Let's rewrite the expression as
[tex] 8x+3ax-5ax = 8-2a \iff 8x - 2ax = 8-2a \iff 2x(4-a) = 2(4-a) [/tex]
So, if [tex] a \neq 4 [/tex], you may divide both sides by [tex] 4-a [/tex], the equation becomes [tex] 2x=2 [/tex], and the only solution is [tex] x=1 [/tex], which means that this is not an identity (an indentity is tautologically true, no matter the value of x).
If instead [tex] a = 4 [/tex], you are multiplying both sides by zero, so the equation becomes
[tex] 2x \cdot 0 = 2 \cdot 0 \iff 0=0 [/tex]
So, it doesn't matter which value for x you choose, because you will always end up with [tex] 0=0 [/tex], which is obviously always true.
