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Grammar in English and Mathematics

Grammar helps us communicate the meaning of a sentence.

Consider the punctuation in the following two sentences:

We can use the exact same words, in the exact same order to write two sentences that have completely different meanings.

In math, our sentences consist of numbers and operations.

We need rules to clarify what we are being asked to calculate.

Lesson Goals

• Review the order of operations.
• Explore when brackets are needed and when they can be removed in an expression.
• Perform calculations involving integers, fractions, and decimals using the order of operations.
• Use the distributive property to simplify calculations.

Introducing Symbols

We introduce symbols to save time and space, and to add precision.

For example,

\(3+5=8\)

"The sum of three and five equals 8."

Can you think about why that is? In this example, it looks just like space saving, but as expressions get more complicated, symbols are essential to ensure precision.

Try This!

Write the following computation using symbols. Your expression should include all needed information so that a friend can calculate the correct answer.

"The sum of three and five is multiplied by the sum of two and four."

Think about this problem, then move on to the next part of the lesson.

BEDMAS

Try This Problem Revisited

Let's go back to our statement for a second from the Try This Problem. It read,

"The sum of three and five is multiplied by the sum of two and four."

Did you come up with 

\((3 +5) \times (2+4)\)

where the \(3+5\) is in brackets and the \(2+4\) is also in brackets? 

Perhaps your expression does not have the same brackets that mine does. How important are these brackets? To answer this question, let's start by looking at a simpler problem.

Example 1

Because we have two operations, there are two ways that we could approach this problem.

You get two different answers depending on the order in which you do the operations. How could you communicate to a friend using only symbols, which of the two calculations you intended for them to compute?

Well, we could underline and number the steps.

We could circle the step we deem to be most important.

If you're like me and have fancy developers, you could have them put birds flying around the calculation you want to do. 

Source: Bird - Ekaterina_P/iStock/Getty Images Plus

But if you've seen the order of operations before, then you will know that the second solution is correct.

What went wrong in the first solution?

It is understood in math that we need to follow a specific order when performing the operations.

The calculation in Solution 1 is breaking this rule. 

BEDMAS

We always need to remember that math is a language. We need to be sure that we consider both the words and the grammar in our language, which means that we need to be aware of the numbers, the operations, and the order in which all of those things appear together.

In order to evaluate complicated expressions, we need to follow the rules for how the words are combined.

BEDMAS is the grammar, and it tells us what operations to compute first when we're evaluating expressions.

BEDMAS stands for

B

Brackets

E

Exponents

D
M

Division and multiplication in the order they appear from left to right

A
S

Addition and subtraction in the order they appear from left to right

There are many people who learn an alternative form of BEDMAS which is PEDMAS, and the P stands for parentheses. Whether you use brackets or parentheses, it doesn't matter. But in this course, going forward we're going to refer to them as brackets.

Recall: Let's think about the expression \(8+2\times 4\) in terms of BEDMAS. We now know that the multiplication should come first, which means Solution 2 is correct.

Solution 2

\(\begin{align*} 8+2 \times 4 &= 8+8\\ &= 16\end{align*}\)

✔

So what do we need to do if we intended for the first calculation to be correct? 

Solution 1

\(\begin{align*} (8+2) \times 4 &= 10\times 4\\ &= 40 \end{align*}\)

✘

We need to use brackets to indicate that you should tackle that addition problem before you multiply.

\(\begin{align*} \class{hl1}{(}8+2\class{hl1}{)} \times 4 &= 10\times 4\\ &= 40 \end{align*}\)

✔

What do we need to do if we intended the first calculation to be correct? 

Brackets are used to indicate that you should tackle the addition before the multiplication.

