### Video Transcript

Rewriting Expressions Using the
Distributive Property

In this video, weβll discuss what
we mean by the distributive property of multiplication over addition and the
distributive property of multiplication over subtraction. Weβll discuss how we can use this
to rewrite algebraic expressions and how we can use this to help us solve
equations.

To start, letβs define what we mean
by the distributive property of multiplication over addition. This equation explains what we mean
by this property. We say that multiplication
distributes over addition because we can multiply each term inside of our
parentheses by π.

One good way of seeing why this is
true is to look at a geometric example. Remember, these letters π, π, and
π are variables; they can represent many different numbers. So one way of thinking about these
is to represent lengths of a rectangle. So letβs let π be the length of
this rectangle. And letβs say that this rectangle
has a length of π. Then we can calculate some useful
pieces of information about our rectangle. For example, we could calculate the
perimeter or the area.

In particular, we want to know the
area of this rectangle. Well, we can calculate it by
multiplying the length by the width. Itβs equal to π multiplied by
π. And we can see that π multiplied
by π appears in the distributive property of multiplication over addition.

Letβs now do exactly the same to
create π multiplied by π. We want to construct a rectangle
whose area is π multiplied by π. One way of doing this is to have a
rectangle with length π and width π. And we can draw this rectangle on
the same diagram we did before. Weβll also add arrows to represent
our lengths π and π. We can see from this diagram the
height or length of our rectangle is going to be the same as π. And we want the area of this
rectangle to be π multiplied by π. So we should choose the width of
our rectangle to be our value of π.

Now the area of this rectangle is
going to be the length multiplied by the width. Itβs going to be π multiplied by
π. So we have π multiplied by π is
the area of our first rectangle. And π multiplied by π is the area
of our second rectangle.

And we can ask the question βWhat
happens when we add these two areas together?β We can see from our diagram that
combining the two rectangles together, we get another rectangle. We can see the height of this
rectangle is π. And we can also see the width of
this rectangle is our two values combined. The total width of this rectangle
is going to be our value of π plus our value of π.

But then we can ask the question
βWhat is the area of our larger rectangle?β Remember, it must be equal to the
area of the two smaller rectangles because we combined these to make our larger
rectangle. The area of this rectangle is going
to be its length multiplied by its width. Itβs going to be π multiplied by
π plus π.

And this is in fact exactly what
weβre saying in the distributive property of multiplication. π multiplied by π plus π is
equal to π multiplied by π plus π multiplied by π. And this is a very useful result
helping us rewrite algebraic expressions and solve equations.

However, before we see how to use
this in examples, there is one thing worth pointing out. We could have also started with π
plus π multiplied by π. This is exactly the same as what we
started before. However, our two factors are in the
opposite order. We could then do exactly the same
we did before, using rectangles to come up with another equation. We would show that a rectangle of
area π plus π all multiplied by π is the same as two rectangles, one of area π
times π and one of area π times π. And both of these represent the
same property: the distributive property of multiplication over addition.

Letβs now see an example of how we
can use the distributive property of multiplication over addition to rewrite an
algebraic expression.

Use the properties of real
numbers to rewrite six multiplied by π₯ plus three as an equivalent expression
that does not contain parentheses.

In this question, weβre given
an algebraic expression six multiplied by π₯ plus three. And we want to find an
equivalent expression to this algebraic expression. However, we want this to not
contain any parentheses. And weβre told that we should
do this by using the properties of real numbers.

The first thing we need to do
is determine which of the properties of real numbers we can use to simplify our
algebraic expression. And to do this, weβre going to
need to look at our algebraic expression. Since our algebraic expression
involves multiplying the sum of two things, we could try simplifying this by
using the distributive property of multiplication over addition. Remember, this means whenever
weβre adding two numbers together and then multiplying the result, we can
instead multiply each of our terms together and then add the result.

Multiplying six by the first
term inside of our parentheses, we get six multiplied by π₯. Then remember we need to add
six multiplied by the second term in our parentheses, which we know gives us six
multiplied by three. Now all we need to do is
simplify this expression. First, remember, we can
simplify six multiplied by π₯ to just be six π₯. Next, we can evaluate six
multiplied by three is equal to 18. And now we can see this is
equivalent to our original expression; however, it doesnβt contain any
parentheses. Therefore, this gives us our
final answer.

We were able to show by using
the distributive property of multiplication over addition that six multiplied by
π₯ plus three is exactly the same as six π₯ plus 18.

So far, weβve only seen
distribution over addition. What would happen if instead we had
subtraction? And to look at this, weβll start
with a very similar expression to what we had before. However, this time inside our
parentheses, weβll now subtract these two terms. Weβll have π multiplied by π
minus π.

We want to know if we can
distribute π over these parentheses. In other words, we want to ask the
question βDoes multiplication distribute over subtraction?β And if it does, weβll end up with a
very similar expression to what we had before. π multiplied by π minus π will
be equal to π times π minus π times π.

