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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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    Your current answer is correct, but numerator and denominator are both negative and you can multiply by $\frac {-1}{-1}$. You have made life rather more complex for yourself by keeping the unnecessary factors of 2 throughout your workings. If you multiply the original equation by 2 and cancel a factor 2 from the first term you get $\frac x {\pi} + 1 - x = 0$. Then clear the $\pi$ to remove the fraction and take the $x$ terms to the opposite side and you get $\pi = x (\pi-1)$ - advice: simplify as much as possible at the beginning.2012-07-16

3 Answers 3

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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 it correct now?2012-07-16
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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 $