$$\frac{1}{x}+\frac{1}{x}-3 = \frac{1}{2}$$

$$ \begin{aligned} \frac{1}{x}+\frac{1}{x}-3 &= \frac{1}{2}&& \text{multiply ALL terms by } \color{blue}{ x\cdot2 }. \\[1 em]x\cdot2\cdot\frac{1}{x}+x\cdot2\cdot\frac{1}{x}-x\cdot2\cdot3 &= x\cdot2\cdot\frac{1}{2}&& \text{cancel out the denominators} \\[1 em]2+2-6x &= x&& \text{simplify left side} \\[1 em]-6x+4 &= x&& \text{move the $ \color{blue}{ x } $ to the left side and $ \color{blue}{ 4 }$ to the right} \\[1 em]-6x-x &= -4&& \text{simplify left side} \\[1 em]-7x &= -4&& \text{ divide both sides by $ -7 $ } \\[1 em]x &= \frac{-4}{-7}&& \\[1 em]x &= \frac{4}{7}&& \\[1 em] \end{aligned} $$

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