Answer
$x_1=\frac{5+3t}{7}$
$x_2=t$
Work Step by Step
The augmented matrix of the system is:
$\begin{bmatrix}
7& -3 |5\\
14 & -6| 10
\end{bmatrix}$
with reduced row-echelon form:
$\begin{bmatrix}
7& -3 |5\\
14 & -6| 10
\end{bmatrix}\approx^1\begin{bmatrix}
7& -3 |5\\
0 & 0| 0
\end{bmatrix} \approx^2\begin{bmatrix}
1& -\frac{3}{7} |\frac{5}{7}\\
0 & 0| 0
\end{bmatrix}$
The equivalent system is:
$7x_1-3x_2=5$
Since $x_2$ does not occur in the system, it is a free variable and therefore not necessarily zero:
$x_2=t$
$7x_1-3t=5 \rightarrow x_1=\frac{5+3t}{7}$
