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((2x)/(4pi))+((1-x))/(2)=0

Guest Jul 26, 2014

Best Answer 

 #1
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$$\begin{array}{rlll}
\frac{2x}{4\pi}+\frac{1-x}{2}&=&0\\\\
\frac{x}{2\pi}+\frac{1-x}{2}&=&0\\\\
2\pi \times \left(\frac{x}{2\pi}+\frac{1-x}{2}\right)&=&2\pi\times 0\\\\
x+\pi(1-x)&=&0\\\\
x+\pi-\pi x &=&0\\\\
x(1-\pi )+\pi &=&0\\\\
x(1-\pi )&=&-\pi \\\\
x&=&\frac{-\pi}{(1-\pi )} \\\\
x&=&\frac{\pi}{(\pi-1 )} \\\\

\end{array}$$

Melody  Jul 26, 2014
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1+0 Answers

 #1
avatar+91436 
+10
Best Answer

$$\begin{array}{rlll}
\frac{2x}{4\pi}+\frac{1-x}{2}&=&0\\\\
\frac{x}{2\pi}+\frac{1-x}{2}&=&0\\\\
2\pi \times \left(\frac{x}{2\pi}+\frac{1-x}{2}\right)&=&2\pi\times 0\\\\
x+\pi(1-x)&=&0\\\\
x+\pi-\pi x &=&0\\\\
x(1-\pi )+\pi &=&0\\\\
x(1-\pi )&=&-\pi \\\\
x&=&\frac{-\pi}{(1-\pi )} \\\\
x&=&\frac{\pi}{(\pi-1 )} \\\\

\end{array}$$

Melody  Jul 26, 2014

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