Example 2

Solution

Using BEDMAS, the first thing that we need to evaluate is the expression in brackets. We get 

\(\begin{align*} & 3\times (5-3) +2\times 7 & \text{Brackets} \\ =\, & 3\times 2 +2\times 7 \end{align*}\)

We have no more brackets, and the next thing in the order of BEDMAS that we need to work with is the multiplication reading from left to right. This simplifies to 

\(\begin{align*} & 3\times (5-3) +2\times 7 & \text{Brackets} \\[1ex] =\, & 3\times 2 +2\times 7 & \text{Multiplication} \\[1ex] =\, &6+ 2\times 7 &\text{Multiplication}\\[1ex] =\, & 6+14 \end{align*}\)

Our only operation left is to add, and we get

\(\begin{align*} & 3\times (5-3) +2\times 7 & \text{Brackets} \\[1ex] =\, & 3\times 2 +2\times 7 & \text{Multiplication} \\[1ex] =\, &6+ 2\times 7 &\text{Multiplication}\\[1ex] =\, & 6+14 &\text{Addition}\\[1ex] =\, & 20 \end{align*}\)

If you didn't get the same answer as me, check each step and determine where you went wrong with your order of operations.

Check Your Understanding 1

Question

Evaluate \(4 \times (12 - 30 \div 3)\).

Answer

\(8\)

Feedback

Evaluate this expression using BEDMAS.

\(\begin{align*} & 4\times (12 - 30 \div 3) & \text{Brackets, Division} \\[1ex] =\, & 4\times (12-30) & \text{Brackets} \\[1ex] =\, & 4\times 2 &\text{Multiplication}\\[1ex] =\, & 8 \end{align*}\)

The Need for an Order of Operations

Do we need a conventional order of operations, e.g., BEDMAS?

Why don't we just perform the operations from left to right like you would read a book in English?

This would work for our first example,

Consider the expression:

\(2\times 4 + 8\times 3\)

If I wanted you to do \(2 \times 4\), then \(8 \times 3\), and then, as a final step, add those two quantities together. There's actually no way for me to rewrite this expression to communicate that information so that it can be solved reading only from left to right.

So already we've shown that reading expressions from left to right is not going to work.

Is there another way we can communicate to perform the multiplication before the addition?

Well, it'd be natural to think that instead of introducing all of the order of operations, why don't we just look at brackets? We could use brackets if the order we want to evaluate the expression is not simply left to right.

In my example, using brackets would communicate to you to do the multiplication first and then the addition. We get

\(\class{hl1 large}{(}2\times 4\class{hl1 large}{)} + \class{hl1 large}{(}8\times 3\class{hl1 large}{)}\)

In a moment we're going to look specifically at operations that involve brackets. But before we move on, I want to ask you one question and leave you to think about it.

Can you think of a reason as to why we choose not to rely solely on brackets to communicate an order if it's different from left to right?

Operations with Brackets

Brackets Versus No Brackets

Brackets are used to give priority.

Brackets say "do this calculation first." 

Now I just asked you to think about a scenario where there was no standard for the order of operations.

In this case, we've determined that we would need to use brackets to indicate which operation to perform any time the brackets don’t agree with the "left to right" approach.

The question I left you with was:

Could you think of a reason why this approach would not be our best strategy?

I'm going to describe a calculation. First, we want to evaluate \(8-2\), then multiply that result by \(3\), then divide that result by \(9\). And finally, we want to add \(3\) to that result again.

Notice how our second expression, which assumes that we know the order of operations, looks a lot less cluttered, and it's significantly easier to read.

Example 3

Solution

We can number the steps in this expression in the order that they are to be performed (as indicated by the brackets).

First, we would need to do \(4-2\).

\((5\times \underbrace{(4-2)}_{\#1})+(3\times 5)\)

Second, is \(5\) times the result of \(4-2\).

\(\underbrace{(5\times \underbrace{(4-2)}_{\#1})}_{\#2}+(3\times 5)\)

Third, is \(3 \times 5\).

\(\underbrace{(5 \times \, \underbrace{(4-2)}_{\#1})}_{\#2}+\underbrace{(3\times 5)}_{\#3}\)

And last, but not least, we would do the addition.

