unit 2 patterns and algebra - jump math for ap book 7-1... · unit 2 patterns and algebra in this...

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Teacher’s Guide for Workbook 7.1

Unit 2 Patterns and Algebra

In this unit, students extend and describe patterns, use T-tables to solve problems, and solve equations using various strategies (guess and check, modelling, and working backwards). For most of this unit, a variable represents an unknown that students try to solve for, as in 2x + 3 = 13. However, in PA7-15, students investigate equations where a variable is used to represent numbers that can change and leave the equation true, as in a + b = b + a.

Meeting your curriculum Teachers following either the Ontario or WNCP curricula can use lessons PA7-1 to PA7-3 as diagnostic and review material. Re-teach the concepts as necessary. Lesson PA7-4 applies the concepts from the first three lessons to deepen understanding. The material in lesson PA7-15, although not on the curriculum, is essential for understanding and applying the distributive property and furthers the understanding developed in NS7-6.

For students following the Ontario curriculum, lessons PA7-12 and PA7-13 are optional.

Some essential practice before starting Before students can create or recognize a pattern in a sequence of numbers, they must be able to tell how far apart the successive terms in the sequence are. Students can count up on their fingers, if necessary, to find the gap between two numbers.

Here is a foolproof method for identifying the gaps between numbers:

EXAMPLE: How far apart are 8 and 11?

Step 1: Say the lower number (8) with your fist closed.

Step 2: Count up by ones, raising your thumb first then one finger at a time, until you reach the higher number (11).

Step 3: The number of fingers you have up when you reach the higher number is the answer. In this case, you have three fingers up, so three is the difference between 8 and 11.

Even the weakest student can find the difference between two numbers using the above method, which you can teach in one lesson. Make sure students say the first number with their fists closed! (Some students will want to put their thumbs up to start; one way to avoid this is to “throw” the starting number to the student and make them “catch” it.)

Eventually, you should wean students off using their fingers to find the gap between a pair of numbers. The exercises in the Mental Math section of this

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C-2 Teacher’s Guide for Workbook 7.1

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manual will help with this.

Here is one approach you can use to help students find larger gaps between larger numbers:

1. Have students memorize the gap between the number 10 and each of the numbers from 1 to 9. EXAMPLE: the gap between 8 and 10 is 2 (you need to add 2 to 8 to get 10).

You could make flash cards to help your students learn these facts:

Front of card Back of card

You could also draw a picture of a number line to help your students visualize the gaps:

2. Have students memorize the gap between 10 and each of the numbers from 11 to 19. Again, you might use flash cards for this:

Front of card Back of card

Point out that the gap between 10 and any number from 11 to 19 is merely the ones digit of the larger number. EXAMPLE: 16 minus 10 is 6, but 6 is just the ones digit of 16. Once students know this, they will have no trouble recognizing the gap between 10 and any number from 11 to 19.

3. Students can now find the gap between a number from 1 to 9 and a number from 11 to 19—say, between 7 and 15—as follows:

Step 1: Find the gap between 7 and 10 (your students will know this is 3).

Step 2: Find the gap between 10 and 15 (your students will know this is 5).

Step 3: Add the two numbers you found in Steps 1 and 2: 3 + 5 = 8. So the gap between 7 and 15 is 8.

Show students why this works with a picture:

4. Students can use the method above to find the gap between any pair of two-digit numbers whose leading digits differ by 1. EXAMPLE: The gap between 47 and 55 is 8—start at 47, add 3 to get to 50, and then add 5 to get to 55.

6 7 8 9 10 11 12 13 14 15 16

+3 +5

10 + ? = 17 10 + 7 = 17

6 7 8 9 10 11 12

8 + ? = 10 8 + 2 = 10

C-3Patterns and Algebra

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This method can ultimately be used to find the gap between any pair of two-digit numbers. EXAMPLE: To go from 36 to 72 on the number line, you add 4 to reach 40, then add 30 to reach 70, then add 2 to reach 72; the gap between 36 and 72 is 4 + 30 + 2 = 36. (NOTE: Before students can attempt questions of this sort, they must be able to find the gap between pairs of numbers that have zeros in their ones place. They can find those gaps by mentally subtracting the tens digits of the numbers. EXAMPLE: the gap between 80 and 30 is 50, since 8 – 3 = 5.)

Do not discourage students from counting on their fingers until they can add and subtract readily in their heads. You should expect students to answer all of the questions in this unit, even if they have to rely on their fingers for help.

46 47 48 49 50 51 52 53 54 55 56

+3 +5


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PA7-1 Extending PatternsPage 22

CUrriCULUM EXPECTATiONSOntario: 7m1, reviewWNCP: review, [r]

VOCAbULAry increasing sequence decreasing sequence

introduce patterns. Use one or both of the Möbius strip activities below (see Activities 1-2). They can be done independently of each other. Then ASK: Why are patterns useful? Explain that patterns allow you to make predictions about things that may be difficult to check by hand. Do you want to try turning the paper in either Activity 100 times? And yet, from the pattern, we can see what will happen without checking.

Extending sequences that were made by adding or subtracting the same number to each term. EXAMPLE: Extend the pattern 3, 6, 9, ... up to six terms.

Step 1: Identify the gap between successive pairs of numbers in the sequence. (Students may count on their fingers, if necessary – see the Introduction.) The gap in this example is three. Check that the gap between successive terms in the sequence is always the same, otherwise you cannot continue the pattern by adding a fixed number. Write the gap between each pair of successive terms above the pairs.

3 , 6 , 9 , , ,

Step 2: Say the last number in the sequence with your fist closed. Count by ones until you have raised the same number of fingers as the gap, in this case, three. The number you say when you have raised your third finger is the next number in the sequence.

3 , 6 , 9 , 12 , ,

Step 3: Repeat Step 2 to continue adding terms to extend the sequence.

3 , 6 , 9 , 12 , 15 , 18

EXTrA PrACTiCE for Question 1: Extend the pattern. a) 6, 9, 12, 15, , bonus d) 99, 101, 103, , b) 5, 11, 17, 23, , e) 654, 657, 660, , c) 2, 10, 18, 26, ,

GoalsStudents will use the gaps between terms to extend patterns.

3 3

3 3

3 3

PriOr KNOWLEDGE rEQUirED

Can add, subtract, and count up to subtract

MATEriALS

5 strips of paper (see details below), a pair of scissors,

and tape for each student

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Patterns and Algebra 7-1

EXTrA PrACTiCE for Workbook Question 2: Extend the pattern. a) 21, 19, 17, 15, , bonus d) 141, 139, 137, 135, ___ b) 34, 31, 28, , , e) 548, 541, 534, 527, ___c) 48, 41, 34, 27, , f) 234, 221, 208, ___

Extend sequences by extending the pattern in the gaps. See Workbook Question 3. In parts a), b), and d), the gaps form sequences similar to those in Questions 1 and 2. In parts c) and e), the gaps form the same sequence as the original.

EXTrA PrACTiCE for Workbook Question 3: Extend the pattern. 1, 3, 6, 10, 15, , , bonus 1, 4, 10, 20, 35, , , Extra bonus 1, 5, 15, 35, 70, , ,

Looking for a pattern

PrOCESS EXPECTATiON

ACTiViTiES 1–2

1. Each student will need 3 strips of paper (11˝ × 1˝), 2 longer strips of paper (say, 22˝ × 1˝), a pair of scissors, and tape (staples do not work for this activity).

Show your students a sheet of paper and ask how many sides it has. (2) Repeat with an 11˝ × 1˝ strip of paper with the ends taped so that it looks like a ring. Trace each side of the ring with your finger, naming them “inside” and “outside.” Point out that you could colour one side and leave the other side blank. Have students create their own rings and colour one side. ASK: If an ant is walking along the coloured side (and never goes over the edge), will it always stay on the coloured side? (yes) Take another strip and tape the ends together as though to make a ring, but this time turn one of the ends once before you tape it.

