$$\frac{8}{x}+3 = \frac{1}{x}+1$$

$$ \begin{aligned} \frac{8}{x}+3 &= \frac{1}{x}+1&& \text{multiply ALL terms by } \color{blue}{ x }. \\[1 em]x\cdot\frac{8}{x}+x\cdot3 &= x\cdot\frac{1}{x}+x\cdot1&& \text{cancel out the denominators} \\[1 em]8+3x &= 1+x&& \text{move the $ \color{blue}{ x } $ to the left side and $ \color{blue}{ 8 }$ to the right} \\[1 em]3x-x &= 1-8&& \text{simplify left and right hand side} \\[1 em]2x &= -7&& \text{ divide both sides by $ 2 $ } \\[1 em]x &= -\frac{7}{2}&& \\[1 em] \end{aligned} $$

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