Under what conditions on $a$ and $b$ will the following linear system have no solutions, one solution and infinitely many solutions?
$$2x-3y = a$$
$$4x-6y = b$$
It is straightforward to see how the system can have infinitely many solutions or no solutions.
In order for the solution to have infinitely many solutions, we require the y-intercept the of the two lines to be equal. This can be accomplished by observing that
$$4x-6y = b \implies 2x-3y = \frac{b}{2}$$
Therefore, the lines will coincide if $b = 2a$.
Consequently, the two lines are parallel when $b \neq 2a$ (it is clear that the two equations have the same slopes, so that does not need to be checked).
Although I can intuitively see that the two lines will never intersect at a single point, my mathematical justification may be weak, but here is my attempt. Thus my question is whether or not the following reasons for the two lines never intersecting at a single point is valid.
If we subtract $2x-3y=a$ from $4x-6y=b$, we get
$$2x-3y = b-a$$
But I don't see how to conclude from this result that the system does not have a single solution.
Best Answer
You are correct! The system can never have exactly one solution.
However, your proof is not good. You have provided no evidence that if $b\neq 2a$, then the system has no solution. Sure, you get that $2x-3y=2a-b$, but why would this mean that the system has no solution?
In fact, if you subtract $2x-3y=a$ from $4x-6y=b$, you do not get $$2x-3y=2a-b,$$ you get $$2x-36=b-a.$$