All we need to do is to have the same denominator
5/x-2 - 2/3x-5
\(\frac{5}{x-2}\)×\(\frac{3x-5}{3x-5}\) -\(\frac{2}{3x-5}\)×\(\frac{x-2}{x-2}\)
\(\frac{15x-25}{(x-2)(3x-5)}\) - \(\frac{2x-4}{(x-2)(3x-5)}\)
Answer = \(\frac{(15x-2x-25-4)}{(x-2)(3x-5)}\)
= \(\frac{13x-29}{(x-2)(3x-5)}\)
