How do you solve `8/3 = 12/n`?

First, multiply each side of the equation by `color(red)(3)/color(blue)(8)`:

`color(red)(3)/color(blue)(8) xx 8/3 = color(red)(3)/color(blue)(8) xx 12/n`

`cancel(color(red)(3))/cancel(color(blue)(8)) xx color(blue)(cancel(color(black)(8)))/color(red)(cancel(color(black)(3))) = 36/(8n)`

`1 = 36/(8n)`

Now, multiply each side of the equation by `color(red)(n)` to solve for `n` while keeping the equation balanced:

`color(red)(n) xx 1 = color(red)(n) xx 36/(8n)`

`n = cancel(color(red)(n)) xx 36/(8color(red)(cancel(color(black)(n))))`

`n = 36/8`

`n = (4 xx 9)/(4 xx 2)`

`n = (color(red)(cancel(color(black)(4))) xx 9)/(color(red)(cancel(color(black)(4))) xx 2)`

`n = 9/2`

Given: `8/3=12/n`

We need to find `n` so lets make it the top value (numerator)

Turn everything upside down (yes you can do this!)#

`3/8=n/12`

We now need to 'get' `n` on its own so we need to 'get rid' of the 12. We do this by turning it into 1 as 1 times anything does not change the value.

Multiply both sides by 12

`color(green)(3/8color(red)(xx12)" "=" "n/12color(red)(xx12))`

`color(green)(" "36/8" "=" "ncolor(red)(xx12/(color(green)(12)))`

But `12/12=1" "` and `" "nxx1=n`

`9/2=n`

`n=9/2" "->" " 4 1/2`