+7 – +4 = A.

To somewhat “simplify” the problem, just get rid of the positive signs to make it a “regular” subtraction problem.

+7 – +4 = A

can be converted to

7 – 4 = A.

To find the value of A, just subtract “normally”.

Now, negatives are the exact opposite of positives. So while subtracting positives you get closer to zero, subtracting negatives you get farther away.

So let’s change the above numbers to negatives.

-7 – -4 = A.

I just imagine it as the negative signs cancel out the “regular” operation of the subtraction sign. So try and imagine adding the two negatives, while keeping the negative sign. It would be 11, since 7+4=11, and now you add your negative sign to 11. Soooo

-7 – -4 = A

can be converted to (in your mind, as if it’s on paper it would demonstrate a different operation)

-7 + -4= A

and we all know that 7+4=11, and with a negative sign, it can be

A = -11.

If you need some elaboration on that, ask.

And finally, the dreaded cross-sign subtraction problem.

Let’s stay with our numbers and say the problem is

+7 – -4 = A.

Well, let’s just change it to

7 – -4 = A

You need to remember that the negative sign cancels out the subtraction sign and changes it to it’s opposite- addition. So, the negative changes subtraction to addition, so you can think of it as….

7 + -4 = A

So basically, your subtracting 4 from seven. Your adding minus 4 to 7. So, to simplify it even further, think of it as..

7 – 4 = A

It’s simple from there.

Now if it’s the opposite…

-7 – +4 = A

We simplify it to

-7 – 4 = A

And remember that the negative sign and the minus sign cancel each other out. So, you can change it to…

-7 + 4 = A

A basic number line can help with that. I’ll draw it out…

(Sorry about the lack of tick marks.)

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0…

You start at -7 and count up four, leading you to -3.

Sooooo

-7 – +4 = -3.

Any elaboration, just post or email me.

if both are positive then x-y = IxI – IyI, and will be positive if x>y, otherwise negative

if both are negative then x – y = -IxI+IyI, and will be positive if y>x, otherwise negative

if x is positive and y negative, then x-y = IxI+IyI, and will always be positive

if x is negative and y is positive then x-y = -IxI-IyI, and will always be negative