Algebra 2 Help Please!

N = 8x +3y + 2n, solve for n. Somehow I need to get:

n = (N - 8x - 3y) / 2

I have the answer in the back of my book but I don't understand how to get it, and I need to because my teacher grades hard on tests, and her grades are based on tests.

N= 8x+ 3y+ 2n
First, you would subtract 8x + 3y from both sides of the equation, which would get you
N -8x -3y = 2n
Then, you would divide both sides by 2.
(N -8x -3y) / 2 = n
Finally, flip the equation around to get
n = (N - 8x - 3y) / 2.

That's how I would do it, anyway.
Good luck :)

you try and isolate the variable you are solving for, n.

subtract (8x + 3y) from both sides of the equation and wind up with

2n = N-8x-3y

then divide both sides of the equation by 2.

8x+3y+2n=N bring all the terms without n to the right
2n=N-(8x+3y)
2n=N+(-1)(8x+3y)
2n=N+(-8x-3y)
2n=N-8x-3y now divide by 2
n=(N-8x-3y)/2

N = 8x +3y + 2n

=> N - 8x - 3y = 2n

=> (N - 8x - 3y) / 2 = 2n / 2

=> (N - 8x - 3y) / 2 = n