
System of Equations
So my linear algebra class is just starting.
So while I realize this isn't a linear algebra thread, it's in a realm of difficultly similar to that of PreCalculus
What are the value(s) of α such that the system has:
a) no solution
b) a unique solution
c) infinite number of solutions
*Solve case c ( with many solutions) parametrically
x + y = 1
αx + 2y = 2
For case b, I'm getting that α can be anything EXCEPT 2. Thus x would be equal to 0, and y equal to 1.
For case c, the only solution I can seem to come up with is α=2.
I'm not sure there is any solution to case a, and I'm unsure of case b and c.
Normally I wouldn't be stuck on a question of system of equations, but this really has me stuck.

Re: System of Equations
I agree with your answer to part b, although you don't need to solve x and y (in fact, their solutions would depend on a).
As for part a), the system of linear equations has no solution when the lines are parallel (and so never cross). What do you know about parallel lines?
And for part c), the system of linear equations has infinitely many solutions when the two lines are in fact, the same. How can you use this piece of information to help you answer the question?

Re: System of Equations
Oh yeah, I was just stating what the unique answers would have been.
Just trying to show that I've actually made an honest effort at the problem before asking.
Parallel lines have the same slope.
Is that a hint?
There are infinitely many solutions when the lines are the same.
That why I assumed α should be equal to 2 in case c.
When simplified, the lines would be the same wouldn't they?
They would only be off by a constant of 2.

Re: System of Equations
The more I think about it, I keep thinking I'm right about case c.
Since the the equations would be
x+y=1
2x+2y=2 which simplifies to 2(x+y=1)
The constant of 2 only increases the rate of change doesn't it?
So the two lines still lay on the same path.
As for part a, I'm still stuck.

Re: System of Equations
Yes, you are correct that when a = 2, you will have infinitely many solutions.
You are also correct that lines are parallel when they have the same gradient. Can you write the lines in the gradient and yintercept form?

Re: System of Equations
As in like y=mx+b?
y=1+x
y=1(ax/2)
I can get the same gradient by saying a=2, but in that case, it becomes the same line as the equation above it, yielding infinite solutions again. The yintercept needs to be different in order to have the same gradient but not cross paths, but we aren't changing the value of that constant, only the value of a.

Re: System of Equations
Is it possible that there is no solution to part a?
If that's the case, I would have been stressing out over nothing (Headbang)