Have students do the same. Have students put a finger somewhere on the strip and ASK: Is your finger on the inside or on the outside of the strip? Ask students to slide their fingers along the strip until everyone has their finger on the outside. Have students continue sliding their fingers along the strip until everyone has their finger on the inside. ASK: How did you slide your finger from the outside to the inside without going over the edge? Could you have done that with the original ring? (no)

Show your own strip and explain that you think it has two sides (point to two opposite “sides” at the same point). Suggest to students that if there were two sides, they should be able to colour only one side. Challenge them to do so. Students will see that if they colour a whole side, they have to colour every part of the paper, even what was originally the other side of the strip!

Explain to the students that when they taped the two ends together after turning one of the ends, they created only one side—they taped the “inside” to the “outside.” Explain that this surface is called a Möbius strip. Show an 11˝ × 1˝ strip of paper with one side coloured.


PrOCESS EXPECTATiON

tape

outside

inside

tape


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Then demonstrate turning it into a Möbius strip and show how the coloured side becomes the white side.

Then ask students what they think will happen if they make two turns instead of one before taping the ends together. Do they end up with one side or two sides? (two sides) Have them predict and then check their prediction. Repeat with three turns (one side), and four turns (two sides), this time using the longer strips of paper. Have students predict what will happen with five turns and six turns. What about 99 turns? (one side) 100 turns? (two sides)

2. Each student will need 3 strips of paper (11˝ × 1˝) and 2 longer strips of paper (say, 22˝ × 1˝) with a line drawn lengthwise in the middle of each strip, a pair of scissors, and tape (staples do not work for this activity). Draw the lines with a marker that bleeds through the paper, so that the lines are visible on the other side of the strip as well.

Ask students to tape the ends of one of the strips of paper to make a ring, so that the ends of the line in the middle meet. SAy: I want to cut this ring along the line (hold up your own ring to illustrate what you mean). What will I get? (two thinner rings). Have students check their predictions by cutting their rings.

Take another strip of paper and tape the ends together, this time turning one of the ends once before you tape them. Make sure the ends of the line meet, as before. Have students do the same. Ask students to predict what will happen when they cut the strip along the line. (Students who have never seen a Möbius strip before will likely predict that there will be two rings. Ask students who have seen or done this before not to reveal the right answer.) Then have students cut their strips to check their prediction. (There will be only one ring!)

Explain that the surface students made before cutting it in half is called a Möbius strip. Then ASK: What will happen to the surface when I cut it if I turn one end two times before taping it to the other end? Have students check their prediction. (There will be two rings linked together.) Repeat with three turns (one ring) and four turns (two rings), this time using the longer strips. Have students predict what will happen with five turns and six turns. What about 99 turns? 100 turns? (For even numbers of turns, there will be two rings. For odd numbers of turns, there will be a single ring.) To explain why this happens, colour half of the strip along the middle line (e.g., colour the bottom half of both sides of the strip). When you turn the end once, the coloured half is taped to the white half, and the resulting ring is half coloured and half white. When you turn the end twice, the coloured half is taped to the coloured half, and the white is taped to the white. This way, when you cut the strip, you separate the coloured ring from the white ring.

tape

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Extensions1. Find the incorrect number in each pattern and correct it. a) 2, 5, 7, 11 add 3 b) 7, 12, 17, 21 add 5 c) 6, 8, 14, 18 add 4 d) 29, 27, 26, 23 subtract 2 e) 40, 34, 30, 22 subtract 6

2. Each pattern was made by adding or subtracting the same number each time. Find the missing number(s) in each pattern and explain the strategy you used. a) 2, 4, , 8 b) 9, 7, , 3 c) 7, 10, , 16 d) 15, 18, , 24, , 30 e) 3, , 11, 15 f) 16, , 8, 4 g) 14, , , 20 h) 57, , , 48

SOLUTiON: In parts a)–f), you can find the gap directly, since two consecutive terms are given, then use the gap to find the missing terms. Parts g) and h) require more work. Here are two possible strategies students can use:

• Guess, check, and revise. For example, for part g), you know you have to add because 20 is more than 14. Try adding 1 each time; this only gets you to 17: 14, 15, 16, 17. Try adding 2 each time; this gets you to 20: 14, 16, 18, 20.

• Find the gap by determining the number of steps needed to get from one given term to the next. For example, in part g), you have to increase 14 by 6 in 3 equal steps, so each step must be an increase of 2. Similarly, for h), you need to decrease 57 by 9 in 3 equal steps, so each step must be a decrease of 3.

bonus 15, , , 24, 59, , , , 71 100, , , , , , 850

[R, C, PS], 7m1, 7m7

PrOCESS ASSESSMENT

Guessing, checking and revising

PrOCESS EXPECTATiON

Using logical reasoning

PrOCESS EXPECTATiON

[R, PS, C], 7m1, 7m7 Workbook Question 4

PrOCESS ASSESSMENT


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PA7-2 Describing PatternsPages 23–24

CUrriCULUM EXPECTATiONSOntario: 5m63, 5m66, reviewWNCP: review

VOCAbULAry increases decreases term Match sequences to descriptions. See Question 2.

EXTrA PrACTiCE for Question 2:Describe each pattern below as one of the following: A: Increases by the same amount B: Decreases by the same amount C: Increases by different amounts D: Decreases by different amounts E: Repeating pattern F: Increases and decreases by different amounts

a) 17, 16, 14, 12, 11 (D) b) 10, 14, 18, 22, 26 (A) c) 54, 47, 40, 33, 26 (B) d) 741, 751, 731, 721 (F) e) 98, 95, 92, 86, 83 (D) f) 210, 214, 218, 222 (A) g) 3, 5, 8, 9, 3, 5, 8, 9 (E) h) 74, 69, 64, 59, 54 (B)

identify terms in patterns. Write these three patterns:1, 5, 10, 1, 5, 10, 1, 5, 10, … red, blue, green, yellow, red, blue, green, yellow, red, blue, green, yellow,… do re mi fa so la ti do re mi fa so la ti ….

ASK: What is the same about all these patterns? (they are all repeating patterns) What is different? (the length of the core—the part that repeats— is different; the patterns consist of different types of things—numbers, colours, and musical notes)

Explain that each thing in a pattern—whether it’s a number, a colour, a musical note, a shape, or anything else—is called a term. Have volunteers identify the third term in each sequence above. (10, green, and mi)

EXTrA PrACTiCE: a) What is the third term of the sequence 2, 4, 6, 8? b) What is the fourth term of the sequence 17, 14, 11, 8? c) Extend each sequence to find the sixth term. i) 5, 10, 15, 20 ii) 8, 12, 16, 20 iii) 131, 125, 119, 113, 107

GoalsStudents will describe increasing, decreasing, and repeating patterns by writing a rule.


Can add, subtract, multiply, and divide Can count up to subtract

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Sequences made by multiplying and dividing each term by the same number. See Questions 4 and 5.

EXTrA PrACTiCE for Workbook Questions 4 and 5: What operation was performed on each term in the sequence to make the next term? a) 2, 4, 8, 16, … (multiply each term by 2) b) 10 000, 1 000, 100, 10, … (divide each term by 10) c) 10 000, 5 000, 2 500, 1 250, … (divide each term by 2) d) 5, 15, 45, 135, … (multiply each term by 3)

introduce rules of the form “Start at ___, add/subtract/multiply by/ divide by ___.” See Workbook Question 6.

