$$\frac{1}{3}(9x+16+2)+2x = 0$$

$$ \begin{aligned} \frac{1}{3}(9x+16+2)+2x &= 0&& \text{multiply ALL terms by } \color{blue}{ 3 }. \\[1 em]3 \cdot \frac{1}{3}(9x+16+2)+3\cdot2x &= 3\cdot0&& \text{cancel out the denominators} \\[1 em]9x+18+6x &= 0&& \text{simplify left side} \\[1 em]15x+18 &= 0&& \text{ move the constants to the right } \\[1 em]15x &= -18&& \text{ divide both sides by $ 15 $ } \\[1 em]x &= -\frac{18}{15}&& \\[1 em]x &= -\frac{6}{5}&& \\[1 em] \end{aligned} $$

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