<s>[INST] 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.
 [/INST] 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 $(a,b,c).$
</s>