The Distributive Property for 7th Grade - Examples, Exercises and Solutions

In order to simplify the calculation , we first break down 94 and 72 into smaller and preferably round numbers.

We obtain the following exercise:

$90+4+70+2=$

Using the associative property, we then rearrange the exercise to be more functional.

$90+70+4+2=$

We solve the exercise in the following way, first the round numbers and then the small numbers.

$90+70=160$

$4+2=6$

Which results in the following exercise:

$160+6=166$

To solve the problem, first we will use the distributive property on the two numbers:

(60+3)-(30+6)

Now, we will use the substitution property to arrange the exercise in the way that is most convenient for us to solve:

60-30+3-6

It is important to pay attention that when we open the second parentheses, the minus sign moved to the two numbers inside.

30-3 = 

27

We will use the distributive law and split the number 143 into a sum of 100 and 43.

The distributive law allows us to distribute, meaning, to split a number into two or more numbers. This actually allows us to work with smaller numbers and simplify the operation.

$(100+43)-43=$

We will operate according to the order of operations

We can remove parentheses and perform addition and subtraction operations in any order since there are only addition and subtraction operations in the equation

$100+43-43=100+0=100$

Therefore, the answer is option C - 100.

And now let's see the solution to the exercise in a centered format:

$143-43= (100+43)-43= 100+43-43=100+0=100$

In order to solve the question, we first use the distributive property for 133:

$(100+33)+30=$

We then use the distributive property for 33:

$100+30+3+30=$

We rearrange the exercise into a more practical form:

$100+30+30+3=$

We solve the middle exercise:

$30+30=60$

Which results in the final exercise as seen below:

$100+60+3=163$

In order to simplify the resolution process, we begin by using the distributive property for 140:

$100+40-70=$

We then rearrange the exercise using the substitution property into a more practical form:

$100-70+40=$

Lastly we solve the exercise from left to right:

$100-70=30$

$30+40=70$

There are several ways to solve the following exercise,

We will present two of them.

In both ways, we begin by decomposing the number 84 into smaller units; 80 and 4.

$\frac{4}{4}=1$

Subsequently we are left with only the 80.

Continuing on with the first method, we will then further decompose 80 into smaller units; $10\times8$

We know that:$\frac{8}{4}=2$

And therefore, we are able to reduce the exercise as follows: $\frac{10}{4}\times8$

Eventually we are left with$2\times10$

which is equal to 20

In the second method, we decompose 80 into the following smaller units:$40+40$

We know that: $\frac{40}{4}=10$

And therefore: $\frac{40+40}{4}=\frac{80}{4}=20=10+10$

which is also equal to 20

Now, let's remember the 1 from the first step and add it in to our above answer:

$20+1=21$

Thus we are left with the following solution:$\frac{84}{4}=21$

We solve each of the options and keep in mind the order of arithmetic operations: calculation of the operation within parentheses, multiplication and division (from left to right), addition and subtraction (from left to right).

a.

$(10+3)\times(30-1)=13\times29$

b.

$10\times3\times30\times1=30\times30\times1=900$

c.

$(10\times3)\times30=13\times30$

d.

$10\times3+29=30+29=59$

Therefore, the answer is option A.

We solve each of the options and keep in mind the order of operations: calculation of the operation within parentheses, multiplication and division (from left to right), addition and subtraction (from left to right).

a.

$36+4=40$

b.

$4\times(30+6)=4\times36$

c.

$40-4+4=36+4=40$

d.

$4\times30+6=120+6=126$

Therefore, the answer is option B.

We solve each of the options and keep in mind the order of arithmetic operations: calculation of the operation within parentheses, multiplication and division (from left to right), addition and subtraction (from left to right).

a.

$3\times8\times3=24\times3=72$

b.

$(2+1)\times(80+3)=3\times83$

c.

$3+(80+3)=3+83$

d.

$3+83=86$

Therefore, the answer is option B.

We solve each of the options and keep in mind the order of operations: calculation of the operation within parentheses, multiplication and division (from left to right), addition and subtraction (from left to right).

a.

$10+4+40+2=14+42=36$

b.

$(10\times4)+(40\times2)=40+80$

c.

$(10+4)\times(40+2)=14\times42$

d.

$10\times4\times40\times2=40\times40\times2=160\times2=360$

Therefore, the answer is option C.

We solve each of the options and keep in mind the order of operations: calculation of the operation inside parentheses, multiplication and division (from left to right), addition and subtraction (from left to right).

a.

$160+6=166$

b.

$(100\times60)\times6=6,000\times6$

c.

$(100+60)+(3+3)=160+6$

d.

$(100+60)\times(3+3)=160\times6$

Therefore, the answer is option D.

We solve each of the options and keep in mind the order of operations: calculation of the operation within parentheses, multiplication and division (from left to right), addition and subtraction (from left to right).

a.

$(30+4)+11=34+11$

b.

$30\times4\times11=120\times11=1,320$

c.

$(30+4)+10+1=34+11$

d.

$(30+4)\times(10+1)=34\times11$

Therefore, the answer is option D.

We solve each of the options and keep in mind the order of operations: calculation of the operation within parentheses, multiplication and division (from left to right), addition and subtraction (from left to right).

a.

$(30×9)+(10×9)= 270+90$

b.

$(30+9)×(10×9)= 39×90$

c.

$(40-1)×(20-1)= 39×19$

d.

$(40-1)+(20×1)= 39+20$

Therefore, the answer is option C.

To make solving easier, we break down 53 into more comfortable numbers, preferably round ones.

We obtain:

$4\times(50+3)=$

We multiply 2 by each of the terms inside the parentheses:

$(4\times50)+(4\times3)=$

We solve the exercises inside the parentheses and obtain:

$200+12=212$

To make solving easier, we break down 11 into more comfortable numbers, preferably round ones.

We obtain:

$(10+1)\times34=$

We multiply 34 by each of the terms in parentheses:

$(34\times10)+(34\times1)=$

We solve the exercises in parentheses and obtain:

$340+34=374$