# Finding a particular solution to a non-homogeneous system of equations

If one asked to solve the set of equation below with the associated homogenous system, I'd know how to do it.

$$S \leftrightarrow \begin{cases} 3x + 5y + z = 8\\\ x + 2y - 2z = 3 \end{cases}$$

$$S' \leftrightarrow \begin{cases} 3x + 4y + z = 0\\\ x + 2y - 2z = 0 \end{cases}$$

You'd find the solution of the homogeneous system $S'$ to be: $$(x, y, z) = \{ k\cdot (-12, 7, 1) | k \in \mathbb{R} \}$$

With the particular solution of $S$... $$(x, y, z) = (1, 1, 0)$$

You can count them up and you'd find: $$(x, y, z) = \{(1 - 12k, 1+ 7k, k)|k \in \mathbb{R}\}$$

And your original system of equations $S$ is solved.

Now I've got one question: how do you find such a particular solution to a non-homogeneous system of equations. How do you find $(1, 1, 0)$ in this case?

Another example:
How do I find one particular solution to this non homogeneous system? \begin{cases} x_1 + x_2 +x_3 =4\\ 2x_1 + 5x_2 - 2x_3 = 3 \end{cases}

-
Give $z$ a particular value, then solve the resulting $2 \times 2$ system. – David Mitra Dec 18 '11 at 15:26

Just set $z=0$, say. With a bit of luck, you'll be able to solve the resulting system: \eqalign{ 3x+5y&=8\cr x+2y&=3 }
The solution of the above system is $y=1 , x=1$; so, a solution to the original equation is $(1, 1 , 0)$.
For your second question, do a similar thing. Set $x_2=0$. Then you can conclude $x_1=11/4$ and $x_3=5/4$.
Yes my particular solution is wrong: I accidentally wrote 4 instead of 5 with the $y$-value of the first equation. – user21385 Dec 18 '11 at 15:55