Question

$\frac{1}{6x} + \frac{3}{2x} = \frac{x-1}{4x^{2} }$

Answer 1

$\dfrac{1}{6x} + \dfrac{3}{2x} = \dfrac{x - 1}{4x^2} \\ \\ \\ \dfrac{2x +6x *3}{12x^2} = \dfrac{x - 1}{4x^2} \\ \\ \\ \dfrac{20x}{12x^2} = \dfrac{x - 1}{4x^2} \\ \\ \\ \dfrac{5}{3x} = \dfrac{x - 1}{4x^2}$

$\dfrac{4x^2 *5 + 3x - 3x^2}{3x * 4x^2} \\ \\ \\ \dfrac{20x^2 + 3x - 3x^2}{12x^3} \\ \\ \\=> \dfrac{17x + 3}{12x^2}$