How to find the value(s) of $a$ and $b$ such that the following system of equations... Find the value(s) of $a$ and $b$ such that the following system of equations,
$\begin{cases}x - 2y = a \\
3x - 6y = b\end{cases}$
(1) is inconsistent. (2) Which values of a and b make this system consistent and further, (3) what can you say about the number of solutions to the system.
I'm not sure if my process for part (1) of the question is correct.
I made an augmented matrix using the system of equations:
$$\begin{bmatrix}
1 & -2 &a \\ 3 & -6 &b
\end{bmatrix}$$
I manipulated the matrix two separate times to find when the system is inconsistent, for b --> R2 = R2 - 3R1,
$$\begin{bmatrix}
1 & -2 &a \\ 0 & 0 &b-3a
\end{bmatrix}$$
and a --> R1 = R1 - (1/3)R2
$$\begin{bmatrix}
0 & 0 &a - {b\over3} \\ 3 & -6 &b
\end{bmatrix}$$
So, the system is inconsistent when/if  $b = 3a$ and $a=b/3$?
Some general advice on how to approach the problem is greatly appreciated, thank you!
 A: 
Some general advice on how to approach the problem...

You can also write,
$$\begin{align}\begin{cases}x-2y=a\\3x-6y=b\end{cases}&\implies \begin{cases}3x-6y=3a\\3x-6y=b\end{cases}\\
&\implies 3a=b.\end{align}$$
We see that,

*

*If $b=3a$, then it is enough to solve $x-2y=a$ or $3x-6y=b$. Because, in this case we have

$$x-2y=a\iff 3x-6y=b$$
This implies, we have infinitely many solutions.

*

*If $b≠3a$, then we immediately get a contradiction. This means, in this case the solution doesn't exist.

A: There is a way to verify if your answer is correct yourself - by substituting values back into the original problem!
Suppose $a=\frac{b}{3}$, for example $a=2$ and $b=6$. Now you have the equations
$$\begin{align*}
x-2y&=2\\
3x-6y&=6
\end{align*}$$
Are these consistent or inconsistent? Solve them and look at the definition of consistency.
A: $\begin{align*}
x - 2y = a \implies 3x-6y&=3a\\
3x - 6y &= b\\
\therefore 3a&=b
\end{align*}$

*

*$b\ne3a$ is inconsistent

*$b=3a$ is consistent

*There are infinite solutions where
$a \in\mathbb{N}$