In fact, just like we did before,
we can see an example of this being true by using a geometric interpretation. This time, weβll start by
constructing a rectangle whose area is π multiplied by π minus π. Remember the area of a rectangle is
its length multiplied by its width. So letβs start by calling the
length of our rectangle π.

Next, remember we want the area of
this rectangle to be π multiplied by π minus π. So the width of our rectangle needs
to be π minus π. But this then gives us an idea. We can compare this to a rectangle
whose area is π multiplied by π. π is going to be bigger than π
minus π because subtracting π from π makes it smaller. So in the same diagram, weβll add a
rectangle whose area is π multiplied by π. This is the outer rectangle.

We can then use this diagram to
find the width of our smallest rectangle. We have π must be equal to π
minus π plus this width. In other words, this width has to
be equal to π. So letβs now see what weβve
shown. The first rectangle we constructed
will have area π multiplied by π minus π. This is because its length is π
and its width is π minus π. Next, we can also add on the area
of the smaller rectangle on the end. This has length π and width π, so
its area is π multiplied by π.

But remember the area of these two
rectangles added together must be equal to the area of our outer rectangle. And we can calculate the area of
this outer rectangle. It has length π and width π. So its area is π multiplied by
π. This gives us π multiplied by π
minus π plus π times π should be equal to π multiplied by π. And we can subtract π multiplied
by π from both sides of the equation. And by doing this, we get π
multiplied by π minus π should be equal to π times π minus π times π. And we call this the distributive
property of multiplication over subtraction.

And before we move on to some
examples, there is one thing worth pointing out. Just like we did before, we could
have instead started with π minus π all multiplied by π. And by using a very similar
argument to what we used above, we could show a very similar result about
distributing multiplication over subtraction. Both of these results are exactly
the same, and theyβre both called the distributive property of multiplication over
subtraction. We have π minus π all multiplied
by π is π times π minus π times π.

Letβs now see an example of how we
can use the distributive property of multiplication over subtraction to simplify an
algebraic expression.

Use the distributive property
to rewrite the algebraic expression four multiplied by π₯ minus nine.

In this question, weβre given
an algebraic expression. And weβre told we need to
rewrite this algebraic expression by using the distributive property. To answer this question, weβre
first going to need to know which of our distributive properties weβre going to
need to use. And we can see the algebraic
expression weβre asked to rewrite is four multiplied by π₯ minus nine. Inside of our parentheses, we
have subtraction. So we need to use the
distributive property of multiplication over subtraction.

So weβll start with our
algebraic expression four multiplied by π₯ minus nine. Then we need to remember when
weβre using the distributive property of multiplication over subtraction,
instead of evaluating four multiplied by π₯ minus nine, we can instead multiply
each term by four. Multiplying the first term by
four, we get four multiplied by π₯. And multiplying the second term
by four, we get four multiplied by nine.

And itβs very important to
remember because weβre subtracting the two numbers inside of our parentheses, we
need to subtract these two values. So we get four times π₯ minus
four times nine. We can evaluate each of these
separately. First, four multiplied by π₯
can be simplified to give us four π₯. Next, we need to subtract four
multiplied by nine. Well, we know four multiplied
by nine is equal to 36. So we need to subtract 36. And this gives us our final
answer.

Therefore, we were able to use
the distributive property to rewrite the algebraic expression four multiplied by
π₯ minus nine to be equal to four π₯ minus 36.

Letβs now see another example of
how we can use the distributive property to rewrite an algebraic expression.

Using the distributive
property, rewrite the expression negative two π₯ minus eight all multiplied by
negative six.

In this question, weβre given
an algebraic expression, and we need to rewrite this by using the distributive
property. To do this, we first need to
determine which of the distributive properties we need to use. We see weβre multiplying an
expression by negative six. And this expression contains
subtraction. So weβre going to use the
distributive property of multiplication over subtraction.

So letβs start by writing out
our algebraic expression in full. Itβs negative two π₯ minus
eight all multiplied by negative six. And in fact, there are two
different methods we could use to answer this question. The first thing we could do is
reorder the two factors given to us in this expression because remember we know
it doesnβt matter which order you multiply two numbers together. It will always give us the same
answer.

However, this is not necessary
because we can directly use the distributive rule of multiplication over
subtraction on this example. We need to multiply both terms
inside of our parentheses by negative six. And remember we need to keep
the subtraction sign. So letβs start with the first
term multiplied by negative six. The first term is negative two
π₯. So we need to multiply this by
negative six. We get negative two π₯
multiplied by negative six.

Next, we need to multiply the
second term by negative six. And this is eight, so we get
eight multiplied by negative six. And remember, we need to
subtract these two values because inside of our parentheses we were
subtracting. This gives us negative two π₯
multiplied by negative six minus eight times negative six.

And we can evaluate each of
these separately. Letβs start with negative two
π₯ multiplied by negative six. Remember, when we multiply an
π₯-term by a constant, we just need to multiply the two numbers together. We need to multiply negative
two and negative six. And remember a negative
multiplied by a negative gives us a positive, so this is equal to 12π₯. Then we need to calculate eight
multiplied by negative six. This is a positive multiplied
by a negative. And we know six times eight is
equal to 48. So this gives us negative
48.

