations - elimination method

The elimination method combines the two equations to eliminate one variable. To do this, you adjust the equations so that one variable has the same coefficient in both equations but opposite signs. Then, when you add the equations, that variable cancels out.

Step 1: Pick the variable to eliminate

Choose a variable that will be easy to eliminate. In this example, $y$ is a good choice because both equations contain $+y$.

Let’s rewrite the second equation so both equations are in a similar form.

Original equations:

$$\begin{array}{rl}x+y& =7\\ y-3& =3x\end{array}$$

Rearranged equations:

$$\begin{array}{rl}x+y& =7\\ -3x+y& =3\end{array}$$

Looking at our equations this way, you can see that both equations contain $+y$.

Step 2: Multiply by a factor (if necessary)

To eliminate $y$ by addition, one equation needs $+y$ and the other needs $-y$. In this example, multiply the entire second equation by $-1$.

Rearranged equations:

$$\begin{array}{rl}x+y& =7\\ -3x+y& =3\end{array}$$

Modified equations:

$$\begin{array}{rl}x+y& =7\\ 3x-y& =-3\end{array}$$

Now the $y$ terms will cancel when you add the equations.

Step 3: Add equations

Add the two equations and combine like terms. This leaves you with one equation in one variable.

$$\begin{array}{rl}(x+y+3x-y)& =(7-3)\\ 4x& =4\\ x& =1\end{array}$$

Now you have the exact value of $x$.

Step 4: Solve for $y$ in an original equation

Substitute $x=1$ into either original equation to find $y$.

$$\begin{array}{rl}x+y& =7\\ 1+y& =7\\ y& =6\end{array}$$

So, our final solutions to the system equations are:

$$\begin{gathered} x=1 \\ y=6 \end{gathered}$$

Other solutions to a system of equations

Sometimes a system of equations does not have exactly one solution. A system can have:

• infinitely many solutions, or
• no real solutions.

You can recognize both situations by what happens when you simplify.

Infinite solutions

A system has infinitely many solutions when the two equations represent the same relationship. This happens when:

• the equations are identical, or
• one equation is a constant multiple of the other.

Below are a few examples of systems with infinite solutions:

Example:

$$\begin{gathered} x+y=3 \\ y+x=3 \end{gathered}$$

Another example:

$$\begin{array}{rl}y-x& =2\\ 2y-2x& =4\end{array}$$

You can often predict infinite solutions before fully solving, but you can also confirm it during solving. When you solve these special systems, you end up with a true statement.

Let’s solve the first example using substitution by solving for $x$:

$$\begin{array}{rl}x+y& =3\\ x& =3-y\\ (3-y)+y& =3\\ 3& =3\end{array}$$

The statement $3=3$ is always true, no matter what values $x$ and $y$ take. That’s why the system has infinitely many solutions.

No real solutions

A system has no real solutions when simplifying leads to a false statement, such as $3=7$. Sometimes you only discover this at the end of solving. Other times, elimination removes both variables at once and reveals the contradiction.

The following is an example of a system with no real solutions:

$$\begin{array}{rl}x+2y& =2\\ 4x+8y& =3\end{array}$$

Using elimination, multiply the top equation by $-4$. Then add the equations. The simplification becomes:

$$\begin{array}{rl}(-4x-8y+4x-8y)& =(-8+3)\\ 0& =-5\end{array}$$

The statement $0=-5$ is always false, regardless of the values of $x$ and $y$. Therefore, this system has no real solutions.