Get the value of $p$ for no solution in equation of first grade

Question

I have this statement:

For what value of $p$ the equation not have solutions ?

$\frac{2-x}{x+4} = p$

This can be lineal,

$2-x = px + 4p$

Well, as equation of first grade:

$px + 4p +x -2 = 0$

It not have solutions when given an equation $ax + b = 0$, $a = 0$, $b \neq 0$, so will not have solutions.

I think that in this case $ax$ is represented by $px + 4p$, and $b$ is represented by $x-2$.

so, px + 4p = 0, but from here, I no longer understand what to do.

Answer 1

Rewrite as: $$ \underbrace{(p + 1)}_{\textsf{Must be } 0}x + \underbrace{(4p - 2)}_{\textsf{Must not be } 0} = 0 $$ which forces $p = -1$.