Solvability of this linear equation system and finding particular solution/ I have been given a task, that involves determining if this lin.eq.system
$$ x_1+2x_2-3x_3+10x_4-x_5=7\\x_1-2x_2+3x_3-10x_4+x_5=9\\x_1+6x_2-9x_3+30x_4-3x_5=5 $$ has a solution by using what our professor calls Kronecker-Cappelli(no idea how it spells) theorem, which states that if rank of coef. matrix is equal to rank of extended matrix, the system has at least on solution. And if both of those ranks are equal to count of variables in system, then system has the one-and-only solution, if it is less than variable count, then system has infinite number of solutions.
The task is to find:
1)If system has a solution
2)Find systems general(?) solution
3)Find at least one particular solution
I have proven, that the system is solvable, and has infinite amount of solutions. Does that stop me from finding the one particular solution and the general solution?
 A: The best way to do this is to put the system into an augmented matrix and use elementary row operations to find its reduced row echelon form. Here your augmented matrix is
$$
\left[
\begin{array}{rrrrr:r}
1&2&-3&10&-1&7 \\
1&-2&3&-10&1&9 \\
1&6&-9&30&-3&5
\end{array}\right]
$$
and its reduced row echelon form is
$$
\left[
\begin{array}{rrrrr:r}
1 & 0 & 0            & 0 & 0            & 8 \\
0 & 1 & -\frac{3}{2} & 5 & -\frac{1}{2} & -\frac{1}{2} \\
0 & 0 & 0            & 0 & 0            & 0
\end{array}
\right]
$$
The leading $1$'s are in columns one and two. This tells us that $x_3$, $x_4$, and $x_5$ are the free variables. To obtain all of the solutions to the system, we need only solve for $x_1$ and $x_2$ in terms of the free variables. The first row gives the relation
$$
x_1=8
$$
and the second row gives the relation
$$
x_2=-\frac{1}{2}+\frac{3}{2}x_3-5x_4+\frac{1}{2}x_5
$$
Hence all of the solutions to the original system are of the form
$$
\begin{bmatrix}
x_1\\x_2\\x_3\\x_4\\x_5
\end{bmatrix}
=
\begin{bmatrix}
8\\-\frac{1}{2}+\frac{3}{2}x_3-5x_4+\frac{1}{2}x_5\\
x_3\\x_4\\x_5
\end{bmatrix}
$$
That is, we can choose any numbers we like for $x_3$, $x_4$, and $x_5$ and the relations above determine $x_1$ and $x_2$ once these numbers are chosen.
A: Just assume any $2$ variables, say, $x_3$ and $x_4$ to be constants. Now you have $3$ equations and $3$ variables. Solve them. You will get values of $x_1,x_2,x_5$ in terms of those. 
For every pair of $(x_3,x_4)$, you can find a particular solution, if exists.
