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!