EXTrA PrACTiCE for Workbook Question 6:1. Write the rule for each pattern. a) 12, 15, 18, 21, … b) 19, 17, 15, 13, … c) 132, 136, 140, 144, … d) 1, 3, 9, 27, …. e) 224, 112, 56, 28, … f) 25, 75, 225, 675, …

2. Use the description of each sequence to find the 4th term of the sequence. a) Start at 5 and add 3. b) Start at 40 and subtract 7. c) Start at 320 and divide by 2. d) Start at 5 and multiply by 4.

Write rules for repeating patterns. See Workbook Question 8.

Extensions1. One of these sequences was not made by adding or subtracting the same number each time. Find the sequence and state the rules for the other two sequences. A. 25, 20, 15, 10 B. 6, 8, 10, 11 C. 9, 12, 15, 18

2. The first term of a sequence of numbers is 2. Each term after the first is obtained by multiplying the preceding term by 5 then subtracting 6. What is the 5th term of the sequence?

3. Match each pattern to its description. 1, 4, 13, 40, 121 A. Multiply by 5 and subtract 1. 1, 4, 7, 10, 13 B. Multiply by 3 and add 1. 1, 4, 19, 94, 469 C. Add 5 and subtract 2.


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PA7-3 T-tablesPages 25–27

CUrriCULUM EXPECTATiONSOntario: 5m63, 5m64, 5m65, 7m1, 7m6, reviewWNCP: review, [r, PS]

VOCAbULAry T-table rule

GoalsStudents will use T-tables to solve word problems.


Can find the gaps between numbers Can extend patterns obtained by doing one operation successively

introduce T-tables using the example on the worksheet. Work through the example at the top of Workbook p.25 together. Point out that this type of chart is called a T-table because the central part of the chart looks like a T.

Teach students how to use and create T-tables by following the progression in Questions 1–4: start by identifying the rules for patterns from completed T-tables (Question 1); then use T-tables to extend patterns (Questions 2 and 3); then create T-tables to extend patterns (Question 4).

EXTrA PrACTiCE: Count the number of toothpicks in each figure. Then use a T-table to determine how many toothpicks make up Figure 5.

Figure 1 Figure 2 Figure 3

Double T-charts. Draw the pattern at left on the board.

Make a double T-chart—a T-chart with 3 columns—with headings Number of Unshaded Squares, Number of Shaded Squares, and Number of Squares. Have students copy the blank chart in their notebooks and fill it in independently. Then have students use the chart to answer these questions: a) How many shaded squares will be needed for a figure with 7 unshaded squares? b) How many squares in total will be needed for a figure with 15 shaded squares?

Show the following double T-chart and ask students to answer the questions below.

Time (min) Fuel (L) Distance from airport (km)

0 1200 525

5 1150 450

10 1100 375

a) How much fuel will be left in the airplane after 25 minutes? b) How far from the airport will the plane be after 30 minutes? c) How much fuel will be left in the airplane when it reaches the airport?


PrOCESS EXPECTATiON

Organizing data

PrOCESS EXPECTATiON

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Students will use double T-charts to solve Workbook p. 27 Question 10.

EXTrA PrACTiCE with Word Problems: 1. The snow is 17 cm deep at 5 p.m. Four centimetres of snow falls each

hour. How deep is the snow at 9 p.m.? (33 cm)

2. Philip has $42 in savings at the end of July. Each month he saves $9. How much will he have by the end of October? ($69)

3. Carol’s plant is 3 cm high and grows 5 cm per week. Ron’s plant is 9 cm high and grows 3 cm per week. How many weeks until the plants are the same height? (3)

4. Rita made an ornament (see margin) using a hexagon (dark grey), pentagons (light grey), and triangles (white). a) How many pentagons does Rita need to make 7 ornaments? (42) b) Rita used 12 hexagons to make ornaments. How many triangles did she use? (144) c) Rita used 12 pentagons to make ornaments. How many triangles did she use? (24)

5. A store rents snowboards at $7 for the first hour and $5 for every hour after that. How much does it cost to rent a snowboard for 6 hours? ($32)

6. a) Look at the pattern in the margin. How many triangles would Ann need to make a figure with 10 squares? (14)

b) Ann says that she needs 15 triangles to make the sixth figure. Is she correct? (No, to make the sixth figure she needs 16 triangles.)

7. Merle saves $55 in August. She saves $6 each month after that. Alex saves $42 in August. He saves $7 each month after that. Who has saved the most money by the end of January? (Merle has $85, whereas Alex has only $77, so Merle has saved the most money by the end of January.)

Extensions1. a) How many 11s are in the sequence 1 3 3 5 5 5 7 7 7 7 …? b) How many 7s are in the sequence 1 2 2 2 2 3 3 3 3 3 3 3 …?

Problem solving

PrOCESS EXPECTATiON

[PS, R, V], 7m1, 7m6

PrOCESS ASSESSMENT

1 2 3

Give each student a set of blocks and ask them to build a sequence of figures that grows in a regular way (according to some pattern rule) and that could be a model for a given T-table. SAMPLE T-tables:

ACTiViTy

Figure # of blocks Figure # of blocks Figure # of blocks

1 4 1 3 1 1

2 6 2 7 2 5

3 8 3 11 3 9


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HiNT: Use a T-chart with headings Number and Number of Times Occurs ANSWErS: a) 6 b) 19

2. Magic Squares. Show students the 3 × 3 array in the margin. Explain that this is a magic square because all the numbers in each row, column, and diagonal add to the same number, in this case 15. Verify this together. (4 + 7 + 4 = 15, 5 + 5 + 5 = 15, and so on)

A pure 3 × 3 magic square places each of the numbers from 1 to 9 exactly once in a 3 × 3 grid in such a way that each row, column, and diagonal adds to the same number. Follow the steps below to make a pure 3 × 3 magic square.

a) By pairing numbers that add to 10, find 1 + 2 + ... + 8 + 9. (45)

b) Your answer to part a) tells you what all 3 rows add to. What does each row add to? (45 ÷ 3 = 15) This is the magic sum.

c) List all possible ways of adding 3 different numbers from 1 to 9 to total 15 (EXAMPLE: 2 + 4 + 9, but not 3 + 3 + 9 or 6 + 9) ANSWEr: 1 + 5 + 9 1 + 6 + 8 2 + 4 + 9 2 + 5 + 8

2 + 6 + 7 3 + 4 + 8 3 + 5 + 7 4 + 5 + 6

d) Look at a 3 × 3 grid. How many sets of numbers that add to 15 must the number in the middle square be a part of? ANSWEr: 4—the middle row, the middle column, and both diagonals. Look at your list from part c) to determine which number must be in the middle. ANSWEr: Only 5 occurs four times, so 5 is in the middle.

e) Which numbers must be corner numbers? Why? ANSWEr: The corner numbers are each part of three sums. This happens for 2, 4, 6, and 8. (The numbers 1, 3, 7, and 9 only occur in two sums, so these must be in the remaining four squares.)

f) Write the numbers in the grid to make a pure 3 × 3 magic square! Compare your magic square with those of other people. What transformations (e.g., rotations or reflections) can you do to a magic square to get another magic square? (SAMPLE ANSWErS: rotate 90° clockwise; reflect vertically using the middle column as a mirror line)

g) Now make a magic square with the numbers 2, 3, 4, 5, 6, 7, 8, 10. What will the new magic sum be? What if you use the numbers 3, 4, 5, 6, 7, 8, 9, 10, 11? Make a T-table of magic sums:

Numbers Used in the Magic Square Magic Sum

1–9 15

2–10 18

Predict the magic sum for a magic square made with the numbers 8, 9, 10, 11, 12, 13, 14, 15, 16. Check your answer. (36)

bonus Make a magic square with numbers 5, 7, 9, 11, 13, 15, 17, 19, 21. What is the magic sum? (39)

4 7 4

5 5 5

6 3 6


PrOCESS EXPECTATiON

Using logical reasoning

PrOCESS EXPECTATiON

Organizing data

PrOCESS EXPECTATiON

Geometry

CONNECTiON

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PA7-4 Patterns (Advanced)Pages 28–29

CUrriCULUM EXPECTATiONSOntario: 7m1, 7m3, 7m5, 7m6, 7m7, reviewWNCP: review, [r, ME, C, CN]

VOCAbULAry even odd

GoalsStudents will investigate patterns in geometrical sequences, the Fibonacci sequence, and Pascal’s triangle.


