$$\frac{81}{4}x+9 = 0\cdot0$$

$$ \begin{aligned} \frac{81}{4}x+9 &= 0\cdot0&& \text{multiply ALL terms by } \color{blue}{ 4 }. \\[1 em]4 \cdot \frac{81}{4}x+4\cdot9 &= 4\cdot0\cdot0&& \text{cancel out the denominators} \\[1 em]81x+36 &= 0&& \text{ move the constants to the right } \\[1 em]81x &= -36&& \text{ divide both sides by $ 81 $ } \\[1 em]x &= -\frac{36}{81}&& \\[1 em]x &= -\frac{4}{9}&& \\[1 em] \end{aligned} $$

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