Here's the question you clicked on:
swin2013
Simplify
\[\frac{ 2x }{ 4\pi }+ \frac{ 1-x }{ 2 }\]
is that really \(\pi\) in the denominator of 1st fraction ?
now we have to make the denominator common , so we need to multiply and divide by 2\(\pi\) in 2nd fraction to get 4\(\pi\)as common denominator . \(\huge\frac{2x}{4\pi}+\frac{2\pi(1-x)}{4\pi}\) now can u solve further ??
i got up to \[\frac{ x-pix }{ 2\pi} = 1/2\]
nopes, \(\huge\frac{2x}{4\pi}+\frac{2\pi(1-x)}{4\pi}=\frac{2x+2\pi(1-x)}{4\pi}=\frac{x+\pi-\pi x}{2\pi}\) u can't simplify it further.
i forgot one important thing. the equation is equal to zero. and i'm solving for x
okay, so it will be \(\huge x+\pi-\pi x=0 \implies x(1-\pi)=-\pi \\\huge x=\frac{-\pi}{1-\pi}or\frac{\pi}{\pi-1}\)
so was my step right so far?
u got -1/2, but its actually, pi/(pi-1)
equal to zero? my teacher did the same thing as me.. but i didn't write the last part leading towards the answer.
i crossed multiply \[\frac{ 4x }{ 8\pi } + \frac{ 4\pi - 4pix }{ 8\pi } = 0\]
i split the equation up also
the i got \[\frac{ x }{ 2\pi } - \frac{ \pi }{ 2\pi } = -1/2 \]
the \[\frac{ \pi }{ 2 reduces to 1/2 and i brought it over...
but u cannot cancel pi, the first term had x in it, not pi.
yes... and i'm solving for it. i haven't found the answer further than that step
what are you talking about cancel? i didn't cancel anything other than the pis... which was pi/2pi... they're in common? this part i know for sure is right. i don't know how to get after that
how did u get -1/2 ...and where did your 8 go from denominator ?
that was a typo. and i reduced \[\frac{ \pi }{ 2\pi } \] to get 1/2 and then i subtracted it on both sides so the equation is equal to -1/2
i'm basically trying to isolate the x as much as possible
\(\huge \frac{ 4x }{ 8\pi } + \frac{ 4\pi - 4\pi x }{ 8\pi } = 0\) u had this correct. then u separated denominator u should get, \(\large \frac{x}{2\pi}+\frac{1}{2}-\frac{x}{2}=0\) but this is not the best way to isolate x.
yea, but it's easier for me to visualize it. instead of clutter i wanted to separate the fractions
ok, so continuing with separating the fraction, you can cancel all the 2's in the denominator(equivalent to multiplying 2 on both sides.) then shifting +1to other side, u get \(\large x(\frac{1}{\pi}-1)=-1\) which again gives, \(\large x=\frac{-1}{1/\pi-1}=\frac{\pi}{\pi-1}\)
ok, so continuing with separating the fraction, you can cancel all the 2's in the denominator(equivalent to multiplying 2 on both sides.) then shifting +1to other side, u get \(\large x(\frac{1}{\pi}-1)=-1\) which again gives, \(\large x=\frac{-1}{1/\pi-1}=\frac{\pi}{\pi-1}\)
so i multiply the top and bottom by 2pi to get rid of it in the denominator?
yes, u can do that also, that will lead to same answer.
as i said, i multiplied both sides by 2
as i said, i multiplied both sides by 2
\[\frac{ 2x }{ 4\pi }+ \frac{ 1-x }{ 2 }\] is not an equation. there is nothing to "solve" for you can add however, by finding the lcd is \(4\pi \) and adding
you would get \[\frac{2x+2(1-x)}{4\pi}\]
i got x = \[\frac{ \pi }{ 1-\pi }\]
- pi / 1-pi = pi / 1+pi
there must be slight error in minus sign, its pi/(pi-1)
no, i subtracted 1/2 from both sides and it was -1/2
then i multiplied both sides by 2pi which made it -pi.
so i got \[x - pix = -pi\]
both those steps are correct. u get x(1-pi)=-pi x=-pi/(1-pi) = pi/(pi-1)
then i factored out the x. \[x(1-\pi) = -\pi \] \[x = \frac{ -\pi }{ 1-\pi } \]
yes so its -pi/(1-pi) which is same as pi/(pi-1)
are you sure? if you take out the negative, wouldn't the denominator be 1+ pi?
nopes it would not be 1+pi. it will be pi-1 -(1-pi) = pi-1 yes, sure.
ok i see you flipped it around. lol