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Question:

Simple math question about brackets?

Hi,I'm trying to relearn math after many years of not using anything more than the very basics and I have forgotten a great deal over the years. I'm in need of clarification about adding and removing brackets.From what I've seen sometimes a positive number becomes negative and vise versa while other times thats not the case, it just stays the same. What are the rules in regard to values changing in some situations but not others.Thanks David

Answer:

Alright, I'll explain this using the variables x, y, and z, where x, y, and z can either be numbers (2, 8, 102) or expressions (7x^3, 9a, 2b, etc.). Brackets segregate one part of an expression or equation from the rest. Lets look at the equation: m 3x - (8y-13z) The easiest way to look at this is to put a 1 in front of the brackets to help us visualize what is going on: m 3x - 1(8y-13z) Since 1 does not change the value of the expression. Next, we can write this equation as m 3x + (-1)(8y-13z) So instead of subtracting the bracketed part, we're adding its negative (which equates to the same thing). Now you just need to remember the simple rules for determining sign: +*+ Positive +*- Negative -*- Positive Simply put, if the signs of the things you are multiplying are the same, then the product is positive, if they're different they're negative. Now, back to the equation: m 3x + (-1)(8y-13z) m 3x + (-1)(8y) + (-1)(-13z) We can write the equation in this way because the -1 is applied to both terms inside the bracket in the same way. Now, simplify by multiplying the -1's through both terms: m 3x + -8y + 13z m 3x - 8y + 13z
Hi okorder
Brackets absolute value, meaning the amount of numbers the number is away from zero. For example [-7] is 7 numbers away from 0 therefore [-7] 7 [7] is 7 numbers away from zero therefore [7] 7. Positive and negative isn't the problem, it's how far away from zero it is.

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