2 - x = (-3 + a)*x - 3 |+3 +x
5 = (-3+a)*x+1*x
5 = (-3+a+1)*x
5=(-2+a)*x
x=5/(-2+a)
\(a\ne 2\Longrightarrow\mathbb L=\{\frac{5}{-2+a}\} \)
Für a=2:
2 - x = (-3 + 2)*x - 3
2-x=-x-3 |+x+3
5=0 Falsche Aussage!
\(a=2\Longrightarrow\mathbb L=\emptyset \)
:-)