$$\frac{3}{x}+1 = \frac{12}{x^2}-1+2$$

$$ \begin{aligned} \frac{3}{x}+1 &= \frac{12}{x^2}-1+2&& \text{multiply ALL terms by } \color{blue}{ xx^2 }. \\[1 em]xx^2\cdot\frac{3}{x}+xx^2\cdot1 &= xx^2\cdot\frac{12}{x^2}-xx^2\cdot1+xx^2\cdot2&& \text{cancel out the denominators} \\[1 em]3+x^3 &= \frac{12}{x^1}-x^3+2x^3&& \text{multiply ALL terms by } \color{blue}{ x^1 }. \\[1 em]x^1\cdot3+x^1\cdot1x^3 &= x^1\cdot\frac{12}{x^1}-x^1\cdot1x^3+x^1\cdot2x^3&& \text{cancel out the denominators} \\[1 em]3x+x^4 &= 12-x^4+2x^4&& \text{simplify left and right hand side} \\[1 em]x^4+3x &= x^4+12&& \text{move all terms to the left hand side } \\[1 em]x^4+3x-x^4-12 &= 0&& \text{simplify left side} \\[1 em]x^4+3x-x^4-12 &= 0&& \\[1 em]3x-12 &= 0&& \text{ move the constants to the right } \\[1 em]3x &= 12&& \text{ divide both sides by $ 3 $ } \\[1 em]x &= \frac{12}{3}&& \\[1 em]x &= 4&& \\[1 em] \end{aligned} $$

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