0x=-2

No solution"

Not true. Let m = 2, then you get the equations: 2x + y = 2, so y = -2x + 2, and y = 2x + 4. A common solution to these is the point (-1/2, 3).

"2. 2. If m+2≠0;m≠2

x=-2/(m+2)

So:

No solution-

m=2

One solution-

m≠2"

The logic here is right, if m = 2 had no solutions, than all lines without that slope intersect the line. Unforutunately, m = 2 has a solution.

"Infinite many solutions-

m∈{}"

Correct. The x-intercepts are off, and they only thing you can change is the slope.

]]>mx+y=2 <=>mx+2x+4=2

(m+2)x+4=2

(m+2)x=-2

1. If m+2=0;m=2

0x=-2

No solution

2. If m+2≠0;m≠2

x=-2/(m+2)

So:

No solution-

m=2

One solution-

m≠2

Infinite many solutions-

m∈{}

sorry for the syntax, I don't know English well.]]>

lol, thanks a lot:)

]]>1. No solution.

2. An infinite number of solutions.

3. One real solution.

I don't know how to solve these kinds of problems :s. please, help someone?

]]>