How does $di+3c=0$ become $d=3ci$

Question

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

Answer 1

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.

Answer 2

YOu have $$d = -{3c\over i}.$$ Since $i\cdot(-i) = 1$, $1/i = -i$. Now you can divide the first equation to get $$d = 3ci.$$

Answer 3

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 $i^2=-1$.

Answer 4

Multiply $i$ on both sides and then multipliy with $-1$ on both sides.

Answer 5

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 $x=8$.