0
$\begingroup$

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!

  • 0
    Hey in the 2nd last line, you multiplied 0 by $4\pi$ and got $4\pi$2012-07-16
  • 0
    You last line contains some errors too.2012-07-16
  • 0
    Fixed the mistake. Anything now?2012-07-16
  • 0
    $\dfrac{2x}{2\pi} \not = 2x$2012-07-16
  • 0
    $4\pi$ times $0$ is zero. Please correct your mistake.2012-07-16
  • 0
    $\dfrac{0}{a}=0$2012-07-16
  • 0
    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

2

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

  • 0
    Is all of my work correct?2012-07-16
  • 0
    @Austin: No - not right now as you have the sign wrong in the final answer.2012-07-16
  • 0
    Is it right now?2012-07-16
  • 0
    @Austin You need to delete the line $2x+(1-x)=\frac{0}{2\pi}=0$2012-07-16
  • 0
    Is it correct now?2012-07-16
3

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, $$\frac{1}{2\pi}(a+b)=\frac{a}{2\pi}+\frac{b}{2\pi}$$

3

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