Can this be solved with any HP CAS?

[SlideRule]

(09-29-2017 05:57 PM)Csaba Tizedes Wrote:  .. this is not true, because we have no information about 'b' when we substract it from both sides of the equation ... \(x=5\) is not a solution. The right solution of the original equation is "no solution".

OK, then expand the equation, collect terms & then simplify.

3x+(1/(x-5)) = 15+(1/(x-5))
expands to
3x² - 15x + 1 = 15x - 75 + 1
collect terms
3x² - 30x + 75 = 0
factor
3(x² - 10x + 25) = 0
factor
3(x-5)² = 0
so either
3=0 or (x-5)=0
since 3 ≠ 0
then
x-5 = 0
solve for x
x = 5

BEST!
SlideRule

ps: added Wolfram Alpha output(s)
[attachment=5208]

[attachment=5207]