I need advice from electrical engineers

I have important projects to build, so I’ve been taking an online engineering course. I’m very new to algebra and barely catching on.
The following problem I know is correct, as it is done by the instructor. However, it’s taught assuming I understand algebra already.
My question is, if I were to do this same problem on my own, how do I know if and when to flip polarities, to find the answer?
Any basic explanation would be greatly appreciated, as I’ve been stuck on this since yesterday.

Not an EE but… I’m sorry, I don’t understand the question. At no point do you flip polarities. If you are meaning the point when -i1+8=0 changes then it is because you are subtracting 8 from each side…

-i1 + 8 - 8 = 0 - 8

Alternatively if you want you can add i1 to both sides…

-i1 + 8 + i1 = 0 + i1

8 = i1

The best maths book I have found (for much more complicated mathematics than this eventually) is Engineering Mathematics by K A Stroud. Check libraries near you for it… It leads you through problems step by step.

For learning algebra, and how to think around algebra, I would highly recommend the game called “dragonbox”.

My son (, who is not particularly good at math,) started solving what I’d define as complex algebra problems for a 6year old, after playing this.

Short video


I mean the part where where the negative is changed to the positive. Where the question mark begins, is where I’m lost, as to how this equation is done.

I think @Zwack explained it well.
Whatever you subtract or add, you have to do that on both sider of the equal sign.

So when you subtract “8” from the left side, you also have to subtract it from the right side, which in turn gives you “-8”

(give dragonbox algebra a try. It really helps to understand as it is a gamification of math)


Basically, you have an equation with one unknown variable in it and you are trying to solve that equation.

So if the equation is x + 10 = 15 and you want to know what x is, you want to end up with x on one side and everything else on the other.

We want to move the + 10 over so we subtract 10 from both sides.

X + 10 - 10 = 15 - 10

+10 - 10 cancel out and we get

X = 15 - 10

Now while we still have an X we also have a sum we can do…

X = 15 - 10 = 5

So X = 5

Does that help at all?

You will also presumably end up with multiplication or division at some point, this is done pretty much the same way

3x = 15

We need to get rid of the 3 on the left so we divide both sides by 3

3x / 3 = 15 / 3


X = 15 / 3


X = 5


I just realized being new to algebra, and circuits, that I was over-complicating simple algebra.


For these problems you can have the arrows pointing in any direction and it doesn’t matter. (As long as not all arrows are pointing inwards, or all pointing outwards. You need in and outs for the nodes.)

If you get a negative number in the end you know the arrow was facing the wrong direction.

I find it helps to look at the context around why you’re doing the calculation to orient yourself throughout the process. In this case it’s the flow of current. Kirchoff’s current law says the current going in equals the current going out. If you know i2 and i3 are both branches off of i1 because of the drawing, it’s a lot harder to get confused about the positives and negatives in the equation. You know there’s no currents “pushing back” so any negative values must just be there to aid solving the equation.