Question
$$\frac{1}{x-3} = 7 \cdot \frac{x}{7x}$$
Answer
$$ x = 0 $$
Explanation
$$ \begin{aligned} \frac{1}{x-3} &= 7 \cdot \frac{x}{7x}&& \text{multiply ALL terms by } \color{blue}{ (x-3)\cdot7x }. \\[1 em](x-3)\cdot7x\cdot\frac{1}{x-3} &= (x-3)\cdot7x\cdot7 \cdot \frac{x}{7x}&& \text{cancel out the denominators} \\[1 em]7x &= 7x^4-21x^3&& \text{move all terms to the left hand side } \\[1 em]7x-7x^4+21x^3 &= 0&& \text{simplify left side} \\[1 em]-7x^4+21x^3+7x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -7x^{4}+21x^{3}+7x = 0 } $, first we need to factor our $ x $.
$$ -7x^{4}+21x^{3}+7x = x \left( -7x^{3}+21x^{2}+7 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ -7x^{3}+21x^{2}+7 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using qubic formulas.
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Equations Solver