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Rewriting addition problems


There are some important properties with whole numbers that you may have been using for a while, but maybe you never knew it.


Commutative Property


The commutative property for addition states that if you take an additive expression like $12+7$12+7 that this is equivalent to $7+12$7+12.  We know that this makes sense as on a number line we can see that $12+7$12+7  and $7+12$7+12 result in the same answer.


Associative Property


The associative property states that it doesn't matter how we group terms when we add them. For example, $2+3+4$2+3+4 is equivalent to $2+3+4$2+3+4. Again, we know this make sense because, when we plot them on a number line, we get the same answer.


Distributive Property

$a\times\left(b+c\right)=a\times b+a\times c$a×(b+c)=a×b+a×c

The distributive property is one that deals with groups.  We can say that $3$3 groups of $4$4 is the same as $3$3 groups of $3$3 and $3$3 groups of $1$1.

We can write this mathematically, remembering that groups of, is another way of representing multiplication.


With one more rearrangement, we can see that 

Mathematically we write this as

$3\times3+3\times1=3\times\left(3+1\right)$3×3+3×1=3×(3+1) $=$= $3\left(3+1\right)$3(3+1)


Let's run through some more examples now.


Question 1


Consider the sum of $42$42 and $39$39.

  1. Complete the gaps such that $42+39$42+39 is rewritten as an equivalent multiplication using the distributive property.






Question 2

Given $12\left(7+6\right)-6\left(7+6\right)$12(7+6)6(7+6):

  1. Complete the gaps to create an equivalent expression in the form $a\left(x+y\right)$a(x+y).

    $12\left(7+6\right)-6\left(7+6\right)$12(7+6)6(7+6) $=$= $\left(\editable{}-6\right)\times\left(7+6\right)$(6)×(7+6)
      $=$= $\editable{}\left(7+6\right)$(7+6)
  2. Using the distributive law, complete the gaps to rewrite the number sentence as the sum of two integers in the form $a+b$a+b.




      $=$= $\editable{}+36$+36

Question 3

Yvonne is splitting a number of students into groups for a treasure hunt. She has $108$108 boys and $48$48 girls and doesn't want any boys or girls left out.

  1. What's the largest number of groups she can make if each group has the same number of boys and the same number of girls?

  2. How many males and females need to go into each group?

    Males Females
    $\editable{}$ $\editable{}$
  3. Rewrite the above problem as an equivalent statement in terms of the number of groups (use the distributive law):

    $108+48$108+48$=$= $12\left(9+\editable{}\right)$12(9+)



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