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Distributive Property Absent copy

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Distributive Property Absent copy

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Distributive PropertyAbsent copy

9/19,20

Simplify. 4(x + 5)

4(x) + 4(5)

4x + 20

Solution

- 1. We can’t add x and 5 why?
- We can not add them because there is a variable and a constant.
- 2. What do we do with the 4 on the outside of the parenthesis?
- We have to distribute the 4 to both the x and the 5. We multiply the 4 to both of them.

4x + 20

Simplify:

6(2y + 3)

6(2y) + 6(3)

12y + 18

Solution

1. We can’t add 2y and 3

why?

- We can not add them because there is a coefficient with the variable and a constant.
2. What do we do with the 6 on the outside of the parenthesis?

We have to distribute the 6 to both the 2y and the 3. We multiply the 6 to both of them.

12y + 18

Simplify:

4(2r + 5) - 3r

4(2r) + 4(5) – 3r

8r + 20 – 3r

1. 8r – 3r = 5r

2. 20

Solution

- Can we add what is in the parenthesis?
Yes or no?

2. What do we do with the 4 that is outside the parenthesis?

- We have to distribute the 4 to both the 2r and the 5. We multiply the 4 to both of them.
3. What is the last step before we get a solution?

- To add or subtract the like terms that are the same.

5r + 20

Simplify:

4(4x + 4) + 3(x + 5)

4(4x) + 4(4) + 3(x) + 3(5)

16x + 16 + 3x + 15

1. 16x + 3x = 19x

2. 16 + 15 = 31

Solution

1. How many groupings do we have?

- We have 2 different groupings.
2. What do we do with the 4 & 3 that are outside the parenthesis ?

- We have to distribute the 4 through the first parenthesis and the 3 through the second parenthesis.
3. What is the last step before we a get solution?

- To add or subtract the like terms that are the same.

19x + 31