So let's look at removing some brackets and starting in the third step. Removing the brackets in the third step does not change the order of operations, because multiplication would already come before addition.

\(\underbrace{(5 \times \, \underbrace{(4-2)}_{\#1})}_{\#2}+\underbrace{\class{hl4}{3\times 5}}_{\#3}\)

In the second step, we can remove the brackets again because \(5\) times the result of \(4-2\) would happen to before \(3 \times 5\), because it is farther to the left. This would all happen before the addition.

\(\underbrace{\class{hl4}{5\; \times} \, \underbrace{(4-2)}_{\#1}}_{\#2}+\underbrace{3\times 5}_{\#3}\)

The final set of brackets are necessary because if they were not present the multiplication would have to happen before the subtraction. These brackets around \(4\) subtract \(2\) must stay.

\(\underbrace{5 \times \, \underbrace{\class{hl1}{(}4-2\class{hl1}{)}}_{\#1}}_{\#2}+\underbrace{3\times 5}_{\#3}\)

Our final expression without the unnecessary brackets is

\(5 \times (4-2) + 3 \times 5\)

Check Your Understanding 2

Question

Select the expression that is equivalent to \(((5 \times (9-4)) \div 6)\) but is written using as few brackets as possible.

1. \(((5 \times (9-4)) \div 6)\)
2. \((5 \times (9-4)) \div 6\)
3. \((5 \times 9-4) \div 6\)
4. \(5 \times (9-4) \div 6\)

Answer

4. \(5 \times (9-4) \div 6\)

Feedback

In this expression, the brackets tell us that we must subtract, multiply, and then divide. We must keep the brackets that force subtraction to be performed first. Thus, option d) \(5 \times (9-4) \div 6\) is the equivalent expression

The rest of the operations follow BEDMAS without the use of brackets.

Example 4

Solution

There is no change because this is already the operation that comes first based on order of operations.

Putting brackets into an expression requires some trial and error. We start to figure this out by putting brackets around some of the expressions within the left-hand side of the equation.

We find to get an answer of \(10\), we need to use brackets to ensure that \(5-2\) is done first in the calculation.

Our conclusion is

\(\begin{align*} 3\times (5-2) +1 &= 3 \times 3 + 1 \\ &= 9+1 \\ &=10 \end{align*}\)

Check Your Understanding 3

Question

Place brackets so that \(16 \times 30 \div 10 - 2\) evaluates to \(60\).

Answer

\(16 \times 30 \div (10-2)\)

Feedback

Without brackets, BEDMAS tells us to multiply, divide, and then subtract. In order for \(16 \times 30 \div 10 - 2\) to evaluate to \(60\), we must subtract, multiply, and then divide. Brackets are used to force subtraction to be the first operation performed.

Operations with Rational Numbers

Example 5

The order of operations applies to the operations of addition (\(+\)), subtraction (\(-\)), multiplication (\(\times\)), and division (\(\div\)), regardless of what numbers are involved.

Solution

By following the order of operations, we know that we have to deal with the expression in brackets first.

\(\begin{align*} & \class{hl1}{\left( \dfrac{2}{3} - \dfrac{1}{3} \right)} +\dfrac{1}{6} \times 3 & \text{Brackets} \end{align*}\)

Let's bring \(\left(\dfrac{2}{3} - \dfrac{1}{3} \right)\) aside. This is a calculation that we know how to do. We have two fractions with the same denominator. So we subtract the numerators, and we find the result is 

\(\begin{align*} \dfrac{2}{3} - \dfrac{1}{3}& =\dfrac{2-1}{3} \\[1ex] &=\class{hl1}{\dfrac{1}{3}}\\ \end{align*}\)

Our entire expression simplifies to be 

\(\begin{align*} & \class{hl1}{\left( \dfrac{2}{3} - \dfrac{1}{3} \right)} +\dfrac{1}{6} \times 3 & \text{Brackets} \\[1ex] =\, & \class{hl1}{\dfrac{1}{3}} + \dfrac{1}{6}\times 3 \end{align*}\)

Next, the order of operations tells us that we need to deal with the multiplication.