Can extend patterns Can use charts and T-tables to display sequences

Finding patterns within patterns. Have students extend this pattern: 1, 4, 7, 10, …. Then tell students that you would like to look for a pattern within this pattern. Review the terms even and odd, then have students identify each term in the pattern as even or odd and record their answers in a table like this one:

Term Number 1 2 3 4 5 6 7 8 9 10

Term 1 4 7 10

Even or Odd O E

Have students predict whether the 100th term will be even or odd and explain their prediction. (The odd-even pattern is “O, E, then repeat.” The 100th term will be even because every even-numbered term is even.)

Now have students extend this pattern: 2, 4, 8, 16, …. Tell students that you would like to know if there is a pattern in the ones digits of this sequence. How about in the tens digits in this sequence? (The ones digits form a repeating pattern: 2, 4, 8, 6, repeat; the tens digits form no easily discernible pattern.)

Finally, have students extend the pattern 1, 4, 9, 16, … and look for an odd-even pattern. (O, E, repeat) bonus Look for a pattern in the ones digits. (1, 4, 9, 6, 5, 6, 9, 4, 1, 0,

repeat; notice the symmetry in the core of this pattern)

Pascal’s triangle. See Workbook Question 3.

EXTrA PrACTiCE: Describe the pattern in the numbers along the 2nd diagonal of Pascal’s triangle. bonus Add the numbers in each row of Pascal’s triangle. For example,

the numbers in the third row add to 1 + 2 + 1 = 4. Use a T-table to find the sum of each of the first five rows, then predict the sum of the 8th row.

bonus Describe the pattern in the numbers along the 4th diagonal of Pascal’s triangle.

Estimating and thinking before solving a problem. Before assigning the Investigation, show students the diagram on Workbook p.29 (the first picture below). If possible, have it reproduced on an overhead transparency so that you can show it for a brief time and remove it quickly (to prevent students


PrOCESS EXPECTATiON

Looking for a pattern, Organizing data

PrOCESS EXPECTATiON


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from counting the lines). Have students guess how many lines there are without counting them. Take all answers. Discuss what makes this problem hard. If students say “Because the lines intersect in so many places”, show them the second picture below:

Challenge students to articulate why this set of lines is easier to count.

Solving the problem in the investigation. Discuss strategies for solving the problem. Direct the students to do a simpler problem first. Suggest that if they find the answer in easier cases first, they might find a pattern. ASK: How can we make this problem easier? PrOMPT: There are 8 dots with all possible lines joining them. What problem can we solve that would be easier? Listen to students’ suggestions and, if they don’t bring it up, point out that if they solve the same problem with 1 dot, 2 dots, 3 dots, and 4 dots, they might find a pattern.

reflecting on other ways to solve the problem. After students finish the Investigation, explain that you noticed another pattern within the pattern of the number of lines for each given number of dots:

Numbers of Dots 1 2 3 4

Number of Lines

Numbers of Dots 5 6 7

Number of Lines

Challenge students to predict the expression for the number of lines for 8 dots. Does their expression give the right answer, 28? Then challenge students to explain why the expression works: What does 7 × 8 tell you? Why do we divide by 2? (There are 8 dots and 7 lines extending from each dot, so 7 × 8 tells you the number of endpoints altogether. Since each line has two endpoints, the total number of lines is 7 × 8 ÷ 2 = 28.)

Reflecting on what made the problem easy or hard

PrOCESS EXPECTATiON

ACTiViTy

Have students get into groups of 2 and shake hands with everyone else in their group. How many handshakes were there? (1) Repeat with groups of 3, groups of 4, and groups of 5. (Ensure that different students are left out, when necessary, of each round.) What do students notice about their answers to this problem and their answers to the Investigation on the worksheet? (They are the same!) Discuss why this happened. Students could arrange themselves in a circle, so that each student represents a point and each handshake represents

Reflecting on what made a problem easy or hard, Organizing data, Connecting

PrOCESS EXPECTATiON

0 =0 × 1

21 =

1 × 2

23 =

2 × 3

26 =

3 × 4

2

15 =5 × 6

221 =

6 × 7

210 =

4 × 5

2

Mental math and estimation

PrOCESS EXPECTATiON

Doing a simpler problem first

PrOCESS EXPECTATiON

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Extensions1. Pick one number from each row in the grid at left; each number must be

in a different column. Add the numbers. Now repeat with a different set of selections. What do you notice about the two sums? (they are the same) Will this always happen? (yes) Can you explain why it happens?

EXPLANATiON: Let’s label each row according to the first number in the row: the 1 row, the 6 row, the 11 row, the 16 row, the 21 row. One number is selected from each row. If you select a number in, say, the 16 row, you can either pick 16 + 0, 16 + 1, 16 + 2, 16 + 3, or 16 + 4. No matter which row you pick from, you are either adding 0, 1, 2, 3, or 4 to the first number in that row. Since you pick one number from each column, you add 0 once, 1 once, 2 once, 3 once, and 4 once, so the sum is 1 + 6 + 11 + 16 + 21 + 0 + 1 + 2 + 3 + 4 = 65.

To help students discover this explanation, ask them to answer the same questions for one or more of the arrays below (or others like them). In the first array, how often are the numbers from the first part of each sum (1, 6, 11, 16, and 21) selected? How often are the numbers 0, 1, 2, 3, and 4 from the second part selected? (once each)

1 + 0 1 + 1 1 + 2 1 + 3 1 + 4 1 + 0 1 + 1 1 + 2 1 + 3 1 + 4

6 + 0 6 + 1 6 + 2 6 + 3 6 + 4 11 + 0 11 + 1 11 + 2 11 + 3 11 + 4

11 + 0 11 + 1 11 + 2 11 + 3 11 + 4 21 + 0 21 + 1 21 + 2 21 + 3 21 + 4

16 + 0 16 + 1 16 + 2 16 + 3 16 + 4 31 + 0 31 + 1 31 + 2 31 + 3 31 + 4

21 + 0 21 + 1 21 + 2 21 + 3 21 + 4 41 + 0 41 + 1 41 + 2 41 + 3 41 + 4

1 + 0 1 + 1 1 + 3 1 + 5 1 + 7 1 + 0 1 + 2 1 + 4 1 + 5 1 + 8

6 + 0 6 + 1 6 + 3 6 + 5 6 + 7 11 + 0 11 + 2 11 + 4 11 + 5 11 + 8

11 + 0 11 + 1 11 + 3 11 + 5 11 + 7 21 + 0 21 + 2 21 + 4 21 + 5 21 + 8

16 + 0 16 + 1 16 + 3 16 + 5 16 + 7 31 + 0 31 + 2 31 + 4 31 + 5 31 + 8

21 + 0 21 + 1 21 + 3 21 + 5 21 + 7 41 + 0 41 + 2 41 + 4 41 + 5 41 + 8

Students can make up their own such 5 × 5 grid and present it as a magic trick to a younger class. One way to present this as a magic trick would be as follows: Each student gives a grid to a younger buddy. The student tells the younger buddy to pick 5 numbers, one from each row and column. (Your student may need to explain to the buddy what a row is and what a column is.) The buddy then adds the numbers but doesn’t tell your student the sum. Your student asks questions that may

1 2 3 4 5

6 7 8 9 10

11 12 13 14 15

16 17 18 19 20

21 22 23 24 25

the line between the two points. Have students arrange themselves into groups of 7 or 8 and to count the handshakes directly. Encourage students to be organized in their counting. Was counting handshakes easier or harder than counting lines between points? Why? (Students might find it easier to keep track of who they have already shaken hands with than which lines they have already counted.)


