Fractional equation with a constant term

Text

Solve the following equation: $\frac{2x}{1-3x}+3-\frac{1}{x} = \frac{23}{12}.$

Solution

Existence conditions: $x\neq 0$ and $x\neq \tfrac13$. Reduce the left side to the common denominator $x(1-3x)$: $\frac{2x\cdot x + 3\,x(1-3x) - (1-3x)}{x(1-3x)} = \frac{-7x^2 + 6x - 1}{x(1-3x)} = \frac{23}{12}.$ Cross-multiplying: $12(-7x^2+6x-1) = 23\,x(1-3x)$, i.e. $-84x^2 + 72x - 12 = 23x - 69x^2 \;\Rightarrow\; 15x^2 - 49x + 12 = 0.$ $x = \frac{49 \pm \sqrt{2401 - 720}}{30} = \frac{49 \pm 41}{30} \;\Rightarrow\; x = 3 \ \text{or}\ x = \frac{4}{15}.$ Both acceptable. $\boxed{x = 3 \ \vee\ x = \tfrac{4}{15}}$