## Intermediate Algebra (6th Edition)

$x=\dfrac{5}{3}$
The factored form of the given equation, $\dfrac{x-2}{x^2-7x+10}=\dfrac{1}{5x-10}-\dfrac{1}{x-5} ,$ is \begin{array}{l}\require{cancel} \dfrac{x-2}{(x-5)(x-2)}=\dfrac{1}{5(x-2)}-\dfrac{1}{x-5} .\end{array} Multiplying both sides by the $LCD= 5(x-5)(x-2) ,$ then the solution to the given equation is \begin{array}{l}\require{cancel} 5(x-2)=(x-5)(1)-5(x-2)(1) \\\\ 5x-10=x-5-5x+10 \\\\ 5x-x+5x=-5+10+10 \\\\ 9x=15 \\\\ x=\dfrac{15}{9} \\\\ x=\dfrac{\cancel{3}\cdot5}{\cancel{3}\cdot3} \\\\ x=\dfrac{5}{3} .\end{array}