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seem relevant, but really are not. (EXAMPLES: Is the number in the third column bigger or smaller than the number in the second column? How far apart are the two biggest numbers?) Students will have to be careful to ask questions that are compatible with their buddies’ level. Students then give the correct answer, to their buddies’ surprise.

2. Look at Pascal’s triangle.

a) Start at the top right of any (right to left) diagonal and move along the diagonal, adding the numbers you encounter. Stop at any point you wish. Where will you find the sum? (ANSWEr: Just below and to the right of the last number you added.) EXAMPLE:

1

3

6

10

b) How can you find the sum 1 + 2 + 3 + 4 + 5 quickly using Pascal’s triangle? HiNT: Use the pattern you found in part a).

c) Extend Pascal’s triangle to 15 rows and shade the even numbers. What patterns do you see?

d) Without extending Pascal’s triangle, can you find the missing numbers in the 8th row?

1 28 708 56 56 28 8 1

HiNT: The first and last numbers in each row are 1. Some students may notice that since the rows are symmetrical, they can reduce their work by half.

[PS, R], 1m1, 1m2

PrOCESS ASSESSMENT

7m1, 7m2, [PS, R] Workbook Investigation

PrOCESS ASSESSMENT

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GoalsStudents will substitute values for the variables in algebraic expressions and translate simple word problems into algebraic expressions.


Can add, subtract, multiply, and divide Knows the order of operations

PA7-5 Constant ratesPages 30–33

CUrriCULUM EXPECTATiONSOntario: 5m68, 6m63, 7m65, 7m1, 7m7, 7m67WNCP: 7Pr5, [C, PS]

VOCAbULAry variable algebraic expression substitution flat fee hourly rate

introduce variables and algebraic expressions. See Workbook Questions 1–3.

A variable represents a changing number. After students do Workbook Question 4, write on the board the cost of renting a pair of skates for 2 hours, 3 hours, 4 hours, and 5 hours. ASK: How much would it cost to rent the skates for 6 hours? 10 hours? 37 hours? h hours? t hours? w hours? r hours?

introduce flat fees and hourly rates. Work through Question 8 a) together, then have students write an expression for the cost of renting a boat (at the same flat fee of $9 and hourly rate of $5) for these times: a) 1 hour b) 3 hours c) 4 hours d) 5 hours e) 11 hours f) 15 hours

Now challenge students to write an expression for the cost of renting a boat for h hours, or m hours, or n hours.

What changes? Have students identify the objects for which the quantity changes in each of the following situations. (This quantity is what the variable in the corresponding algebraic expression represents.)

a) Poppies are on sale for 5¢ each. (poppies) b) An Internet café charges $2 for each hour. (hours) c) A grocery store charges 5¢ for each plastic bag. (plastic bags)

Introduce the terms flat fee and hourly rate. Then have students decide what quantity in the following situations must be represented by a variable.

a) A skate rental company charges a $2 flat fee and then $3 for each hour. (hours)

b) A boat rental company charges a $10 flat fee and then $5 per hour. (hours)

c) A taxi company charges a $5 flat fee and then $2 for each kilometre. (kilometres)

bonus A bus company charges 10¢ per kilometre and $5 per passenger. (both kilometres and passengers)

Substituting values for the variable in expressions involving only addition. See Workbook Question 12.


PrOCESS EXPECTATiON


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The need for brackets when substituting. See Workbook Question 13.

Using brackets as another notation for multiplication. See Workbook Question 14.

EXTrA PrACTiCE: Evaluate these expressions.a) 3(5) + 4 b) 7 + 2(3) c) 7 – 2(3) d) 3(4) – 5 bonus 7(6) – 2(5) + 3(3) – 8(5)

Substituting in a context. See Question 16.

interchangeable expressions. Using different variables in the same expression, or writing the terms of the expression in a different order, doesn’t change the meaning of the expression. Students will discover this by doing Questions 18–22.

Substituting for 2 variables. See Question 23. EXTrA PrACTiCE: Find the value of each expression for x = 2 and y = 3. ANSWErS:a) 5x + 4y 22 b) 6x – 2y 6 c) 9x – y 15bonus 9xy 54

Extensions1. A rectangle has area xy. Find the area if x = 5 and y = 7.

ANSWEr: 35

2. A triangle has area 12

bh. Find the area if b = 8 and h = 3. ANSWEr: 12

3. Write an expression for the cost of a pizza (x) divided among 4 people. ANSWEr: x ÷ 4

4. Give students a copy of a times table. Ask them to write an expression that would allow them to find the numbers in a particular column of the times table given the row number. For example, to find any number in the 5s column of the times table, you multiply the row number by 5; each number in the 5s column is given by the algebraic expression 5 × n, where n is the row number. Ask students to write an algebraic expression for the numbers in a given row.

7m1, [PS] Workbook Question 24

PrOCESS ASSESSMENT

7m1, [R]

PrOCESS ASSESSMENT

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GoalsStudents will solve equations of the form ax + b = c by guessing small values for x, checking by substitution, and then revising their answer.


Can substitute a number for a variable Can read charts Can verify equations

CUrriCULUM EXPECTATiONSOntario: 6m66, 7m1, 7m6, 7m7, 7m69 WNCP: 5PR2, [r, C]

VOCAbULAry solving for the variable (e.g., solving for x)

introduce using a chart to solve equations. Review Question 24 from PA7-5: The cost, in dollars, of renting the bike is 8h + 5, where h is the number of hours the bike is rented for. We want to know how many hours it will take for the cost to reach $61. ASK: How can we express that using an equation? (we want to find h so that 8h + 5 = 61; this will be the largest amount of hours we can rent the bike for; if h was higher, the price would be more than $61) Show students how to solve 8h + 5 = 61 by using a chart and plugging h = 1, h = 2, h = 3, and so on, into the expression. See Questions 1 and 2 on the worksheet for this lesson.

introduce the guess and check method to solve equations. Show the equation 7h – 2 = 61. Tell students that you are going to solve this equation by guessing and checking. Start by guessing h = 5. ASK: If h = 5, what is 7h – 2? (33) What does this tell me? Should h be higher or lower to make 7h – 2 = 61? (higher) What would your next guess be? Continue in this way until students see that h = 9.

Compare the two methods of solving equations. ASK: Which method takes less work? Which method is quicker? (the guess and check method is quicker) Which method is more like looking up a word in the dictionary using alphabetical order? (guess and check) Which method is more like looking up a word in the dictionary without knowing or using alphabetical order? (using the chart) Have students explain the connection. (In a dictionary, each page you turn to tells you whether to look to the right or to the left.)

Equations that mean the same thing. When solving Question 7, students might prefer to rewrite the equation in the form ax + b = c. They can do this and know they will still get the same answer because of their work in Questions 5 and 6.

Look for equations that mean the same thing. See Question 8. Discuss strategies for finding an equation that means the same thing as 2 + 7x = 23. If the equations mean the same thing, would they have the same answer? (yes, the same number will solve both) Would you look for

Guessing, checking and revising

PrOCESS EXPECTATiON

Connecting

PrOCESS EXPECTATiON

Organizing data

PrOCESS EXPECTATiON

PA7-6 Solving Equations—Guess and CheckPages 34–35


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an equation that has an x? (no, 7x means the same thing as 7r or 7u) Which equations are solved by the same number as part e)? (a, d, I, m, n, and p) What part or parts of the equation should you look for? (a 7 in front of the variable and a 23 or a 2 in the equation). ASK: Should the 23 be on the same side of the equal sign as the variable or on the opposite side? (the side opposite the variable) What about the 2? (the same side as the variable)

HiNT for Bonus: Try splitting the problem into two easier problems: 2x + 1 = 7 and 4y – 1 = 7.