\(\begin{align*} & \left( \dfrac{2}{3} - \dfrac{1}{3} \right) +\dfrac{1}{6} \times 3 & \text{Brackets} \\[1ex] =\, & \dfrac{1}{3} + \class{hl4}{\dfrac{1}{6}\times 3} & \text{Multiplication}\\ \end{align*}\)

So again, if we look at this expression, we're being asked to calculate \(\dfrac{1}{6} \times 3\). Well, this is something else that we know how to deal with. To solve \(\dfrac{1}{6} \times 3\), we simplify this to be

\(\begin{align*} \class{hl4}{\dfrac{1}{6} \times 3} &= \dfrac{1 \times 3}{6} \\[1ex] &= \class{hl4}{\dfrac{3}{6}} \end{align*}\)

So when we put that back, we find that the expression we're trying to evaluate simplifies to be

\(\begin{align*} & \left( \dfrac{2}{3} - \dfrac{1}{3} \right) +\dfrac{1}{6} \times 3 & \text{Brackets} \\[1ex] =\, & \dfrac{1}{3} + \dfrac{1}{6}\times 3 & \text{Multiplication}\\[1ex] =\, & \dfrac{1}{3} + \class{hl4}{\dfrac{3}{6}} \end{align*}\)

Now we just need to add. And what's even better is if we look at the denominators, we see that we can find a common denominator of \(6\) quite easily.

\(\begin{align*} & \left( \dfrac{2}{3} - \dfrac{1}{3} \right) +\dfrac{1}{6} \times 3 & \text{Brackets} \\[1ex] =\, & \dfrac{1}{3} + \dfrac{1}{6}\times 3 & \text{Multiplication}\\[1ex] =\, & \dfrac{1}{3} + \dfrac{3}{6} & \text{Denominator}\\[1ex] =\, & \class{hl3}{\dfrac{2}{6}+\dfrac{3}{6}} \end{align*}\)

Adding the numerators, we find that our final answer is

\(\begin{align*} & \left( \dfrac{2}{3} - \dfrac{1}{3} \right) +\dfrac{1}{6} \times 3 & \text{Brackets} \\[1ex] =\, & \dfrac{1}{3} + \dfrac{1}{6}\times 3 & \text{Multiplication}\\[1ex] =\, & \dfrac{1}{3} + \dfrac{3}{6} & \text{Denominator}\\[1ex] =\, & \class{hl3}{\dfrac{2}{6}+\dfrac{3}{6}} & \text{Addition}\\[1ex] =\, & \class{hl3}{\dfrac{5}{6} } \end{align*}\)

Keep in mind that even though we're introducing fractions and things that make these expressions look harder to evaluate, look at them step by step, whole things aside, and we can work through each line one operation at a time.

Check Your Understanding 4

Question

Evaluate \(\left( \dfrac{5}{6} - \dfrac{1}{2}\right) + 5 \times \dfrac{1}{6}\).

Answer

\(\dfrac{7}{6}\)

Feedback

Evalulate this expression using BEDMAS. 

\(\begin{align*} & \left( \dfrac{5}{6} - \dfrac{1}{2} \right) + 5 \times \dfrac{1}{6} & \text{Brackets} \end{align*}\)

Think about the question \(\dfrac{5}{6} - \dfrac{1}{2}\) on it's own: 

\(\begin{align*} \dfrac{5}{6} - \dfrac{1}{2} &= \dfrac{5}{6} - \dfrac{3}{6} \\ &= \dfrac{2}{6}\end{align*}\)

Now, put this answer back into the original problem. 

