<s>[INST] Solve the system of equation with answers as x=___+___s Solve the system:
\begin{cases}
-4x_1 + x_2=3 \\
8x_1-2x_2=-6
\end{cases}
I need the solution in the form of:
\begin{cases}
x_1=\underline{\hspace{1cm}}+\underline{\hspace{1cm}}\times s\\[0.3em]
x_2=\underline{\hspace{1cm}}+\underline{\hspace{1cm}}\times s\\[0.3em]
\end{cases}
I've attempted to due this multiple times and end with the same answer each time.
\begin{array}{ccc|c}
-4 & 1 &  & \phantom13 \\
8 & -2 &  &         -6 \\
\end{array}
Multiplying row 2 by $\frac12$:
\begin{array}{ccc|c}
-4 & 1 &  & \phantom13 \\
4 & -1 &  &         -3 \\
\end{array} 
Adding row 1 to row 2:
\begin{array}{ccc|c}
-4 & 1 &  & \phantom13 \\
0 & 0 &  &         0 \\
\end{array}
thus $x_2$  is a free variable, and:
\begin{align*}
x_1 ={}& \frac34-\frac14s \\
x_2 ={}& 0 + 1s
\end{align*}
Is what I get for my answer.
 [/INST] Reading off from your last matrix equation, you have
$$-4x_1 + x_2 = 3 \Rightarrow x_1 = 1/4 x_2 - 3/4$$
Now choose $x_2 = s$. Then $x_2 = 0 + 1s$ is equivalent to making that choice, and substituting $x_2 = s$ gives $x_1 = 1/4s - 3/4$ in the first equation.
$$x_2 = 0 + 1s$$
$$x_1 = - 3/4 + 1/4s$$
Now if $x$ is a vector, $$x = (x_1, x_2) = (-3/4 + 1/4s, 0 + s) = (-3/4, 0) + (1/4, 1)s$$
or written more suggestively,
$$x = 
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix} =
\begin{bmatrix}
-3/4 + 1/4s \\
0 + s
\end{bmatrix} = 
\begin{bmatrix}
-3/4 \\
0 
\end{bmatrix} + 
\begin{bmatrix}
1/4s \\
s
\end{bmatrix} =  
\begin{bmatrix}
-3/4 \\
0 
\end{bmatrix} + 
\begin{bmatrix}
1/4 \\
1
\end{bmatrix}s $$ 
Which is in the form requested.
</s>