In math, you can loosely think of an inverse as the number or operation that "undoes" another number or operation. For example, multiplication and division are inverse operations because what one does, the other undoes; if you multiply and then divide by the same amount, you'll end up right back where you started. An additive inverse, on the other hand, only applies to addition as the name suggests, and it's the number you add to another to get zero.
TL;DR (Too Long; Didn't Read)
The additive inverse of any number is the same number with the opposing sign. For example, the additive inverse of 9 is −9, the additive inverse of −z is z, the additive inverse of (y – x) is -(y – x) and so on.
Defining the Additive Inverse
You may intuitively see that the additive inverse of any number is the same number with its opposite sign. To really grasp this, it helps to envision a line of numbers and work through a few examples.
Imagine that you have the number 9. To "get" to that spot on the number line, you start at zero and count back up to 9. To get back to zero, you count 9 spaces backwards on the line, or in the negative direction. Or, to put it another way, you have:
Thus, the additive inverse of 9 is −9.
What if you start by counting backwards on the number line, in the negative direction? If you count backwards by 7 places, you'll end up at −7. To get back to zero you'll have to count forwards by 7 spots, or to put it another way, you'll have to start at −7 and add 7. So you have:
This means that 7 is the additive inverse of −7 (and vice versa).
The additive inverse is a relation that works both ways. In other words, if a number x is the additive inverse of a number y, then y is automatically the additive inverse of x.
Using the Additive Inverse Property
If you're studying algebra, the most obvious application for the additive inverse property is solving equations. Consider the equation
If you've been asked to solve for x, you must first isolate the variable term on one side of the equation.
The additive inverse of 3 is −3 and, knowing that, you can add it to both sides of the equation, which has the same effect as subtracting 3 from both sides. So, you have:
which simplifies to:
Now that the variable term is by itself on one side of the equation, you can continue solving. Just for the record, you would apply a square root to both sides and reach the answer x = 4; however, this is only possible because you first used your knowledge of the additive inverse property to isolate the x2 term.