B2.1 Use the properties of operations, and the relationships between operations, to solve problems involving whole numbers, decimal numbers, fractions, ratios, rates, and whole number percents, including those requiring multiple steps or multiple operations.
Activity 1: Only Threes!
The following activity introduces the use of brackets to specify the order of operations in a number sentence. The activity is designed for students who are not familiar with the order of operations.
If students already know the order of operations or know how to use brackets, the activity could be done differently. As a class, tell the students that it is possible to represent the number 0 by a number sentence using only the number 3 and one or more of the mathematical operations. Give them an example by writing \(0\; = \;3\; - \;3\) on the board.
Then introduce the number sentence \(3\; \times \;3\; - \;3\; \times \;3\) and show them that if you do the operations in the order they appear, you get 18:
\(\begin{array}{l}3\; \times \;3\; - \;3\; \times \;3 &= \;9\; - \;3\; \times \;3\\ &= \;6\; \times \;3\\ &= \;18\end{array}\)
However, if we perform both multiplications first, we get 0:
\(\begin{array}{l}3\; \times \;3\; - \;3\; \times \;3 &= \;9\; - \;9\\ &= \;0\end{array}\)
It is also possible to obtain 0 by first performing the subtraction:
\(\begin{array}{l}3\; \times \;3\; - \;3\; \times \;3 &= \;3\; \times \;0\; \times \;3\\ &= \;0\end{array}\)
Explain that, by convention, brackets can be used to give priority to one or other of the operations. For example, if you want to give priority to both multiplications, you can write \(\left( {3\; \times \;3} \right)\; - \;\left( {3\; \times \;3} \right)\). However, if you want to give priority to subtraction, you can write \(3\; \times \;\left( {3\; - \;3} \right)\; \times \;3\).
To ensure that students have understood, ask them to represent the number 1 in number sentences using only the number 3 and one or more operations. Point out that they should use brackets to give priority to one or more operations, if necessary. Ask a few students to write a sentence on the board and ask the others to check its accuracy.
Examples of possible answers include:
\(\begin{align} 1 = \;3\; \div \;3\;{\rm{or}}\;\frac{3}{3}\end{align}\)
\(\begin{align}1 &= \;3\; - \;\left( {3\; \div \;3} \right)\; - \;\left( {3\; \div \;3} \right)\ &= 3-1-1 \ &= 2-1 \ &= 1 \end{align}\)
\(\begin{align} \; 1 &= \frac{{\left( {3\; + \;3} \right)}}{{\left( {3\; + \;3} \right)}} \\ &= \frac{6}{6}\ \\ &= \ 1 \end{align}\)
Group students in pairs and ask them to present the numbers 2 to 10 in different number sentences using the number 3, the four operations and brackets. Once the task is completed, ask a few students to take turns writing their number sentences on the board and grouping them according to the number in question. Facilitate a mathematical discussion and invite other students to observe the number sentences and check their accuracy.
Note: It is possible that some students may use brackets within brackets. In such situations, it should be mentioned that priority is given first to the brackets inside the others. For example:
\(\begin{align}3\; - \;\left( {\left( {3\; + \;3} \right)\; \div \;3} \right) &= 3\; - \;\left( {6\; \div \;3} \right)\\ &= 3\; - \;2\\ &= 1\end{align}\)
Here are some examples of possible answers:
\(2\; = \;\left( {3\; + \;3} \right)\; \div \;3\;{\rm{or}}\;\left( {3\; \div \;3} \right)\; + \;\left( {3\; \div \;3} \right)\)
\(3\; = \;\left( {3\; \times \;3} \right)\; - \;\left( {3\; + \;3} \right)\;{\rm{or}}\;\left( {3\; + \;3} \right)\; - \;3\)
\(4\; = \;3\; + \;\left( {3\; \div \;3} \right)\;{\rm{or}}\;\left( {3\; + \;3} \right)\; - \;\left( {\left( {3\; + \;3} \right)\; \div \;3} \right)\)
\(5\; = \;3\; + \;3\; - \;\left( {3\; \div \;3} \right)\;{\rm{or}}\;\left( {3\; \times \;3} \right)\; - \;3\; - \;\frac{3}{3}\)
\(6\; = \;3\; + \;3\;{\rm{or}}\;\left( {3\; \times \;3} \right)\; - \;3\)
\(7\; = \;\left( {3\; \times \;3} \right)\; - \;\frac{{3\; + \;3}}{3}\;{\rm{or}}\;\left( {3\; + \;3} \right)\; + \;\left( {3\; \div \;3} \right)\)
\(8\; = \;3\; + \;3\; + \;3\; - \;\left( {3\; \div \;3} \right)\;{\rm{or}}\;3\; + \;3\; + \;\left( {3\; \div \;3} \right)\; + \;\left( {3\; \div \;3} \right)\)
\(9\; = \;3\; + \;3\; + \;3\;{\rm{or}}\;3\; \times \;3\)
\(10\; = \;3\; + \;3\; + \;3\; + \;\left( {3\; \div \;3} \right)\;{\rm{or}}\;\left( {3\; \times \;3} \right)\; + \;\left( {3\; \div \;3} \right)\)
Source: translated from Guide d’enseignement efficace des mathématiques de la 4e à la 6e année, Numération et sens du nombre, Fascicule 1, Nombres naturels, p 209-211.
