3 Equations, 4 Variables, $\infty$ Solutions

In this solution, we will use Row Reduction Techniques to simplify the system.

First, we will begin by writing the system as an augmented matrix.

$\left[ \begin{array}{c c c c | c} 2 & 4 & 0 & 1 & 7\\ 5 & 2 & 1 & 2 & 0\\ 3 & 3 & 1 & 1 & 8\\ \end{array}\right]$

We will make the pivot in the 1st column by dividing the 1st row by 2.

$\left[ \begin{array}{c c c c | c} 2 & 4 & 0 & 1 & 7\\ 5 & 2 & 1 & 2 & 0\\ 3 & 3 & 1 & 1 & 8\\ \end{array}\right] \ce{->[\large \text{Divide the first row by } 2]} \left[ \begin{array}{c c c c| c} 1 & 2 & 0 & 1/2 & 7/2\\ 5 & 2 & 1 & 2 & 0\\ 3 & 3 & 1 & 1 & 8\\ \end{array} \right]$

$\left[ \begin{array}{c c c c | c} 1 & 2 & 0 & 1/2 & 7/2\\ 5 & 2 & 1 & 2 & 0\\ 3 & 3 & 1 & 1 & 8\\ \end{array}\right] \ce{->[\large R_{2} - 5R_{1} \text{ and } R_{3} - 3 R_{1}]} \left[ \begin{array}{c c c c| c} 1 & 2 & 0 & 1/2 & 7/2\\ 0 & -8 & 1 & -1/2 & -35/2\\ 0 & -3 & 1 & -1/2 & -5/2\\ \end{array} \right]$

Now, we make the pivot in the second row by dividing the second row by -8

$\left[ \begin{array}{c c c c | c} 1 & 2 & 0 & 1/2 & 7/2\\ 0 & -8 & 1 & -1/2 & -35/2\\ 0 & -3 & 1 & -1/2 & -5/2\\ \end{array}\right] \ce{->[\large \text{Divide the second row by } -8]} \left[ \begin{array}{c c c c| c} 1 & 2 & 0 & 1/2 & 7/2\\ 0 & 1 & -1/8 & 1/16 & 35/16\\ 0 & -3 & 1 & -1/2 & -5/2\\ \end{array} \right]$

$\left[ \begin{array}{c c c c | c} 1 & 2 & 0 & 1/2 & 7/2\\ 0 & 1 & -1/8 & 1/16 & 35/16\\ 0 & -3 & 1 & -1/2 & -5/2\\ \end{array}\right] \ce{->[\large R_{1} - 2R_{2} \text{ and } R_{3} + 3 R_{2}]} \left[ \begin{array}{c c c c| c} 1 & 0 & 1/4 & 3/8 & -7/8\\ 0 & 1 & -1/8 & 1/16 & 35/16\\ 0 & 0 & 5/8 & -5/16 & 65/16\\ \end{array} \right]$

Now, we make the pivot in the third row by dividing the third row by 5/8

$\left[ \begin{array}{c c c c | c} 1 & 0 & 1/4 & 3/8 & -7/8\\ 0 & 1 & -1/8 & 1/16 & 35/16\\ 0 & 0 & 5/8 & -5/16 & 65/16\\ \end{array}\right] \ce{->[\large \text{Divide the third row by } 5/8]} \left[ \begin{array}{c c c c| c} 1 & 0 & 1/4 & 3/8 & -7/8\\ 0 & 1 & -1/8 & 1/16 & 35/16\\ 0 & 0 & 1 & -1/2 & 13/2\\ \end{array} \right]$

$\left[ \begin{array}{c c c c | c} 1 & 0 & 1/4 & 3/8 & -7/8\\ 0 & 1 & -1/8 & 1/16 & 35/16\\ 0 & 0 & 1 & -1/2 & 13/2\\ \end{array}\right] \ce{->[\large R_{1} - \frac{1}{4}R_{3} \text{ and } R_{2} + \frac{1}{8} R_{3}]} \left[ \begin{array}{c c c c| c} 1 & 0 & 0 & 1/2 & -5/2\\ 0 & 1 & 0 & 0 & 3\\ 0 & 0 & 1 & -1/2 & 13/2\\ \end{array} \right]$

The solutions to this system of equations are

$\left[ \begin{array}{c} w \\ x \\ y \\ z \end{array} \right] = \left[ \begin{array}{c} -5/2 \\ 3 \\ 13/2 \\ 0 \end{array} \right] + t \left[ \begin{array}{c} -1 \\ 0 \\ 1 \\ 2 \end{array} \right]$

Here, $t$ is the parameter. As we change the parameter $t$ , we see that the value of $w, y,$ and $z$ change but $x$ is always equal to 3. Therefore the fixed variable is $\boxed{ x }$ . $_\square$