Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
7. The Distributive Property
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Exercise 6 Page 49

Practice makes perfect
a Recall the Distributive Property.
a(b+c)=a* b+a* c.We want to verify whether or not the given equation demonstrates the Distributive Property. -2(x+1)=-2x-2 We can do that by distributing -2 on the left-hand side to the terms inside the parentheses. Then we will compare the obtained result to the right-hand side of the equation.
-2(x+1)? =-2x-2
â–Ľ
Distribute -2
-2* x+(-2)* 1? =-2x-2
- 2x+(- 2)? =-2x-2
- 2x-2=-2x-2 âś“
Since the left-hand side now matches the given right-hand side, we know that the Distributive Property was shown.
b Consider the given equation.

(s-4)8=8(s-4) This equation demonstrates the Commutative Property of Multiplication which shows that the factors of a product can be rearranged without changing the product's value. a* b&= b* a (s-4) 8&= 8 (s-4) Note, the 8 was not distributed to the terms inside the parentheses. Therefore, we did not use the Distributive Property.

c Consider the given equation.
5n-45=5(n-9)This equation demonstrates factoring out a common factor.
5n-45? =5(n-9)
5* n-5* 9? =5(n-9)
5(n-9)=5(n-9) âś“
Note that if we rearrange the equation, we will get a perfect example of the use of the Distributive Property. 5n-45=5(n-9) ⇕ 5(n-9)=5n-45 Therefore, the given equation demonstrates the Distributive Property but in the opposite direction.
d Consider the given equation.

8+(t+6)=(8+t)+6 This equation demonstrates the Associative Property of Addition which shows that we can add terms in different orders without changing the sum. a+ ( b+ c)&=( a+ b)+ c 8+ ( t+ 6)&=( 8+ t)+ 6 In the equation does not occur any distribution, so it does not demonstrate the Distributive Property.