Given

\dfrac{x-4}{x-5}+\dfrac{x-6}{x-7} = \dfrac{10}{3}

\dfrac{x-5+1}{x-5}+\dfrac{x-7+1}{x-7} = \dfrac{10}{3}

\dfrac{1}{x-5}+\dfrac{1}{x-7} =\dfrac{4}{3}

2x-12= \dfrac{4}{3}(x^2-12x+35)

3x-18 = 2(x^2-12x+35)

2x^2-27x+88 = 0

(x-8)(2x-11) = 0

x = 8 or x = \dfrac{11}{2}