Chapter 4 - Polynomials - 4.1 Exponents and Their Properties - 4.1 Exercise Set - Page 236: 93

$x=-1$

Multiplying both sides by the $LCD=6$, then the solution to the given expression, $ \dfrac{1}{2}x+\dfrac{1}{3}=\dfrac{1}{6}x $, is \begin{array}{l}\require{cancel} 3(1x)+2(1)=1(1x) \\\\ 3x+2=x \\\\ 3x-x=-2 \\\\ 2x=-2 \\\\ x=-\dfrac{2}{2} \\\\ x=-1 .\end{array}