Answer
Solution set = $\displaystyle \{(x,y)\ \ | \ \ x=\frac{7}{3}+\frac{1}{3}y, \ \ y\in \mathbb{R}\}$
Work Step by Step
Reduce the augmented matrix $[A|B]$ to reduced row echelon form and interpret the result
$\left[\begin{array}{llll}
3 & -1 & | & 7\\
9 & -3 & | & 21
\end{array}\right]\rightarrow\left(\begin{array}{l}
R_{1}=r_{1}\div 3.\\
R_{2}=r_{2}-3r_{1}
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & -1/3 & | & 7/3\\
0 & 0 & | & 0
\end{array}\right]$
The last equation is always true, the system is consistent (dependent).
Take $y\in \mathbb{R}.$
Equation 1 $\Rightarrow x=\displaystyle \frac{7}{3}+\frac{1}{3}y$
Solution set = $\displaystyle \{(x,y)\ \ | \ \ x=\frac{7}{3}+\frac{1}{3}y, \ \ y\in \mathbb{R}\}$