Activity 2: What a Great Challenge!
The following activity challenges students to apply their knowledge of the order of operations and the effect of operations on numbers. Students are asked to use each of the numbers in a given series only once, along with some of the four operations to get as close as possible to a target number. For example, if the sequence of numbers is 2, 4, 5, 7, 10, and 25 and the target is 433, students might solve as follows:
Solution 1
\(\begin{array}{l}10\; \times \;25\; = \;250\\5\; \times \;7\; = \;35\\250\; - \;35\; = \;215\\215\; \times \;2\; = \;430\\430\; + \;4\; = \;\bf{434}\end{array}\)
Solution 2
\(\begin{array}{l}10\; \times \;7\; = \;70\\70\; \times \;5\; = \;350\\350\; + \;25\; = \;375\\4\; + \;2\; = \;6\\375\; + \;6\; = \;\bf{381}\end{array}\)
Solution 3
It is also possible to represent the total number of hours
The level of difficulty of the activity can vary according to certain modalities, namely:
- the choice of numbers – it can be useful to provide at least four or five numbers less than 10 and two or three numbers that facilitate the calculations (for example, 15, 25, 40, 50, 75, 100);
- use of numbers – whether a number can be used once or repeatedly;
- choice of calculation strategy - on paper, mentally or with a calculator.
If the order of operations has been studied in class, students can summarize their solution using a number sentence. For example, they could summarize their solution as \(\left( {5\; \times \;7} \right)\; \times \;\left( {10\; + \;2} \right) \; + \;25\; - \;4\; = \;441\) for Solution 3.
Source: translated from Guide d’enseignement efficace des mathématiques de la 4e à la 6e année, Numération et sens du nombre, Fascicule 1, Nombres naturels, p 110-111.
Activity 3: Distributive Property of Multiplication
Write each of the following sets of related operations on the board or a large sheet of paper, one at a time, and ask students to perform the operations in it.
Once a set is completed, draw out the various mental math strategies by asking questions such as:
- How did you perform the last operation?
- To perform the last operation, did you use any of the previous operations?
- How would you describe the order in which you performed the operations?
If necessary, before doing the same with the next math string, present other examples of similar related computations.
If students have difficulty seeing and applying the distributive property of multiplication to solve the last operation in each set, support them to do so by representing these operations as area models.
Example
Findings for each series
Series 1
This set allows us to review the distributive property of multiplication over addition at its simplest. The idea is to recognize that to obtain the product of \(30 \times 46\), it is possible to perform an operation on a sum of terms and obtain the same result as if the operation had been performed on each term, that is:
\(\begin{align}3\; \times \;46\; &= \;3\; \times \;\left( {40\; + \;6} \right)\\ &= \;\left( {3\; \times \;40} \right)\; + \;\left( {3\; \times \;6} \right)\end{align}\)
Series 2
This set allows us to see that the decomposition related to distributive property can be performed on the second term as well as on the first, that is:
\(22\; \times \;13\; = \;\left( {20\; \times \;13} \right)\; + \;\left( {2\; \times \;13} \right)\)
or
\(22\; \times \;13\; = \;\left( {22\; \times \;3} \right)\; + \;\left( {22\; \times \;10} \right)\)
Series 3
This series shows that decomposing a number to apply distributive property can be done in more than two parts. For example:
\(\begin{align}4\; \times \;77\; &= \;4\; \times \;\left( {50\; + \;25\; + \;2} \right)\ \ &= \;\left( {4\; \times \;50} \right)\; + \;\left( {4\; \times \;25} \right)\; + \;\left( {4\; \times \;2} \right)\end{align}\)
Series 4
This set allows us to see that we can solve the last operation using the distributive property of multiplication over subtraction. For example:
\(\begin{align}4\; \times \;77\; &= \;4\; \times \;\left( {50\; + \;25\; + \;2} \right)\\ &= \;\left( {4\; \times \;50} \right)\; + \;\left( {4\; \times \;25} \right)\; + \;\left( {4\; \times \;2} \right)\end{align}\)
It also shows that this operation can be solved by using the distributive property of multiplication over addition and other properties. For example:
\(5\; \times \;95\; = \;3\; \times \;\left( {5\; \times \;30} \right)\; + \;\left( {5\; \times \;5} \right)\)
Source: translated from Guide d’enseignement efficace des mathématiques de la 4e à la 6e année, Numération et sens du nombre, Fascicule 1, Nombres naturels, p 221-222.