#### Answer

$\begin{cases}x_1=-9 \\ x_2=4 \\ x_3 \text{ is free} \end{cases}$

#### Work Step by Step

To find the general solution, we use row operations on the augmented matrix $\left[
\begin{array}{cccc}
1 & 4 & 0 & 7\\
2 & 7 & 0 & 10
\end{array}
\right]$ to obtain an equivalent matrix in reduced row echelon form. Then we solve the corresponding system of equations.
First, we want to eliminate the 2 in the second row since it is below the leading term of the first row. To do this, we use row-replacement. So we replace row 2 with -2*(row 1)+(row 2), which gives
$$\left[
\begin{array}{cccc}
1 & 4 & 0 & 7\\
0 & -1 & 0 & -4
\end{array}
\right].$$
Next, we need to eliminate the 4 above the leading one of the second row. We will again use row-replacement. We replace row 1 with 4*(row 2) + (row 1) to get
$$\left[
\begin{array}{cccc}
1 & 0 & 0 & -9\\
0 & -1 & 0 & -4
\end{array}
\right].$$
Thirdly, we multiply row 2 by -1 to obtain the equivalent matrix
$$\left[
\begin{array}{cccc}
1 & 0 & 0 & -9\\
0 & 1 & 0 & 4
\end{array}
\right].$$
Now we solve the corresponding system of equations, which is
$$x_1=-9\\ x_2=4.$$
From the reduced row echelon form of the matrix, we see that $x_1$ and $x_2$ are the leading variables (since they correspond to the pivot positions of the matrix) and $x_3$ is free. Our system is already solved for the leading variables, and we know $x_3$ is free. So our general solution is
$$\begin{cases}x_1=-9 \\ x_2=4 \\ x_3 \text{ is free} \end{cases}.$$