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Section 6.4 Special Cases of Systems of Equations

In the last sections, we used two different strategies to combine the equations to get rid of one of the variables so we could solve for the other. In this section, we will look at what happens when both variables cancel out during that process.

In this section, we will use substitution in the examples because that is the author's preferred strategy. However, these two special cases will also happen when you use elimination, and the way you identify them is exactly the same.

Example 6.14.

Let's try to solve the system of equations

\begin{equation*} \begin{cases} y \amp= 3x-1\\ 5 \amp= -3x+y\\ \end{cases} \end{equation*}

Since we already have \(y\) by itself in the first equation, we will plug that into the second equation:

\begin{align*} 5 \amp= -3x+y\\ 5 \amp= -3x+(3x-1)\\ 5 \amp= -3x + 3x - 1\\ 5 \amp= -1 \end{align*}

When we tried to simplify this equation, both of the \(x\)'s canceled out, and we are left with a nonsense statement. Since \(5\) is never equal to \(-1\text{,}\) there is no way to make both equations true at the same time. Thefore, this system has no solution.

Example 6.15.

Let's try to solve the system of equations

\begin{equation*} \begin{cases} x \amp= 3-y\\ -2x \amp= 2y-6 \end{cases} \end{equation*}

Since we already have \(x\) by itself in the first equation, we will plug that into the second equation:

\begin{align*} -2x \amp= 2y-6\\ -2(3-y)\amp=2y-6\\ -6 + 2y\amp=2y-6\\ -6 \amp=-6 \end{align*}

When we tried to simplify this equation, both of the \(y\)'s canceled out, and we are left with a true statement. Therefore, there are lots of ways to make both equations work at the same time. In fact, this system has infinitely many solutions.

Checkpoint 6.17.

Solve the following system of equations.

\begin{equation*} \begin{cases} 5 \amp= 2x-y\\ -10 \amp= 2y-4x \end{cases} \end{equation*}
Answer.

Infinitely many solutions

Solution.

Let's start by solving the first equation for \(y\text{:}\)

\begin{align*} 5 \amp= 2x-y\\ 5 - 2x\amp= -y\\ 2x-5 \amp= y \end{align*}

Now, let's plug that into the second equation:

\begin{align*} -10 \amp= 2y - 4x\\ -10 \amp= 2(2x-5) - 4x\\ -10 \amp= 4x - 10 - 4x\\ -10 \amp= -10 \end{align*}

All of our variables canceled out, and we are left with a true statement. Therefore, this system has infinitely many solutions.

Checkpoint 6.18.

Solve the following system of equations:

\begin{equation*} \begin{cases} x \amp= 5y-4\\ 9 \amp= 5y - x \end{cases} \end{equation*}
Answer.

No solution

Solution.

Since the first equation already has \(x\) by itself, let's plug that into the second equation:

\begin{align*} 9 \amp= 5y-x\\ 9 \amp= 5y-(5y-4)\\ 9 \amp= 5y - 5y + 4\\ 9 \amp=4 \end{align*}

All of our variables canceled out, and we are left with a nonsense statement. Therefore, this system has no solution.