If

\begin{align*}
3x+4y-12z&=10,\\
-2x-3y+9z&=-4,
\end{align*}

compute $x$.
To solve for $x$, I need to eliminate $y$ and $z$ from the equations. I notice that the coefficients of $y$ and $z$ are proportional in both equations, so I can use that to eliminate them. If I multiply the first equation by 3 and the second by 4, I get

\begin{align*}
9x+12y-36z&=30,\\
-8x-12y+36z&=-16.
\end{align*} Now I can add the two equations and get 9x-8x+12y-12y-36z+36z=30-16, x=14. 