Originally posted by ASeeker343
reply to post by Cuervo
You are using the distributive property incorrectly. See my previous posts.
Originally posted by lifeoflyman
reply to post by Cuervo
Your exactly right you have to implement the distributive law!
THE DISTRIBUTIVE LAW
For all real numbers a, b, and c:
a(b+c)=ab+ac
The distributive law offers two different orders of operation that always give the same result:
The expression a(b+c) specifies this order:
add b to c; then multiply a by this sum.
The expression ab+ac specifies this order:
multiply a and b; multiply a and c; then add these results.
One effective visual way to understand the statement of the distributive law is to use areas (see below).
So 6/2(1+2) Lets break it down! 6 divided by 2(1+2)
then gives you 6 divided by (2+4)
Now we are left with 6 dived by 6= 1
Originally posted by prolific
reply to post by Itop1
You wouldn't need Google if you knew that 2(3) is not distribution but its actually multiplication.
Heres an example:
edit on 1-5-2011 by prolific because: (no reason given)
Originally posted by snowspirit
6÷2(1+2)=?
Order of operations: BEDMAS
Brackets
Exponents
Division
Multiplication
Addition
Subtraction
1+2=3
2x3=6
6/6=1edit on 1-5-2011 by snowspirit because: spelling
Originally posted by lifeoflyman
reply to post by ASeeker343
6÷2(1+2)=
so according to the order of operations if you have 6*2÷3= You would do the multiplication first! According to your statement you say that since 2÷3 are together you can not break them apart there for that in of itself violates the orders of operation because you are multiplying only the 6 and the 2.
I think the problem is in how it is written. I would say that 6/2 and 6÷2 could be interpreted differently. as ÷ shows the action that is need to be done as / could be used for division or a fraction.
What if the problem was 6*2(1+2) what would be the first action you would take?
Originally posted by Cataclysmo
This is hilarious:
6÷2(1+2)=1
6 ÷ 2(1+2)
c ÷ b(a+b) (a+b) = d
c ÷ b(d)
c÷ bd
Brackets first, then if you remember how algebra works "bd" together means you multiply those two. And if I remember correctly because it's written as b(d) that means this has to be resolved first before you finish the rest. It's like exponents in that they are attached to each other. Would you do 6 ÷ 2^(3) with 6 divided by 2 first or complete finding out what 2 to the third power was?