Equation with parameters and discussion

Question

Given the following equation: $\frac{3x+p}{3x-1} = \frac{x+1}{x-p}$.

The domain of definition is: $ \mathbb R \setminus$ { $\frac{1}{3}; p $}.

Now, if I solve the equation normally, I receive $x = \frac{1-p}{2}$.

But under which condition for $p$ is this true? And what are the other conditions for $p$ ?

Thanks for any help!

Answer 1

Now, delete solutions for $\frac{1-p}{2}=p$ and for $\frac{1-p}{2}=\frac{1}{3}.$

Id est, for $p=\frac{1}{3}$ our equation has no solutions and for $p\neq\frac{1}{3}$ we obtain $x=\frac{1-p}{2}.$