Extensions1. How many digits does the solution to 3x + 5 = 8000 have? Explain.

ANSWEr: To determine the number of digits in the solution, we need to determine the first power of 10 (10, 100, 1000, etc.) that is greater than the solution. So, substitute increasing powers of 10 for the variable until the answer is larger than 8000: 3(10) + 5 = 353(100) + 5 = 3053(1 000) + 5 = 3 0053(10 000) + 5 = 30 005So x is between 1 000 and 10 000, which means that it has 4 digits. Indeed, x = 2665.

2. Students who do the Bonus will see that the solution to 2x + 1 = 7 = 4y – 1 is x = 3 and y = 2. How many solutions can students find to 2x + 1 = 4y – 1 (the same expressions but not necessarily both equal to 7) if x and y are whole numbers? ANSWEr: Find 2x + 1 for various values of x:

x 1 2 3 4 52x + 1 3 5 7 9 11

Now find 4y – 1 for various values of y:

y 1 2 3 4 54y – 1 3 7 11 15 19

Look for numbers that are the same in the second rows: 2x + 1 = 3 = 4y – 1 when x = 1 and y = 12x + 1 = 7 = 4y – 1 when x = 3 and y = 22x + 1 = 11 = 4y – 1 when x = 5 and y = 3

Students might continue the pattern to find more solutions (x = 7 and y = 4 is the next one).

Splitting into simpler problems

PrOCESS EXPECTATiON

7m1, 7m7, [PS, C]

PrOCESS ASSESSMENT

7m7, [C] Workbook Question 5

PrOCESS ASSESSMENT

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GoalsStudents will use pictures to model and solve equations.


Can use variables to represent an unknown value Can solve an equation to find an unknown value Can solve equations by guessing and checking

MATEriALS

paper bags and counters

PA7-7 Modelling EquationsPage 36

CUrriCULUM EXPECTATiONSOntario: 7m1, 7m2, 7m6, 7m7, 7m69WNCP: 7Pr4, 7Pr6, 7Pr7, [C, r, V]

VOCAbULAry expression equation variable

Finding the unknown in a concrete model. Divide students into pairs and have them play the following game:

Step 1: Player 1 takes a small number of paper bags and puts an equal number of counters in each bag. Player 1 also selects some counters to be left outside the bags.

Step 2: Player 1 tells Player 2 the total number of counters placed both in the bags and outside the bags. EXAMPLE: If Player 1 places 2 counters in each of 5 bags and has 3 counters outside, Player 1 tells Player 2 “I’ve placed 13 counters altogether”.

Step 3: Player 2 has to figure out (without looking in the bags) how many counters are in each bag. Players then switch roles.

Students can use guessing, checking, and revising as a strategy. In the example in Step 2, a student might start by guessing 1 counter in each bag and then check how many counters there would be in total (only 8, so this guess is too small). Students might also discover the strategy of working backwards. EXAMPLE: If there are 13 counters altogether and I see 3 counters outside the bags, there must be 10 counters in the bags. There are 5 bags, so there must be 2 in each bag.

relate the model to algebra. ASK: What is the unknown you were looking for? (the number of counters in each bag) Represent that by x. ASK: How can you get the number of counters in all the bags from x? (multiply x by the number of bags) How can you get the total number of counters from x? (x × the number of bags + the number of counters outside of bags)

Draw a model on the board and build the corresponding algebraic expression step by step:

There are x counters in each bag. There are 4x counters in all the bags because there are 4 bags.There are 4x + 3 counters altogether.

Modelling

PrOCESS EXPECTATiON


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Then tell students that there are 35 counters altogether. ASK: What equation can we write? (4x + 3 = 35) Challenge students to determine the number of counters in each bag by solving the equation. (x = 8)

Draw a model to verify an answer. EXAMPLE: For the equation 2x + 5 = 11, start by drawing 2 containers and 5 counters, then add 1 counter to each container until you have 11 counters altogether.

Students can solve equations by guessing, checking, and revising, and then verify their answers by drawing a model. Students should draw their models so that the items inside the bags (e.g., counters, apples) are clearly visible.

ExtensionFind as many solutions as you can with x and y whole numbers: 10x + 4y = 74. HiNT: Use guessing and checking. ANSWEr: (x, y) = (7, 1), (5, 6), (3, 11), (1, 16)

bonus Find solutions that are not whole numbers. SAMPLE ANSWEr: (0, 18.5), (2, 13.5), (4, 8.5), (6, 3.5), (0.5, 17.25)

Modelling

PrOCESS EXPECTATiON

7m1, [PS]

PrOCESS ASSESSMENT

7m6, [V], Workbook Question 4

7m2, 7m7, [R, C], Workbook Question 5

PrOCESS ASSESSMENT

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GoalsStudents will “undo” many operations to get back where they started when starting with a number, and will “undo” one operation when starting with a variable.


Knows that multiplication and division undo each other Knows that addition and subtraction undo each other Can substitute numbers for variables

PA7-8 Solving Equations—Preserving EqualityPages 37–38

CUrriCULUM EXPECTATiONSOntario: 7m1, 7m6, 7m69WNCP: 7Pr3, 7Pr6, 7Pr7, [r]

VOCAbULAry variable

Undoing one operation. Have students pair up. Player 1 chooses a secret number. Player 2 gives Player 1 an operation—either multiplication or addition—to do to the secret number (EXAMPLE: multiply by 3, or add 7). Player 1 carries out the operation and tells Player 2 the answer. Player 2 has to find the secret number. Partners can trade roles and repeat.

Discuss how students “undid” operations to find—or get back to—the numbers their partners started with. For example, ASK: How did you get back to the original number if your partner multiplied the number by 3? (divided the answer by 3) How did you get back to the original number if you instructed your partner to add 7? (subtracted 7 from the answer)

Have students do Workbook p. 37 Questions 1–4.

Undoing more than one operation. EXAMPLE: If I start with 16, and my partner tells me to add 6 then multiply by 3, I will tell my partner the answer is 66. Now my partner has to undo both adding and multiplying! Have students discuss whether they should undo the adding first or the multiplying. Then discuss ways to check if they got the right answer without getting confirmation from you. For example, if they try to undo “add 6 then multiply by 3” by subtracting 6 then dividing by 3, they will get: 60 ÷ 3 = 20. They can check this answer by starting with 20, adding 6 (26), and multiplying by 3. This gives 78 when the answer is supposed to be 66, so 20 is incorrect; you undo “add 6 then multiply by 3” by dividing by 3 then subtracting 6. Write out the steps as on the worksheet:

Start with 16 16 Get back to 16 Add 6 22 Subtract 6 16 Multiply by 3 66 Divide by 3 22

Tell students that you started with a number, added 4, divided by 3, multiplied by 2, and then subtracted 1. You ended up with 9. Write out the steps on the board and challenge students to determine your original number. (11) Tell students that this is like a parcel wrapped in many layers: you start by unwrapping the last layer wrapped.

Have students make game cards for a partner. On the front of the card, students write a sequence of operations and the final answer obtained.

7m1, [R]

PrOCESS ASSESSMENT


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Each sequence should include all four operations, and nowhere along the way should a decimal number be obtained (e.g., don’t start with 7 and divide by 2). On the back of the card, students write the original number. Partners trade cards and undo the operations on the front of the card to find the original number. EXAMPLE:

Front of Card back of Card Start with ?. ? = 7 Multiply by 2. Subtract 1. Add 5. Divide by 3. Get 6.

