Proving values of m for which there are no solutions

[Asx4Life]

Consider the system of simultaneous linear equations given by
mx+y=2
2x+(m-1)y=m
Find the value(s) of m for which there is no solution.

[Greatness]

If there are no solutions then the graphs dont intersect. So the gradients are the same but the y intercepts are different.
rearrange those equations into linear ones y=.....
equate gradients equal then find for what values the y intercept are different,

[Asx4Life]

ah yes thanks, I got it. 1 more question, can I find the determinant of the matrix and equate it to 0?

[Aurelian]

Alternately you can solve the determinate for the equivalent matrix representation of those two simultaneous equations for zero.

You may get a quadratic and hence two solutions for m, one of which will yield infinite solutions (i.e., the same line) rather than none, so you'll have to sub both back in to the original equations and use your judgment

EDIT: Oops, beaten... but hopefully this post answers that question haha

[onur369]

That is what I do, Swarley and Aurelian are correct. It took me a while to understand this.
mx+y=2
2x+(m-1)y=m
Make y the subject for both.
y=2-mx
y= (m-2x)/(m-1)
Get the gradients make them equal to each other, -m=(-2)/(m-1) , Solve for m, you will get 2 and -1. After this you get those two numbers and then substitute them back into the original equations.
When you sub in 2, you get 2x+y=2 and 2x+y=2 which shows that it is infinitely many solutions, but we are looking for no solution so the equations must not be the same, so you do the same with -1, you get -x+y=2 and 2x-2y=-1.
Therefore there is no solution when m=-1.