Math 7 (Period 4)

Chapter 3.4 Subtracting Integers
NEVER SUBTRACT—Add the Opposite

I call it "Double Check method" because you always change 2 signs

1) Change the subtraction sign to an addition sign (check)

2) Change the subtrahend’s sign (the 2^nd number’s sign) to its opposite
(if it was negative change it to positive,
if it had no sign then put a negative because no sign meant it was positive) (double check)

3) Follow the rules of integer addition from  Section 3.3

Example

5 –(-10) = 5 +(+10) =15

-5 –(-10) = -5 +(+10) = +5

-5- 10 = -5 +(-10) = -15

More than 1 subtrahend?  Double check each one!

By the way NEVER EVER EVER CHANGE THE FIRST NUMBER”S SIGN!!

Identifying terms:

Terms are separated by ADDITION
(Remember there is no such thing as subtraction)

2xy is only ONE term—but it has 3 factors

2xy + 3  is 2 two terms

2xy – 3z- (-10) is 3 terms made up of the following

2xy, -3z, and +10
(you must put the terms in Addition Format to determine their signs)

ALWAYS SIMPLIFY BEFORE EVALUTATING

11x + 14 – 21x + 6 + 12x  when x = 5

You will get the same answer if you plug and chug x = 5 in all of the terms as when you SIMPLIFY FIRST and then plug and chug just once.

11(5) + 14 – 21(5) + 6 12(5)

55 + 14 -105 + 6 + 60

135 + (-105)

30

VS

(11x -21x + 12x) + ( 14 + 6)

2x + 20

2(5) + 20

10 + 20

30

Which way do you want to do these types of problems?

Simplifying first is usually the least work!

Quick review of Coefficient and Constants

Remember that a coefficient goes along with a variable and EVERY VRIABLE MUST HAVE  COEFFICIENT

so

2a – 3b –(-c) –d – 12

has 5 terms, 4 variables, 4 coefficeints and 1 constant

The coefficients are 2, 3, 1, -1

The IDENTITY Property of Multiplication (IDx) says that you can sneak in the “1” by multiplication in front of any variable that has no other coefficient.  

Fractional Coefficients

2x

3

can be written 2/3(x)

2 x

3

So the Coefficient of this one term is 2/3

Review of Absolute Value with subtraction inside

Absolute value symbols are similar to parentheses  in that you must simplify inside using Order of Operations (O³) BEFORE applying the absolute value at the end

│2∙3²-30│

You need to do the power first, then multiply, then subtract and THEN absolute value at the very end.

│2∙9-30│= │18-30│=│-12│ = 12