\(\dfrac{1}{3}-\dfrac{3}{2}x=\dfrac{4}{3}x-1\\ \dfrac{4}{3}x+\dfrac{3}{2}x=\dfrac{1}{3}+1\\ x\left(\dfrac{4}{3}+\dfrac{3}{2}\right)=\dfrac{4}{3}\\ x\cdot\dfrac{17}{6}=\dfrac{4}{3}\\ x=\dfrac{4}{3}:\dfrac{17}{6}\\ x=\dfrac{4}{3}\cdot\dfrac{6}{17}\\ x=\dfrac{8}{17}\)