Solve for $x$: $$\dfrac{2x}{4\pi}+\dfrac{1-x}{2}=0$$
$$\dfrac{2x}{4\pi}+\dfrac{2\pi(1-x)}{2\pi(2)}=0$$ $$\dfrac{2x+2\pi (1-x)}{4\pi}=0$$ $$2x+2\pi (1-x)=0$$ $$2x+2\pi -2\pi x=0$$ $$2x-2\pi x=-2\pi$$ $$2x(1-\pi )=-2\pi$$ $$2x=\dfrac{-2\pi}{1-\pi}$$ $$\left(\frac{1}{2}\right)\cdot (2x)=\left(\dfrac{-2\pi}{1-\pi}\right)\cdot \left(\frac{1}{2}\right)$$ $$x=\dfrac{-\pi}{1-\pi}$$
I believe this is correct so far, but my thoughts about myself have been off tonight. I do not understand what to do next, either because I honestly don't know or I made a mistake in my work. Please hints only!