Answer
Solution set: $\{( -2t+2$ , $2t+\displaystyle \frac{1}{2}$ , $t )$ , $t\in \mathbb{R}\}$
Work Step by Step
Start with the augmented matrix.
Use matrix row operations to get ls down the main diagonal from upper left to lower right, and 0s below the $1\mathrm{s}$ (row-echelon form).
$\left[\begin{array}{lllll}
5 & 8 & -6 & | & 14\\
3 & 4 & -2 & | & 8\\
1 & 2 & -2 & | & 3
\end{array}\right]$
Swapping R1 and R3 brings a 1 into the first diagonal position
$\left[\begin{array}{lllll}
1 & 2 & -2 & | & 3\\
5 & 8 & -6 & | & 14\\
3 & 4 & -2 & | & 8
\end{array}\right]\left\{\begin{array}{l}
.\\
-5R1+R2\rightarrow R2.\\
-3R1+R3\rightarrow R3.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 2 & -2 & | & 3\\
0 & -2 & 4 & | & -1\\
0 & -2 & 4 & | & -1
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
-R2+R3\rightarrow R3.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 2 & -2 & | & 3\\
0 & -2 & 4 & | & -1\\
0 & 0 & 0 & | & 0
\end{array}\right]$
Last row: all zeros,
the system is consistent.
We take z to be any nuber $t\in \mathbb{R}.$
Substituting into row 2:
$-2y+4(t)=-1$
$-2y=-4t-1\qquad/\div(-2)$
$y=2t+\displaystyle \frac{1}{2}$
Substituting into row 1,
$x+2(2t+\displaystyle \frac{1}{2})-2(t)=3$
$x+4t+1-2t=3$
$x=-2t+2$
Solution set: $\{( -2t+2$ , $2t+\displaystyle \frac{1}{2}$ , $t )$ , $t\in \mathbb{R}\}$