\(\begin{align*} & \left( \dfrac{5}{6} - \dfrac{1}{2} \right) +5 \times \dfrac{1}{6} & \text{Brackets} \\[1ex] =\, & \dfrac{2}{6} + 5 \times \dfrac{1}{6} & \text{Multiplication} \end{align*}\)

Think about the question \(5 \times \dfrac{1}{6}\) on its own: 

\(\begin{align*} 5 \times \dfrac{1}{6} &= \dfrac{5 \times 1}{6} \\ &= \dfrac{5}{6} \end{align*}\)

Now, put this answer back into the original problem.

\(\begin{align*} & \left( \dfrac{5}{6} - \dfrac{1}{2} \right) +5 \times \dfrac{1}{6} & \text{Brackets} \\[1ex] =\, & \dfrac{2}{6} + 5 \times \dfrac{1}{6} & \text{Multiplication} \\ =\, & \dfrac{2}{6} + \dfrac{5}{6} & \text{Addition} \\ =\, & \dfrac{7}{6} \end{align*}\)

Example 6

Let's look at another example.

Solution

The order of operations tells us that we need to deal with the brackets first.

\(\begin{align*} & 7 \div \class{hl1}{(4\times 0.7)} - (-3) & \text{Brackets} \end{align*}\)

Again, if we look at this calculation on its own, we know that \(4 \times 0.7\) is very similar to the question, \(4 \times 7\), which equals \(28\). Since \(7\) is \(10\) times larger than \(0.7\), we know that \(28\) is \(10\) times larger than the answer to \(4 \times 0.7\). We get

\(\begin{align*} \class{hl1}{4 \times 0.7} & = 4 \times \dfrac{7}{10} \\[1ex] &= \dfrac{4 \times 7}{10} \\[1ex] &= \dfrac{28}{10} \\[1ex] &= \class{hl1}{2.8} \end{align*}\)

We can put that back in to the expression we're trying to evaluate.

\(\begin{align*} & 7 \div (4\times 0.7) - (-3) & \text{Brackets}\\[1ex] =\, & 7 \div 2.8 - (-3) \end{align*}\)

We now have a division and a subtraction problem, and the order of operations tells us that we must deal with the division first.

\(\begin{align*} & 7 \div (4\times 0.7) - (-3) & \text{Brackets}\\[1ex] =\, & \class{hl4}{7 \div 2.8}- (-3) & \text{Division} \end{align*}\)

We now want to evaluate \(7 \div 2.8\), so let's look at this problem on its own. Using our strategy for dividing a whole number by a decimal, \(7 \div 2.8\) is equal to \(70 \div 2.8\), which you can use long division to show that

\(\begin{align*} \class{hl4}{7 \div 2.8} &\;= 70 \div 28 \\[1ex] &\;= \class{hl4}{2.5} \end{align*}\)

So we can substitute this back in and we get

\(\begin{align*} & 7 \div (4\times 0.7) - (-3) & \text{Brackets}\\[1ex] =\, & 7 \div 2.8 - (-3) & \text{Division}\\[1ex] =\, & 2.5 - (-3) \end{align*}\)

The last operation we have to deal with is the subtraction and we know that when we're dealing with integers, to subtract a negative integer is equal to adding the opposite.

\(\begin{align*} & 7 \div (4\times 0.7) - (-3) & \text{Brackets}\\[1ex] =\, & 7 \div 2.8 - (-3) & \text{Division}\\[1ex] =\, & \class{hl3}{2.5 - (-3)} & \text{Subtraction} \\[1ex] =\, & \class{hl3}{2.5+3} \end{align*}\)

Completing the addition, our final answer is

\(\begin{align*} & 7 \div (4\times 0.7) - (-3) & \text{Brackets}\\[1ex] =\, & 7 \div 2.8 - (-3) & \text{Division}\\[1ex] =\, & 2.5 - (-3) & \text{Subtraction} \\[1ex] =\, & 2.5+3 \\[1ex] =\, & 5.5 \end{align*}\)

Check Your Understanding 5

Question

Evaluate \(10 \div (3.4 + 3) -1\). Write your answer as a decimal.

