A fractional equation reduced to second degree

Statement

Solve the fractional equation $\dfrac{x-5}{x-3} = \dfrac{2}{x}$.

Solution

Existence conditions: $x \neq 3$ and $x \neq 0$.

Cross-multiply: $x(x-5) = 2(x-3) \implies x^2 - 5x = 2x - 6.$ Move everything to the left: $x^2 - 7x + 6 = 0 \implies (x-1)(x-6) = 0,$ so $x = 1$ or $x = 6$. Both satisfy the existence conditions. Check: for $x=1$: $\dfrac{-4}{-2} = 2$ and $\dfrac{2}{1} = 2$; for $x=6$: $\dfrac{1}{3}$ and $\dfrac{2}{6} = \dfrac13$. Correct. $\boxed{x = 1 \ \vee\ x = 6}$