Math 6 Honors (Period 6 and 7)

Quotients of Integers 11-6

We all remember 2⋅ 5 = 10
and we know corresponding information
10÷ 5 = 2

The quotient of two positive OR two negative integers is POSITIVE!!

The quotient of a positive AND a negative integer is NEGATIVE!!

The same rules of multiplication apply to division. The same life story!! :-)

Although we talked about the sum of two integers always being an integer and
the difference of two integers always being an integers...

the QUOTIENT of two integers is NOT ALWAYS an integer!!

10/4 = 2 1/2 --> which is NOT an integer!!

We also reviewed:

10/0 --> is undefined!!
whereas,
0/10 = 0

98/-14 = -7

Simplify the following:
6 × 8 + -3 × 5
7 × 5 + -3 × 8

Make sure you perform operations using PEMDAS
(also called Aunt Sally's rules or even Order of Operation O³)

6 × 8 + -3 × 5
7 × 5 + -3 × 8

= 3

Solving Equations 11-7

Now that we have learned about negative integers, we can solve an equation such as
x + 7 = 2
We need to subtract 7 from both sides of the equation
x + 7 = 2
- 7 = - 7

to do this use a side bar and use the rules for adding integers
Notice the signs are different so
ask yourself... Who wins? and By How Much?
stack the winner on top and take the difference
so
x + 7 = 2
- 7 = - 7
x = -5

y -^- 6 = 4
add the opposite and you get
y + 6 = 4
now you need to subtract 6 from both sides of the equation
y + 6 = 4
- 6 = - 6
Again the signs are different -- ask your self those all important questions
"Who Wins? and "By How Much?"
Use a side bar, stack the winner on top and take the difference. Make sure to use the winner's sign in your answer!!
y = 2

2- STEP EQUATIONS

What about
3z - ^- 15 = 9
add the opposite first and you get
3x + 15 = 9
In order to solve this 2 step equation
we need to do the reverse of PEMDAS-- as we did with unwrapping the present so many months ago
3x + 15 = 9
subtract 15 from both sides of the equation
3x + 15 = -9
- 15 = - 15

This time the sides are the same-- so just add them and use their sign
3x + 15 = -9
- 15 = - 15
3x = -24
Now divide both sides by 3
3x = -24
3 3

x = -8

Make sure to BOX your answer!!
What about this one
(1/2)(x) + 3 = 0
subtract 3 from both sides
(1/2)x = -3

Multiple by the reciprocal of 1/2 which is 2/1
(2/1)(1/2)x = -3(2/1)
x = -6
Again box your answer.

3u - 1 = -7
+ 1 = + 1
3u = -6
divide both sides by 3 ( or multiple by the reciprocal of 3 which is 1/3)
3u/3 = -6/3
u = -2

What about x = -6 + 3x
OH dear... we have variables on BOTH sides of the equations... we need to get the variables on one side all the constants on the other.
We need to isolate the variable!!

x = -6 + 3x
What if we add six to both sides
x = -6 + 3x
+6 = + 6
x + 6 = 3x
now we need to subtract x from both sides
x + 6 = 3x
- x - x
6 = 2x
so now divide both sides by 2
6/2 = 2x/2
3 = x

How about this one
3 - r = -5 + r
- 3 = - 3
-r = -8 + r
if subtract r from both sides, I will get rid of the +r on the right side

-r = -8 + r
- r = -r
-2r = -8
Now divide by -2 on both sides

-2r/-2 = -8/-2
r = 4