$$ \begin{aligned} 3x\cdot(4+5)-\frac{15}{3} &= 0&& \text{multiply ALL terms by } \color{blue}{ 3 }. \\[1 em]33x\cdot(4+5)-3\cdot\frac{15}{3} &= 3\cdot0&& \text{cancel out the denominators} \\[1 em]81x-15 &= 0&& \text{ move the constants to the right } \\[1 em]81x &= 15&& \text{ divide both sides by $ 81 $ } \\[1 em]x &= \frac{15}{81}&& \\[1 em]x &= \frac{5}{27}&& \\[1 em] \end{aligned} $$

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