# Solve a linear system of three variables and represent the solution in parametric form

Solve the system \begin{cases} \phantom6x_1 + \phantom7x_2 + 3x_3 = 5,\\[0.3em] 6x_1 + 7x_2 - 3x_3 = 5; \end{cases}

I need the solution in the form of \begin{cases} x_1=\color{gray}{\text{blank}}+\color{gray}{\text{blank}}\times s;\\[0.3em] x_2=\color{gray}{\text{blank}}+\color{gray}{\text{blank}}\times s;\\[0.3em] x_3=\color{gray}{\text{blank}}+\color{gray}{\text{blank}}\times s. \end{cases}

Very similar to a question I have previously asked with two variables. My attempt is to set it up in a matrix and eliminate:

\begin{align} &\left[\begin{array}{ccc|c} 1 & 1 & \phantom-3 & 5 \\ 6 & 7 & -3 & 5 \\ \end{array}\right] % \begin{array}{}\\+\text{row}_1\end{array}\\ \iff&\left[\begin{array}{ccc|c} 1 & 1 & 3 & \phantom15 \\ 7 & 8 & 0 & 10 \\ \end{array}\right] % \begin{array}{}\\-7\text{row}_1\end{array}\\ \iff&\left[\begin{array}{ccc|c} 1 & 1 & \phantom{-2}3 & \phantom{-2}5 \\ 0 & 1 & -21 & -25 \\ \end{array}\right] % \begin{array}{}-\text{row}_2\\{}\end{array}\\ \iff&\left[\begin{array}{ccc|c} 1 & 0 & \phantom-24 & \phantom-30 \\ 0 & 1 & -21 & -25 \\ \end{array}\right] % \begin{array}{}\\.\end{array} \end{align}

Would I then end up with the following?

\begin{cases} x_1 = \phantom- 30 - 24s,\\[0.3em] x_2 = -25 + 21s,\\[0.3em] x_3 = \phantom{-2}0 + \phantom20s. \end{cases}

You should set $x_3 = s$.
Think of $x_3$ as a free variable. You started with two equations and three unknowns; this means you have (at least) one free variable. It didn't have to be $x_3$, but it might as well be.
When you write $x_3 = 0+0s$, you are saying $x_3$ has to be zero for the system to hold; while in fact $x_3$ can take on any value, and the system will be satisfied as long as $x_1$ and $x_2$ behave correctly according to $x_3$ (or $s$).
• $x_3 = s$ is the same as $x_3 = 0+1\cdot s$, which is the correct format @davidscidmore – GaussTheBauss Feb 19 '16 at 1:05