But remember, weβre subtracting
this value. And when we subtract a negative
value, thatβs the same as just adding a positive value. So instead of subtracting
negative 48, weβll just instead add 48. And this gives us our final
answer. Therefore, by using the
distributive property of multiplication over subtraction, we were able to
rewrite the expression negative two π₯ minus eight all multiplied by negative
six as 12π₯ plus 48.

Letβs now look at a question asking
us to identify when the distributive property has been used.

Which of the following
demonstrates the distributive property? Option (A) negative four plus
π₯ plus three is equal to three plus π₯ minus four. Option (B) negative four π₯
plus three is equal to three minus four π₯. Option (C) negative four π₯
plus three plus zero is equal to negative four π₯ plus three. Or option (D) negative four π₯
multiplied by π₯ plus three is equal to negative four π₯ minus 12.

In this question, weβre given
four options. And we need to determine which
of these demonstrates use of the distributive property. Remember when weβre using the
distributive property, we need something multiplied by the sum or difference of
two values. And just by using this, we can
see options (A), (B), and (C) donβt contain this. Thereβs nothing multiplied by
the sum or difference of two terms. So we could already just
eliminate these three options. However, for due diligence,
letβs go through what they represent.

Letβs start with option
(A). In option (A), the only
difference between both sides of our equation is weβve rearranged our
expression. Weβve swapped negative four and
three. And thereβs a few different
ways we could arrive at this. For example, we could use our
order of operations and the fact that we know when you add two numbers together,
you can add them in either order. It wonβt change the result. This would give us that these
two sides of the equation are equal. However, this does not use the
distributive property. Therefore, option (A) does not
use the distributive property.

We can say something very
similar about option (B). The only difference between
both sides of the equation is weβre swapping the order of the terms. And once again, this does not
use the distributive property.

In option (C), the only
difference between our two sides of the equation is on the left-hand side weβre
adding zero. And we know that these are
equal because adding zero to a number doesnβt change its value. However, this does not use the
distributive property.

Finally, letβs look at option
(D). On the left-hand side of the
expression in option (D), we have negative four multiplied by π₯ plus three. And we can see weβre indeed
multiplying the sum of two values. So we could do this by using
the distributive property of multiplication over addition. We multiply each of the terms
inside of our parentheses by negative four and then add the result together. We get negative four π₯ plus
negative four multiplied by three. And if we simplify this
expression, weβll see that we get a negative four π₯ minus 12. And this is exactly what is
shown in option (D). Therefore, option (D) correctly
demonstrates the distributive property of multiplication over addition.

And this gives us our final
answer. Of the four options, only
option (D) demonstrates the distributive property.

Letβs now go through one final
example of how we can use the distributive property to help us solve a real-world
problem.

Liam had π dollars in his
savings account and then he deposited 11 dollars. Seven months later, his balance
had doubled. Which of the following is
equivalent to his new balance of two multiplied by π plus 11 dollars? Option (A) two π plus 11. Option (B) π plus 22. Option (C) two π plus 13. Option (D) two π plus 22. Or option (E) π plus 13.

In this question, weβre given
the information to find the new balance of Liamβs savings account. However, weβre also given this
value. Itβs two multiplied by π plus
11. And weβre given five options,
and we need to determine which of those five options is equivalent to this
value.

And to do this, we need to
notice that the balance of Liamβs savings account is given in a very useful
form. Weβre multiplying the sum of
two values. So we can rewrite this by using
the distributive property of multiplication over addition.

So letβs start with our
expression of two multiplied by π plus 11. We need to use the distributive
property of multiplication over addition. Remember, this means we
multiply each term inside of our parentheses by two. The first term inside of our
parentheses is π. So we multiply this by two to
get two π. Then the second term inside of
our parentheses is 11. So we multiply this by two to
get two times 11.

And itβs very important to
remember because weβre adding the two values inside of our parentheses, we need
to add these two values together. So we have two π plus two
times 11. Of course, we can simplify
this. We know that two times 11 is
equal to 22. So this simplifies to give us
two π plus 22. And this is exactly the same as
option (D) given to us in the question. Therefore, we were able to show
the balance of two multiplied by π plus 11 is equal to two π plus 22, which
was option (D).

Letβs now go over the key points of
this video. First, we showed that we can use
the distributive property to rewrite certain algebraic expressions without
parentheses. Next, we saw that multiplication
can be distributed over both addition and subtraction.

In fact, we saw four different ways
we could use this. First, we saw that π multiplied by
π plus π is π times π plus π times π. We also saw that π plus π all
multiplied by π is π times π plus π times π. And in both of these cases, we turn
a product involving a sum into a sum of products. We also saw π multiplied by π
minus π is π times π minus π times π. Similarly, we also saw π minus π
all multiplied by π is π times π minus π times π. And both of these results turn a
product involving a difference into the difference of products.