Find $(a, b, c)$ that satisfies the system of equations $2a + 3b -4c = 7, a-b+2c=6$ $\begin{align} 
2a + 3b -4c = 7, \\
a - b + 2c = 6
\end{align}$
I haven't seen a three-variable system of equations with only two equations before. I try to make the problem into a two-variable system.
Solving for $a$ in the first equation
$\begin{align} 
a = \frac{-3b +4c+7}{2}
\end{align}$.
Solving for $a$ in the second equation:
$\begin{align} 
a = b-2c+6
\end{align}$.
Then substitute the equations into $a$ for the original equations.
$\begin{align} 
2(b-2c+6) + 3b -4c = 7 \\
2b-4c+12+3b-4c=7 \\
5b-8c = -5
\end{align}$
$\begin{align} 
\frac{-3b +4c+7}{2} - b + 2c = 6 \\
-3b+4c+7-2b+4c=12 \\
-5b+8c=5
\end{align}$
This create a new system of equations:
$\begin{align} 
5b-8c =-5 \\
-5b+8c=5
\end{align}$
This leads to the solution $0=0$. So I assume this means I can select any $b$ and $c$. This however isn't the case, I can't find $a$ that satisfies both equations with this premise.
The problem additionally asks to solve for $b$ in terms of $a$ and solve for $c$ in terms of $a$. I did so using elimination.
$\begin{align} 
b = -4a +19 \\
c = \frac{-5a+25}{2}
\end{align}$
So we have both $b$ and $c$ depending on $a$. There is an infinite number of choices for $a$ so there are infinitely many solutions to this system.
Could I not find one or more ordered triples that satisfy both equations if I did not (following the instruction) create this final system of equations? Until now I thought the number of equations in the system was simply related to the number of variables.

First, see my comment, following your query.
Since you are trying, not only to solve the equations, but also to express
$b$ and $c$ in terms of $a$, I would have approached it differently.
$$2a + 3b - 4c = 7. \tag1$$
$$a - b + 2c = 6. \tag2$$
Multiplying equation (2) above by $(3)$ gives
$$3a - 3b + 6c = 18.\tag3$$
Adding equations (3) and (1):
$$5a + 2c = 25 \implies c = \frac{25 - 5a}{2}. \tag4$$
Now, repeat the process to express $b$ in terms of $a$. 
Multiplying equation (2) by $(2)$ gives
$$2a - 2b + 4c = 12. \tag5$$
Adding equations (5) and (1):
$$4a + b = 19 \implies b = 19 - 4a. \tag6$$
At this point, you are actually done.  $(a)$ can be chosen to be any number.
Once $(a)$ is chosen, $(b)$ and $(c)$ are computed via equations (4) and (6) above.
Then, you have generated a solution $(a,b,c).$
