Question
$$5x^0 = 4x^0+7x^1$$
Answer
$$ x = \dfrac{1}{7} $$
Explanation
$$ \begin{aligned} 5x^0 &= 4x^0+7x^1&& \text{move the $ \color{blue}{ 7x } $ to the left side} \\[1 em]5-7x &= 4&& \text{ move 5 to the right } \\[1 em]-7x &= 4-5&& \\[1 em]-7x &= -1&& \text{ divide both sides by $ -7 $ } \\[1 em]x &= \frac{-1}{-7}&& \\[1 em]x &= \frac{1}{7}&& \\[1 em] \end{aligned} $$
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