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Can I not use distributive property when dealing with multiplication and division?

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You can for three of them, as long as you follow the order of operations (BEDMAS/PEDMAS).  The third one is the exception since we have to divide by 2 before multiplying by the (2+1).
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First, the distributive law is distribution of MULTIPLICATION over ADDITION. I've seen people try to use multiplication over multiplication, exponentiation over addition, and other combinations that are not multiplication over addition. Don't do that.

Now let's take a look at your problems.

>6×2(2+1) =36

You don't need the distributive law. This is a product of three numbers: 6, 2 and (2 + 1) = 3. The associative and commutative laws (of multiplication) apply. You can multiply the 6, 2 and 3 in any order, obtaining 36.

But if you want to use the distributive law, you can do 12(2 + 1) = 12*2 + 12*1 = 24 + 12 = 36.

Or 6 \* (2\*2 + 2\*1) = 6 \* 6 = 36.

>6+2(2×1) =10

The distributive law doesn't apply here because nothing is being multiplied by the sum (6 + 2). By order of operations you have the product 2(2 x 1) = 4 and that is being added to 6 to give 6 + 4 = 10.

>6÷2(2+1) =9 There is a multiplication of a sum here, but the question is what is being multiplied by it?

By order of operations, you do 6/2 = 3 first and then the (2 + 1) is multiplied by that. So you are evaluating 3(2 + 1). If you want to use the distributive law it's valid, and this is equal to 3\*2 + 3 \* 1 = 6 + 3 = 9.

>6+2(2÷1) =10

Again, nothing is being multiplied by the sum, so there's no distributive law here. 6 is being added to the product 2(2/1) which is 4. 6 + 4 = 10
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You can.

What exactly do you mean by "use the distributive property" here? What are you doing that leads you to a different answer?

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