How does $di+3c=0$ become $d=3ci$ I understand that you can move the $3c$ to the other side of the $=$ sign and you get $-3c$. But how can you move the $i$ so that the equation becomes $d=3ci$.
Thanks in advance
 A: You multiply both sides by $i$ : $di^2+3ic=3ic-d=0$. Thus $d=3ci$.
To make an $i$ vanish you multiply it by $i$ because $i^2=-1$.
A: First, as you mention,
\begin{align*}
di=-3c.
\end{align*}
Then, we want to multiply $di$ by the inverse of $i$ (to remove $i$). The inverse of $i$ is $-i$ because $-i\times i=1$. If we multiply the left-hand side by $-i$ we also have to multiply the right-hand side by $-i$ which gives us
\begin{align*}
d=(-i)\times di=(-i)\times -3c=3ci,
\end{align*}
as desired.
A: YOu have 
$$d = -{3c\over i}.$$
Since $i\cdot(-i) = 1$, $1/i = -i$.
Now you can divide the first equation to get
$$d = 3ci.$$
A: Multiply $i$ on both sides and then multipliy with $-1$ on both sides.
A: Others have answered but I want to point out something that will save you a lot of trouble later.
$i^1 = i $
$i^2 = -1$
$i^3 = -i  $
$i^4 = -i*i=-(-1)=1$
$i*(-i)=1$ so  $1/i = -i = i^3$ and $1/-i = i $
So if you ever have $xi = y $ you can always do $x= xi/i = y/i  = y (-i)=-yi $.
Also if you get $5i^{27} + 3i^3 = xi + 2xi^3$
You know $i^4=1$ So $i^{27}=(i^4)^8*i^3=1^8*i^3=i^3=-i$.
So you know $-5i-3i=xi-2xi$ so $-8i = -xi $ so $x=8$.
