The system has a corresponding augmented matrix
[tex]\left[\begin{array}{ccccc|c}2&-4&3&-1&2&0\\3&-6&5&-2&4&0\\5&-10&7&-3&18&0\end{array}\right][/tex]
Let's get this into reduced row echelon form.
Add 3(row 1) to -2(row 2), and add 5(row 1) to -2(row 3):
[tex]\left[\begin{array}{ccccc|c}2&-4&3&-1&2&0\\0&0&-1&1&-2&0\\0&0&1&1&-26&0\end{array}\right][/tex]
Multiply through row 2 by -1:
[tex]\left[\begin{array}{ccccc|c}2&-4&3&-1&2&0\\0&0&1&-1&2&0\\0&0&1&1&-26&0\end{array}\right][/tex]
Add -1(row 2) to row 3:
[tex]\left[\begin{array}{ccccc|c}2&-4&3&-1&2&0\\0&0&1&-1&2&0\\0&0&0&2&-28&0\end{array}\right][/tex]
Multiply through row 3 by 1/2:
[tex]\left[\begin{array}{ccccc|c}2&-4&3&-1&2&0\\0&0&1&-1&2&0\\0&0&0&1&-14&0\end{array}\right][/tex]
Add row 3 to row 2:
[tex]\left[\begin{array}{ccccc|c}2&-4&3&-1&2&0\\0&0&1&0&-12&0\\0&0&0&1&-14&0\end{array}\right][/tex]
Add -3(row 2) and row 3 to row 1:
[tex]\left[\begin{array}{ccccc|c}2&-4&0&0&24&0\\0&0&1&0&-12&0\\0&0&0&1&-14&0\end{array}\right][/tex]
Multiply through row 1 by 1/2:
[tex]\left[\begin{array}{ccccc|c}1&-2&0&0&12&0\\0&0&1&0&-12&0\\0&0&0&1&-14&0\end{array}\right][/tex]
This tells us
[tex]\begin{cases}x_1-2x_2+12x_5=0\\x_3-12x_5=0\\x_4-14x_5=0\end{cases}[/tex]
The matrix has rank 3, so we can pick 2 free variables, say [tex]x_1[/tex] and [tex]x_2[/tex]. Then the general solution to the system is
[tex]\begin{cases}x_1=s\\x_2=t\\x_3=2t-s\\x_4=\frac{7(2t-s)}6\\x_5=\frac{2t-s}{12}\end{cases}[/tex]
where [tex]s,t\in\Bbb R[/tex].