An analogy. Remind students that when they put on their socks and shoes, they put their socks on first and then their shoes. ASK: How do you undo these two operations - which do you do first? (Undo the operations in reverse order - take off your shoes first, then your socks.) This is how we undo operations in math too. If you add first and then multiply, you undo multiplying first and then undo adding.

Treating variables like numbers. Tell students that to add 3 to 4, you would write 4 + 3. ASK: What would you write to add 3 to x? (x + 3) Continue with other operations, as on Workbook p. 38 (top).

Writing in words what was done to the variable. See Question 8.

Undoing operations done to variables. You undo operations done to variables in the same way you undo operations done to numbers. See Questions 1, 2, 3, and 9.

Preserving equality. Give students equations with one operation. Have students describe what was done to the variable, x, and how to undo that operation to find x. EXAMPLES:a) 3x = 12 b) x + 3 = 12 c) x ÷ 3 = 5 d) x – 3 = 5ANSWErS: a) x was multiplied by 3 to get 12, so divide 12 by 3 to get x (x = 12 ÷ 3 = 4)b) 3 was added to x to get 12, so subtract 3 from 12 to get x (x = 12 – 3 = 9)c) x was divided by 3 to get 5, so multiply 5 by 3 to get x (x = 5 × 3 = 15)d) 3 was subtracted from x to get 5, so add 5 to 3 to get x (x = 5 + 3 = 8)

When students are comfortable describing how to undo operations to find x, show them how to do it with the equation:

3x = 123x ÷ 3 = 12 ÷ 3 (if 3x and 12 are the same number, then dividing them both by 3 will still result in the same number) x = 4

Explain that you divided both sides by 3 because you wanted to undo the multiplying by 3 and get back to where you started.

7m6, [R] Workbook Question 10

PrOCESS ASSESSMENT

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GoalsStudents will “undo” many operations to get back to the variable they started with.


Knows that multiplication and division undo each other Knows that addition and subtraction undo each other Can substitute numbers for variables

PA7-9 Solving Equations—Two OperationsPages 39–40

CUrriCULUM EXPECTATiONSOntario: 7m1, 7m2, 7m6, 7m69WNCP: 7Pr3, 7Pr6, 7Pr7, [r, PS, C]

VOCAbULAry variable substitute solve for

Writing the expression that shows what was done to the variable. Have students write the expression that results from each series of operations: a) Start with x. Multiply by 3. Add 4 3x + 4 b) Start with x. Multiply by 4. Add 3. 4x + 3 c) Start with x. Multiply by 5. Subtract 2 5x – 2

Writing the equation when you know the final result of the operations. Tell students that the final result, after doing the operations above, was always 43. Write the equation for each question above. ANSWErS: a) 3x + 4 = 43 b) 4x + 3 = 43 c) 5x – 2 = 43 d) 5x – 7 = 43

Undoing each operation in turn to find x. Now have students undo the operations in reverse order to find x. ANSWEr for a): 3x + 4 = 43 3x + 4 – 4 = 43 – 4 Undo adding 4 by subtracting 4. 3x = 39 Write the new equation. 3x ÷ 3 = 39 ÷ 3 Undo multiplying by 3 by dividing by 3. x = 13 Write the new equation.

Start with an expression and have students say what operations were done, and in what order. EXAMPLE: 3x + 4 has multiplying by 3 and adding 4, but which was done first—multiplying by 3 or adding 4? (multiplying by 3) What would starting with x, adding 4, and then multiplying by 3 look like? ANSWEr: 3(x + 4).

Have students write what was done to x to get each answer and then undo those operations in reverse order to solve for x:a) 3x + 7 = 31 b) 2x – 1 = 11 c) 5x – 2 = 48 d) 7x – 1 = 48ANSWEr for a): Start with x. Multiply by 3. Then add 7. Get 31. 3x + 7 = 31Undo adding 7 by subtracting 7: 3x + 7 – 7 = 31 – 7 Write the new equation: 3x = 24 Undo multiplying by 3 by dividing by 3: 3x ÷ 3 = 24 ÷ 3Write the new equation: x = 8

Checking your answer. Encourage students to verify their answers by substituting them into the original equations. ANSWEr for a): 3(8) + 7 = 24 + 7 = 31. It works!

Working backwards

PrOCESS EXPECTATiON


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More word problems. 1. a) A grocery store charges 5¢ for each grocery bag. Write an expression for the cost of buying n grocery bags. (5n¢) b) Substitute for n to find out how much it costs to buy i) 3 bags ii) 5 bags iii) 10 bags ANSWErS: i) 5(3) = 15¢ ii) 5(5) = 25¢ iii) 5(10) = 50¢

2. a) A telephone company charges 25¢ per minute. Write an expression for talking on the phone for m minutes. (25m¢) b) Substitute for m to find out how much it costs to talk for

i) 3 minutes ii) 10 minutes iii) 13 minutes

ANSWErS:25(3) = 75¢25(10) = 250¢ or $2.5025(13) = 325¢ = $3.25 (or just add the costs for 3 minutes and 10 minutes)

bonus How much would it cost to talk for 1 hour? (25¢ × 60 minutes = 1500¢ = $15)

c) Sara paid $1.50 = 150¢ to talk on the phone. Write an equation and solve for m to determine how long she talked for.

3. It costs $8 to rent a pair of skis and $4 an hour to use the ski hill. a) Write an expression for how much it costs to rent a pair of skis and use the ski hill for h hours. b) How much will it cost to rent the skis for 5 hours? c) How long can you ski for if you have $36? d) For which question—b) or c)—did you need to solve for h? For which question did you need to substitute for h?

4. Another ski hill charges $20 to rent a pair of skis but only $2 an hour to use the ski hill. a) Sara wants to ski for 5 hours. On which ski hill will she pay less—this

one or the one from Question 3? b) Bob wants to ski for 7 hours. On which ski hill will he pay less? bonus Jeff calculates that the two ski hills will charge him the same

amount. How long does he plan to ski for? Justify your answer. (ANSWEr: 6, since 5 hours cost less on the ski hill from Question 3, and 7 hours cost less on this ski hill. Students can check directly that it costs $32 to ski for 6 hours on each hill.)

Extensions1. Kyle paid $22 for a taxi ride. The initial charge was $2 and he rode for

5 minutes. What was the charge per minute? (2 + 5x = 22 so 5x = 20, so x = 4)

2. See More word problems, Question 2 (above). How much would it cost to talk for 24 seconds? (10¢; either solve the ratio 25¢ : 60 sec = ? : 24 sec or multiply 25¢ per min × 24/60 min.)

7m1, 7m6, [PS, C]

PrOCESS ASSESSMENT


PrOCESS ASSESSMENT

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GoalsStudents will use a balance model to model the process of solving equations of the form ax + b = c.


Can solve equations of the form ax + b = c by working backwards

PA7-10 Modelling Equations—AdvancedPage 41

CUrriCULUM EXPECTATiONSOntario: 7m1, 7m6, 7m69WNCP: 7Pr3, 7Pr6, 7Pr7, [V, PS]

VOCAbULAry none Using mass to model equations. Draw a triangle and two circles.

Tell students that each circle has mass 1 kg but you don’t know the mass of the triangle. Let’s call its mass x, an unknown, because we don’t know what it is.

balancing any combination of 1 triangle and circles (corresponds to equations with addition only). Show a situation where 1 triangle and 2 circles balance 7 circles. ASK: How can we determine the mass of the triangle? (find out how many circles balance the triangle)

x + 2 = 7

Remove all the circles from the left-hand side and an equivalent number of circles from the right-hand side.

x + 2 – 2 = 7 – 2

The triangle has the same mass as 5 circles.

x = 5

Have students show the equation each scale represents.

a) b)

Then have students solve the equations. ANSWErS: a) x = 7, b) x = 3.

