Solve the equation: $$ \frac{1}{x}+\frac{2}{3 x^{2}}=\frac{1}{2}+\frac{1}{3 x} $$

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Solve the equation $$ \frac{1}{x}+\frac{2}{3x^{2}}=\frac{1}{2}+\frac{1}{3x} $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\frac{1}{x}+\frac{2}{3x^{2}}=\frac{1}{2}+\frac{1}{3x}$$
- step1: Find the domain:
  $$\frac{1}{x}+\frac{2}{3x^{2}}=\frac{1}{2}+\frac{1}{3x},x
eq 0$$
- step2: Multiply both sides of the equation by LCD:
  $$\left(\frac{1}{x}+\frac{2}{3x^{2}}ight)	imes 6x^{2}=\left(\frac{1}{2}+\frac{1}{3x}ight)	imes 6x^{2}$$
- step3: Simplify the equation:
  $$6x+4=3x^{2}+2x$$
- step4: Move the expression to the left side:
  $$6x+4-\left(3x^{2}+2xight)=0$$
- step5: Calculate:
  $$4x+4-3x^{2}=0$$
- step6: Factor the expression:
  $$\left(2-xight)\left(2+3xight)=0$$
- step7: Separate into possible cases:
  $$\begin{align}&2-x=0\&2+3x=0\end{align}$$
- step8: Solve the equation:
  $$\begin{align}&x=2\&x=-\frac{2}{3}\end{align}$$
- step9: Check if the solution is in the defined range:
  $$\begin{align}&x=2\&x=-\frac{2}{3}\end{align},x
eq 0$$
- step10: Find the intersection:
  $$\begin{align}&x=2\&x=-\frac{2}{3}\end{align}$$
- step11: Rewrite:
  $$x_{1}=-\frac{2}{3},x_{2}=2$$
The solutions to the equation $$ \frac{1}{x}+\frac{2}{3x^{2}}=\frac{1}{2}+\frac{1}{3x} $$ are $$ x = -\frac{2}{3} $$ and $$ x = 2 $$.
The solutions are $$ x = -\frac{2}{3} $$ and $$ x = 2 $$.

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