The given equation is
[tex]\frac{1}{3}x=\frac{1}{3}+\frac{6}{x}[/tex]First, we subtract 6/x on each side
[tex]\frac{1}{3}x-\frac{6}{x}=\frac{1}{3}+\frac{6}{x}-\frac{6}{x}\rightarrow\frac{1}{3}x-\frac{6}{x}=\frac{1}{3}[/tex]Second, we solve the difference in the left side of the equation
[tex]\frac{x^2-18}{3x}=\frac{1}{3}[/tex]Third, we simplify numbers 3, and multiply the equation by x
[tex]\frac{x(x^2-18)}{x}=x\rightarrow x^2-18=x[/tex]Fourth, we subtract x on each side to get the quadratic equation
[tex]x^2-18-x=x-x\rightarrow x^2-18-x=0[/tex]Additionally, we need to find two number which product is 18, and which difference is 1. Those numbers are no integers, so, we must use a calculator.
Using a calculator, the solutions are
[tex]x_1=4.77,x_2=-3.77[/tex]