Answer

\(0.5625\)

Feedback

Evaluating this expressiong using BEDMAS, we have the following: 

\(\begin{align*} & 10 \div (3.4+3) - 1 & \text{Brackets}\\[1ex] =\, & 7 \div (6.4)- 1 & \text{Division} \end{align*}\)

Think about the question \(10 \div 6.4\) on its own: 

\(\begin{align*} 10 \div 6.4 &= 10 \div \dfrac{64}{10} \\ &= 10 \times \dfrac{10}{64} \\ &= \dfrac{100}{64} \\ &=\dfrac{25}{16}\end{align*} \)

Now, put this answer back into the original problem. 

\(\begin{align*} & 10 \div (3.4+3) - 1 & \text{Brackets}\\[1ex] =\, & 7 \div (6.4)- 1 & \text{Division} \\[1ex] =\, & \dfrac{25}{16} - 1 &\text{Subtraction}\\[1ex] =\, & 1\dfrac{9}{16} -1 \\[1ex] =\, & \dfrac{9}{16} \\ =\, & 0.5625 \end{align*}\)

The Distributive Property—Multiplication

Example 7

Solution 1

We can solve this using the order of operations. The order of operations tells us that we need to address the brackets first. So our first instinct is to add and we get

\(\begin{align*} 6\times (20+7) &=6\times 27 \\[1ex] &= 162 \end{align*}\)

But dealing with the brackets first, does not mean that you have to evaluate the expression inside the brackets. Dealing with the brackets means that you must remove the brackets first.

Solution 2

This seems like a problem for which the distributive property would be useful.

What's nice about this is once we've used the distributive property, our mental calculations or our mental math can actually be done easier.

\(\begin{align*} 6\times (20+7) & = 6\times 20 + 6\times 7 \\[1ex] &= 120 + 42 \\[1ex] & = 162 \end{align*}\)

Now it doesn't matter which method you use to solve the problem, we always want to make sure that we check our work.

Check Your Work

To do this, we can divide our final answer by \(6\). The result should be \(27\), which it is.

\(162 \div 6 = 27\) ✔

The Distributive Property—Multiplication

In our previous example, we showed that the distributive property could be used to distribute multiplication over an addition statement. Well, now let me state that this can always be done.

What this statement means is that we can distribute multiplication over an addition problem such as

We can also distribute multiplication over a subtraction problem such as

Recall

Multiplication is commutative, which means that changing the order gives the same result. 

It makes sense to ask ourselves that if we flip the order, and that means put the expression in brackets first, does that distributive property still hold?

I encourage you to pick your own numbers to put in the boxes and solve your problem using both the distributive property and using the order of operations to show that the distributive property holds.

\((\square + \square) \times \square\)

Check Your Understanding 6

Question

Select the expression that is equivalent to \((9-7) \times 11\).

1. \(9-11 \times 7\)
2. \(11 \times 9 - 7\)
3. \(11 - 9 \times 11 - 7\)
4. \(11 \times 9 - 11 \times 7\)

Answer

4. \(11 \times 9 - 11 \times 7\)

Feedback

Using the distributive property, we multiply \(11\) with both \(9\) and \(7\).

Explore This 1

Description

Use the following investigation to explore the distributive property further.

In Example 1, we used diagrams to show that \(4 \times (4+2)\) is equal to \(4 \times 4 + 4 \times 2\). Specifically, in Option 1, we show what it looks like to evaluate inside the brackets first. So visually, we're showing you that \(4 \times (4+2)\) is equal to \(4 \times 6\), which is equal to \(24\).

In Option 2, we distribute first and visually show that \(4 \times (4+2)\) is equal to \(4 \times 4 + 4 \times 2\), which is also equal to \(24\).

