Question

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

Answer 1

$\dfrac{x + 3}{x} = 1 + \dfrac{1 + 3x}{2x} \\ \\ \\ \dfrac{x + 3}{x} = \dfrac{2x + 1 - 3x}{2x} \\ \\ \\ \dfrac{x + 3}{x} = \dfrac{-x - 1}{2x} \\ \\ \\ \dfrac{x + 3}{x} = -\dfrac{x - 1}{2x}$

$\dfrac{2x^2 + 6x + x^2 - x}{2x *x} \\ \\ \\ \dfrac{2x^2 + 6x + x^2 - x}{2x^2} \\ \\ \\ => \dfrac{3x+5}{2x}$