Why use elimination method vs. substitution method in linear algebra? What is the motivation to use the elimination method as opposed to the substitution method? I've always found the latter much easier. Are there any examples where the substitution method fails or when it is inferior to the elimination method?
For an example of the substitution method see:
Substitution method review (systems of equations)
 A: There are two aspects which makes the substitution method inferior to the elimination method, and vice versa.


*

*Many people prefer one method to another method due the personal preferences. 

*The original problem  is made to be solved with one method.
In my case, I do prefer the elimination method. So I first of all I look whether the linear equation system can be solved with this method easily or not. An example:
$2x+y=5$
$4x+y=9$
Here it is easy to see that this system can be solved with the elimination method. We just subtract the first equation from the second equation in order to eliminate $y$. 
$2x=4\Rightarrow x=2$
But in another case I would use the substitution method due the structure of the equations:
$\frac23u+4v=14$
$u+v=6$
The second equation can be easily solved for u: $u=6-v$. This equation system might be a case where I would apply the substitution method, although I generally prefer the elimination method.
$\frac23\cdot (6-v)+4v=14$
$4-\frac23v+4v=14$
$-\frac23v+\frac{12}3v=10$
$\frac{10}3v=10\Rightarrow v=3$
A: Note that the elimination method is equivalent to the substitution method, where the choice of the variable for which we solve is constrained.  
Example: Consider the $3 \times 3$ system
$$
\begin{cases}
x + y + z = 2\\
x + 2y + 3z = 5\\
x + 4y + 9z = 15
\end{cases}
$$
With elimination, we would proceed as follows:
$$
\pmatrix{1&1&1&2\\
1&2&3&5\\
1&4&9&15} \leadsto
\pmatrix{1&1&1&2\\
0&1&2&3\\
0&3&8&13} \leadsto
\pmatrix{1&1&1&2\\
0&1&2&3\\
0&0&2&4} \leadsto 
\pmatrix{1&1&1&2\\
0&1&2&3\\
0&0&1&2}.
$$
At this point, we see that the original system of equations can be rewritten as
$$
\begin{cases}
x + y + z = 2\\
\quad \ \  y + 2z = 3\\
\quad \ \ \quad \ \ z = 2.
\end{cases}
$$
If we solve the first equation for $x$, substitute, solve the first of the resulting two equations for $y$, then the result is exactly the same. In particular,
$$
x+y+z = 2 \implies x = 2-y-z.
$$
Substituting this into the second and third equations yields
$$
\begin{cases}
(2-y-z) + 2y + 3z = 5\\
(2-y-z) + 4y + 9z = 15
\end{cases} \implies
\begin{cases}
y + 2z = 3\\
3y + 8z = 13.
\end{cases}
$$
This is exactly what we have after the first step of elimination. Solving the second equation gives us
$$
y + 2z = 3 \implies y = 3 - 2z,\\
3(3-2z) + 8z = 13 \implies 2z = 4.
$$
This is exactly what we have after the second step of elimination.
