$$ \begin{aligned} \frac{1}{3}(10y-2)+\frac{1}{12}(8y+13) &= 0&& \text{multiply ALL terms by } \color{blue}{ 12 }. \\[1 em]12 \cdot \frac{1}{3}(10y-2)+12\frac{1}{12}(8y+13) &= 12\cdot0&& \text{cancel out the denominators} \\[1 em]40y-8+8y+13 &= 0&& \text{simplify left side} \\[1 em]48y+5 &= 0&& \text{ move the constants to the right } \\[1 em]48y &= -5&& \text{ divide both sides by $ 48 $ } \\[1 em]y &= -\frac{5}{48}&& \\[1 em] \end{aligned} $$

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