# Solve for $x$ ; $\dfrac{2x}{4\pi}+\dfrac{1-x}{2}=0$

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!

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Hey in the 2nd last line, you multiplied 0 by $4\pi$ and got $4\pi$ – Chris Dugale Jul 16 '12 at 5:22
You last line contains some errors too. – Henry Jul 16 '12 at 5:23
Fixed the mistake. Anything now? – Austin Broussard Jul 16 '12 at 5:24
$\dfrac{2x}{2\pi} \not = 2x$ – Henry Jul 16 '12 at 5:24
$4\pi$ times $0$ is zero. Please correct your mistake. – Hassan Muhammad Jul 16 '12 at 5:26

Although, it is a homework problem, but use the comments above to correct your mistakes and you will find your answer as $$x=\frac{\pi}{\pi-1}$$

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Is all of my work correct? – Austin Broussard Jul 16 '12 at 5:36
@Austin: No - not right now as you have the sign wrong in the final answer. – Henry Jul 16 '12 at 5:38
Is it right now? – Austin Broussard Jul 16 '12 at 5:40
@Austin You need to delete the line $2x+(1-x)=\frac{0}{2\pi}=0$ – Henry Jul 16 '12 at 5:49
Is it correct now? – Austin Broussard Jul 16 '12 at 5:54

You have mixed up multiplying both sides by $4\pi$ with adding $4\pi$.

You also forgot to distribute your multiplication by $\frac{1}{2\pi}$ over the left side in the last line; just as with multiplication by any other number, $$\frac{1}{2\pi}(a+b)=\frac{a}{2\pi}+\frac{b}{2\pi}$$

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$$2x+2\pi (1-x)= 2x+2\pi-2\pi x$$

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