The Distributive Property for 7th Grade: Introducing factorization

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.

Given that we have two exercises in parentheses, we know that each of them represents a number.

Therefore, what we will do is solve the exercises in parentheses to reveal which two numbers we are multiplying:

$5+1=6$

$500+30=530+2=532$

Now we get the exercise:

$6\times532$

Let's keep in mind that each set of parentheses represent a number.

That is, if we solve the exercises within parentheses we will get a multiplication exercise of two numbers.

We solve each of the exercises within parentheses to reveal what the numbers are:

$10+6=16$

$70+3=73$

That is, the exercise we obtained is the multiplication between two numbers:

$16\times73$$$

Let's consider that each set of parentheses represent a number.

That is, if we solve the exercises within parentheses we will obtain a multiplication exercise of two numbers.

We solve each of the exercises within parentheses to reveal what the numbers are:

$400+20=420$

$4+1=5$

That is, the exercise we obtained is the multiplication between two numbers:

$420\times5$

Let's keep in mind that we have been given a multiplication exercise between two parentheses.

Each set of parentheses represents a number. If we solve the exercises within the parentheses, we will reveal what the two numbers are between which there is a multiplication exercise:

$2+1=3$

$60+2=62$

That is, the multiplication exercise we obtained is:

$3\times62$

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.

$50\times2\times60=100\times60=6,000$

b.

$(50+2)+(2+60+2)=52+64$

c.

$(2\times3)+60+60=6+120$

d.

$(50+2)\times2\times(60+2)=52\times2\times62$

Therefore, the answer is option D.

Let's keep in mind that each set of parentheses represents a number.

That is, if we solve the exercises within the parentheses, we will get a multiplication exercise of two numbers.

We solve each of the exercises within the parentheses to reveal what the numbers are:

$30+2=32$

$10+4=14$

That is, the exercise we obtained is the multiplication between two numbers:

$32\times14$

Let's consider that each set of parentheses represents a number.

That is, if we solve the exercises within the parentheses, we will get a multiplication exercise of three numbers. We can't simplify the 3 so it does not get factored.

We solve each of the exercises within the parentheses to reveal what the numbers are:

$20+3=23$

$20+5=25$

That is, the exercise we obtained is the multiplication between three numbers:

$3\times23\times25$

Note that all parentheses represent a number for us.

This means that if we solve the expressions in parentheses, we will get an exercise of multiplication between three numbers.

We will leave 2 as it is since it is not a number inside parentheses but stands on its own.

Let's solve each of the expressions in parentheses to discover what the numbers are:

$10+2=12$

$20+2=22$

Therefore, the exercise we got is multiplication between three numbers:

$2\times12\times22$

Let us begin by decomposing 19 into a subtraction exercise:

$(20-1):8=$

Next we divide each term inside of the parentheses by 8:

$20:8-1:8=$

We can see that answers C+D can be discarded since we divided 20 by 8.

Now we write the division exercise on the right as a fraction and we get:

$20:8-\frac{1}{8}$

First, we break down 74 into a subtraction exercise as follows:

$(80-6)\times13=$

Next we multiply each of the terms inside the parentheses by 13 and obtain the following:

$80\times13-6\times13$

We begin by breaking down 29 into a subtraction exercise:

$(30-1)\times28=$

We then multiply each term inside the parentheses by 28 and we obtain the following:

$30\times28-1\times28=$

Let's observe the multiplication exercise on the right and remember that each term multiplied by 1 is equal to the term itself.

That is:

$30\times28-28$