$$ \begin{aligned} \frac{1}{x}+2+\frac{2}{x} &= \frac{3}{x}+2&& \text{multiply ALL terms by } \color{blue}{ x }. \\[1 em]x\cdot\frac{1}{x}+x\cdot2+x\cdot\frac{2}{x} &= x\cdot\frac{3}{x}+x\cdot2&& \text{cancel out the denominators} \\[1 em]1+2x+2 &= 3+2x&& \text{simplify left and right hand side} \\[1 em]2x+3 &= 2x+3&& \text{move the $ \color{blue}{ 2x } $ to the left side and $ \color{blue}{ 3 }$ to the right} \\[1 em]2x-2x &= 3-3&& \text{simplify left and right hand side} \\[1 em]2x-2x &= 0&& \\[1 em]0 &= 0&& \\[1 em] \end{aligned} $$
Since the statement $ \color{blue}{ 0 = 0 } $ is TRUE for any value of $ x $, we conclude that the equation has infinitely many solutions.
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