We labeled the equations this time to be able to keep track of things a little more easily. The most obvious first step here is to eliminate [latex]y[/latex] by adding equations (2) and (3).

Now we can substitute the value for [latex]z[/latex] that we obtained into equation [latex](2)[/latex].

[latex]\begin{array}{rrr}-2y+(6)=6\\-2y=6-6\\-2y=0\\\,\,\,\,y=0\end{array}[/latex]

Be careful here not to get confused with a solution of [latex]y = 0[/latex] and an inconsistent solution.  It is ok for variables to equal [latex]0[/latex].

Now we can substitute [latex]z = 6[/latex] and [latex]y = 0[/latex] back into the first equation.

[latex]\begin{array}{rrr}x-y+z=5\\x-0+6=5\\x+6=5\\x=5-6\\x=-1\end{array}[/latex]

Solution: [latex]{(x,y,z)}={(-1,0,6)}[/latex]

There will always be several choices as to where to begin, but the most obvious first step here is to eliminate [latex]x[/latex] by adding equations (1) and (2).

[latex]\begin{align}x - 2y+3z&=9\\ -x+3y-z&=-6 \\ \hline y+2z&=3 \end{align}[/latex][latex]\hspace{5mm}\begin{gathered}\text{(1})\\ \text{(2)}\\ \text{(4)}\end{gathered}[/latex]

The second step is multiplying equation (1) by [latex]-2[/latex] and adding the result to equation (3). These two steps will eliminate the variable [latex]x[/latex].

[latex]\begin{align}−2x+4y−6z&=−18 \\ 2x−5y+5z&=17 \\ \hline −y−z&=−1\end{align}[/latex][latex]\hspace{5mm}\begin{align}&\text{(2) multiplied by }−2\\&\left(3\right)\\&(5)\end{align}[/latex]

In equations (4) and (5), we have created a new two-by-two system. We can solve for [latex]z[/latex] by adding the two equations.

[latex]\begin{align}y+2z&=3 \\ -y-z&=-1 \\ \hline z&=2 \end{align}[/latex][latex]\hspace{5mm}\begin{align}(4)\\(5)\\(6)\end{align}[/latex]

Choosing one equation from each new system, we obtain the upper triangular form:

[latex]\begin{align}x - 2y+3z&=9 && \left(1\right) \\ y+2z&=3 && \left(4\right) \\ z&=2 && \left(6\right) \end{align}[/latex]

Next, we back-substitute [latex]z=2[/latex] into equation (4) and solve for [latex]y[/latex].

[latex]\begin{align}y+2\left(2\right)&=3 \\ y+4&=3 \\ y&=-1 \end{align}[/latex]

Finally, we can back-substitute [latex]z=2[/latex] and [latex]y=-1[/latex] into equation (1). This will yield the solution for [latex]x[/latex].

[latex]\begin{align} x - 2\left(-1\right)+3\left(2\right)&=9\\ x+2+6&=9\\ x&=1\end{align}[/latex]

The solution is the ordered triple [latex]\left(1,-1,2\right)[/latex].