Under what conditions on a and b will the following linear system have no solutions, one solution, infinitely many solutions?
first row: 2x − 3y = a second row: 4x − 6y = b
My work: I wrote the system as an augumented matrix and then multiply the first row by -2 and add it to the second row which makes the second row 0 0 -2a + b.
My Answer: It is infinitely many solutions when 2a = b because you can set a to be any arbitary value and it will solve b correctly. There is no solutions when it is the opposite: 2a is not equal to b.
Now I am stuck figuring out if it is even possible for it to have only one solution. This seems rather difficult since I shown it has many solutions. The only thing I can think of was to graph it and find the intersection of these two equations of a line. But can I replace a with x and b with y something like this: 2x = y?
Best Answer
The system will never have only one solution.
If $x=x_0, y=y_0$ is a solution to this system, then $x = x_0+3, y = y_0 + 2$ will also be a solution.
The other way to think of this is that two lines with the same slope can't intersect each other at only one point.