Can this be solved with any HP CAS?

[SlideRule]

(09-30-2017 09:10 AM)Nigel (UK) Wrote:  ... is only equivalent to the original equation when \(x\neq5\) ...

EXACTLY! BUT how do you know this is true until you solve the expression? It's not so much that the expression can not be solved as the solution requires understanding. You have two polynomials set in an equivalency expression that appear to have been divided by an expression which is equivalent to zero WHEN the variable common to both expressions is set to 5 (one interpretation), where 5 is also the solution to the quadratic expression prior to a synthetic division. This is a useful process and analysis, wherein the results need interpretation. Wolfram does this, look at the attached Jpg's in my previous post.
The solution reveals the undefined operation.

BEST!
SlideRule