Example 1

\(\class{hl4}{4}\times(\class{hl1}{4}+\class{hl2}{2})\)

Choose different numbers and work through multiple examples on your own. Complete enough examples to convince yourself of the validity of the distributive property.

Example 2

\(\class{hl4}{2}\times(\class{hl1}{3}+\class{hl2}{5})\)

Example 3

\(\class{hl4}{5}\times(\class{hl1}{4}+\class{hl2}{1})\)

Created with GeoGebra. Author: University of Waterloo. CC BY-NC-SA 4.0. https://ggbm.at/BRg5Afbc

In the next example, we will look at how the distributive property can be used to make mental arithmetic easier.

Online Version

https://ggbm.at/BRg5Afbc

Example 8

The distributive property is helpful when performing calculations without a calculator.

Solution

\(19\) can be written as \(20-1\). So we get 

\(\begin{align*} 19 \times 8 & = \underbrace{(20-1)}_{19} \times 8 \end{align*}\)

Using the distributive property, this is equal to

\(\begin{align*} 19 \times 8 & = \underbrace{(20-1)}_{19} \times 8 \\[1ex] &\; = 20 \times 8 - 1 \times 8 &\text{Distributive Property}\\[1ex] & \; = 160-8 \\[1ex] & \; = 152 \end{align*}\)

In fact, if you look at the second line, \(20 \times 8 - 1 \times 8\), you've probably used this strategy to help you with mental math before today. All we're doing is putting the distributive property name with the strategy.

Check Your Understanding 7

Question

Evaluate \(2 \times 26\) using the distributive property.

Answer

\(52\)

Feedback

To evaluate \(2 \times 26\), we split \(26\) into \(20 + 6\) and use the distributive property.

\(\begin{align*} 2 \times 26 & = 2 \times (20+6) \\ &= 2 \times 20 + 2 \times 6 \\ &=40 + 12 \\ &=52\end{align*}\)

The Distributive Property—Division

Example 9

Again, it doesn't matter which method that you use to solve a problem, we always want to check our work.

Check Your Work

To do this, we can multiply our final answer by \(5\), and the result should be \(515\), which if you check your work, you'll find that it is.

\(5\times 103 =515\) ✔

Can Division Always Distribute Over Brackets?

Recall \((500+15)\div 5\).

In this case, we apply the distributive property and show that it gave the same answer as evaluating inside the brackets first.

Our solution used the fact that no matter which numbers you try, the distributive property will always hold for division when the expression in the brackets comes first.

 What happens if we have the expression in the brackets to the right of the division sign? That is, does

\(\square \div (\square + \square) = \square \div \square + \square \div \square\)

One strategy that mathematicians use to answer general questions such as these is to try some numbers. Let's try the numbers \(1\), \(2\), and \(3\):  

Now with one example, we've shown that these answers are not the same. One example is enough to conclude: 

The distributive property doesn’t always hold when the expression is is to the right of the division sign.

\(\square \div (\square + \square)\)

Go ahead and try as many examples as you want until you're convinced that the result is not always true.

The Distributive Property—Division

Check Your Understanding 8

Question

Evaluate \(770 \div 7\) using the distributive property. 

Answer

\(110\)

Feedback

We can evaluate \(770 \div 7\) by splitting \(770\) into two numbers that we can easily divide by \(7\). One of the ways to do this is by splitting \(770\) into \(700+70\).

\(\begin{align*} 770 \div 7 & = (700+70) \div 7 \\ &= 700 \div 7 + 70 \div 7 \\ &= 100 + 10 \\ &= 110\end{align*}\)

Take It With You

In this lesson, we showed that we cannot distribute \(16\) over the addition expression \(6+2\).

\(16 \div (6+2) \neq 16 \div 6 + 16 \div 2\)

Use a real-life example to explain why you cannot distribute division over subtraction in the following example:

\(\square \div (\square - \square)\)