Lesson Video: Rewriting Expressions Using the Distributive Property | Nagwa Lesson Video: Rewriting Expressions Using the Distributive Property | Nagwa

# Lesson Video: Rewriting Expressions Using the Distributive Property Mathematics

In this video, we will learn how to rewrite algebraic expressions using the distributive property.

17:58

### Video Transcript

Rewriting Expressions Using the Distributive Property

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

To start, letβs 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 letβs let π be the length of this rectangle. And letβs 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. Itβs equal to π multiplied by π. And we can see that π multiplied by π appears in the distributive property of multiplication over addition.

Letβs 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. Weβll 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. Itβs 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 βWhat 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 βWhat 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. Itβs going to be π multiplied by π plus π.

And this is in fact exactly what weβre 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.

Letβs 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, weβre 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 weβre 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, weβre 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 weβre 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 doesnβt 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, weβve only seen distribution over addition. What would happen if instead we had subtraction? And to look at this, weβll start with a very similar expression to what we had before. However, this time inside our parentheses, weβll now subtract these two terms. Weβll 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 βDoes multiplication distribute over subtraction?β And if it does, weβll 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, weβll 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 letβs 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, weβll 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 letβs now see what weβve 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 theyβre 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.

Letβs 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, weβre given an algebraic expression. And weβre told we need to rewrite this algebraic expression by using the distributive property. To answer this question, weβre first going to need to know which of our distributive properties weβre going to need to use. And we can see the algebraic expression weβre 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 weβll start with our algebraic expression four multiplied by π₯ minus nine. Then we need to remember when weβre 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 itβs very important to remember because weβre 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.

Letβs 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, weβre 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 weβre multiplying an expression by negative six. And this expression contains subtraction. So weβre going to use the distributive property of multiplication over subtraction.

So letβs start by writing out our algebraic expression in full. Itβs 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 doesnβt 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 letβs 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. Letβs 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, weβre subtracting this value. And when we subtract a negative value, thatβs the same as just adding a positive value. So instead of subtracting negative 48, weβll 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.

Letβs 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, weβre given four options. And we need to determine which of these demonstrates use of the distributive property. Remember when weβre 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) donβt contain this. Thereβs nothing multiplied by the sum or difference of two terms. So we could already just eliminate these three options. However, for due diligence, letβs go through what they represent.

Letβs start with option (A). In option (A), the only difference between both sides of our equation is weβve rearranged our expression. Weβve swapped negative four and three. And thereβs 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 wonβt 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 weβre 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 weβre adding zero. And we know that these are equal because adding zero to a number doesnβt change its value. However, this does not use the distributive property.

Finally, letβs 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 weβre 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, weβll 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.

Letβs 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, weβre given the information to find the new balance of Liamβs savings account. However, weβre also given this value. Itβs two multiplied by π plus 11. And weβre 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 Liamβs savings account is given in a very useful form. Weβre multiplying the sum of two values. So we can rewrite this by using the distributive property of multiplication over addition.

So letβs 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 itβs very important to remember because weβre 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).

Letβs 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.

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