We have the equation:
[tex]\begin{gathered} \frac{x}{3}+1=5x-3 \\ \end{gathered}[/tex]We can isolate x passing all the terms with x to one side and the numbers to the other side:
[tex]\begin{gathered} \frac{x}{3}-5x=-3-1 \\ (\frac{1}{3}-5)\cdot x=-4 \\ (\frac{1-5\cdot3}{3})\cdot x=-4 \\ (\frac{1-15}{3})\cdot x=-4 \\ -\frac{14}{3}x=-4 \\ x=(-4)\cdot(-\frac{3}{14})=\frac{3\cdot4}{14} \\ x=\frac{12}{14}=\frac{6}{7} \end{gathered}[/tex]