Have students draw the scale for each of these equations. a) n + 2 = 8 b) n + 3 = 10 c) n + 5 = 9

Ask students to write equations for the following scales using the letter n as the unknown.

a) b)

Modelling

PrOCESS EXPECTATiON

NOTE: The pan balances pictured here and on the worksheet are called scales throughout.


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ASK: What do you notice about the operations in all of the equations we have seen so far? (there is only addition) What do you notice about all of the scales we have seen so far? (there is only one triangle)

balancing any combination of triangles only (corresponds to equations with multiplication only). Have students write the equations represented by these scales.

a) b)

(3x = 12) (2x = 12)

c) (4x = 12)

Draw a scale where 1 triangle balances 3 circles. ASK: How many circles will 2 triangles balance? 3 triangles? 4 triangles? 5 triangles? Display the pictures and corresponding equations for each situation.

Tell students that you know that 5 triangles balance 10 circles. ASK: How many circles will 1 triangle balance? How do you know? Explain that if a certain number of triangles balances another number of circles, and you divide both sets into the same number of groups, then each single group of triangles will balance a single group of circles. For example, if 5 triangles balance 10 circles, then 1/5 of the triangles will balance 1/5 of the circles, so 1 triangle balances 2 circles. If 3 triangles balance 12 circles, then 1/3 of the triangles balance 1/3 of the circles, so 1 triangle balances 4 circles. Illustrate how this affects the equations:

5x = 10 ? = ?

x = 2

Tell students that you asked three people to write an equation for the middle scale and they gave you three different answers: A: 5x – 4 = 10 – 8 B: 5x – 4x = 10 – 8 C: 5x ÷ 5 = 10 ÷ 5

ASK: Is Student A’s equation correct? Does Student A get the correct answer if he solves his equation? (5x – 4 = 2, so 5x = 6, so x = 6/5 = 1.2. This is not correct, since the last scale clearly shows that x = 2.) What went wrong? First, if 5x = 10, then subtracting 4 from 5x cannot possibly give the same result as subtracting 8 from 10. Also, subtracting 4 triangles is not the same as subtracting 4; we are subtracting 4x from the left-hand side, not 4!Student B took this into account and wrote 5x – 4x = 10 – 8, which gives x = 2. This is the correct answer. However, it is not at first clear that subtracting 4x is the same as subtracting 8. It is only because you are

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taking away the same fraction of each side that this is true. Student C also got the right answer. She divided each side into 5 equal groups and kept one of the groups. So one fifth of each side still balances. This equation is particularly convenient because it is clear that you are doing the same thing to both sides.

Write equations for models that include both addition and multiplication, and solve for x.Step 1: Write the equation that represents the model.Step 2: Remove all circles from the side that has the triangle(s) and remove the same number of circles from the other side. Write the new equation. Step 3: Divide the circles into the number of groups given by the number of triangles. Keep only one group of circles and one triangle. Write the new equation. This will be the solution!

EXTrA PrACTiCE: Draw the scales for each equation below (don’t show students the equations) and have students do Steps 1, 2, and 3 to solve for x. a) 3x + 4 = 16 b) 2x + 5 = 11 c) 5x + 3 = 18 d) 4x + 7 = 23

ACTiViTy

In the magic trick below, the magician can predict the result of the sequence of operations performed on any number. Try the trick with students: ask them to pick a number but not tell you what it is, have them perform the operations in sequence, then tell them the answer, 3. No matter what mystery number students choose, after performing the operations in the trick, they will always get the number 3. Encourage students to figure out how the trick works by drawing a model (and give them lots of hints!).

The trick A model for the algebra

Pick any number Use a square to represent the mystery number.

Add 4 Use 4 circles to represent the 4 ones that were added.

Multiply by 2 Create 2 sets of shapes to show the doubling.

Subtract 2 Take away 2 circles to show the subtraction.

Divide by 2 Remove one set of shapes to show the division.

Subtract the Remove the square. mystery number The answer is 3!

Encourage students to make up their own similar trick.7m1, [PS]

PrOCESS ASSESSMENT

7m6, [V] Workbook Question 5

PrOCESS ASSESSMENT


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Extensions1. Scale A is balanced. Draw the number of circles needed to balance Scale B.

a)

A B

b)

A B

c)

A B

d)

A B

e)

A B

f)

A B

2. Which part of Extension 1 shows that 2x + 1 = 7 is solved by x = 3? (Part e))

3. Scales A and B are balanced. Draw the number of circles needed to balance Scale C. Explain how you know.

a)

A B

C

b)

A B

C

7m1, 7m6, 7m7, [PS, V, C]

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GoalsStudents will group like terms and cancel opposite terms to simplify and solve equations.


Can solve equations of the form ax + b = cCan substitute for the variable

PA7-11 Solving Equations—AdvancedPage 42

CUrriCULUM EXPECTATiONSOntario: 7m1, 7m2, 7m69WNCP: 7Pr7, [PS, r]

VOCAbULAry cancel simplify group like terms Multiplication as a short form for addition. Remind students that the

expression 3 × 5 is short for repeated addition: 3 × 5 = 5 + 5 + 5. Similarly, 3x is short for x + x + x.

Have students write the following expressions as repeated addition. a) 5n (n + n + n + n + n) b) 4x (x + x + x + x) c) 7y (y + y + y + y + y + y + y)

Have students write each sum as a product. a) x + x + x + x + x (5x) b) n + n + n (3n) c) m + m + m + m + m + m (6m)

Grouping x’s to simplify an expression. Challenge students to write 2x + 3x as a single term.ANSWEr: 2x + 3x = x + x + x + x + x = 5x

Explain that we can do this because each x represents the same number. Show students how to verify that 2x + 3x and 5x are equal for various values of x. x = 1 2(1) + 3(1) = 2 + 3 = 5 and 5(1) = 5 so 2x + 3x = 5x for x = 1x = 2 2(2) + 3(2) = 4 + 6 = 10 and 5(2) = 10 so 2x + 3x = 5x for x = 2x = 3 2(3) + 3(3) = 6 + 9 = 15 and 5(3) = 15 so 2x + 3x = 5x for x = 3x = 4 2(4) + 3(4) = 8 + 12 = 20 and 5(4) = 20 so 2x + 3x = 5x for x = 4

Have students substitute x = 5 and x = 6 into the two expressions to verify equality. Explain that this works simply because 2 anythings plus 3 anythings is 5 anythings: 2 ones + 3 ones = 5 ones 2 twos + 3 twos = 5 twos2 threes + 3 threes = 5 threes 2 x’s + 3 x’s = 5 x’s

ASK: How many x’s are there in 3x + 4x? (7) So 3x + 4x = 7x. Explain that grouping all the x’s together is called simplifying. Then have students simplify these expressions: ANSWErS:a) 8x + 2x 10x b) 9x + 4x + 3x 16x c) 3x + 3x + 4x 10x bonus x + 2x + 3x + 4x + 5x + 6x 21x


PrOCESS EXPECTATiON

3x2x


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Grouping with word problems. Tell students that pizza costs $5 per student and drinks cost $2 per student. Have students write an expression for: a) the cost of x students buying pizza (5x)b) the cost of x students buying drinks (2x) c) the total cost of x students buying pizza and drinks (5x + 2x or 7x)

Explain the two ways of getting the answers in part c): 5x + 2x is the sum of the two separate costs; 5 + 2 is the cost per student, so the total cost to all students will be (5 + 2)x = 7x.

Cancelling. Explain as on the worksheet. See Workbook Question 5.

EXTrA PrACTiCE for Workbook Question 6:a) 8x – 4x b) 3x – 2x c) 6x – 3x d) 9x – 5x e) 7x – 2x f) 8x – 6x

Have students subtract the variables and then the numbers, and compare the answers. a) 7x – 2x = and 7 – 2 = b) 3x – x = and 3 – 1 = c) 6x – 4x = and 6 – 4 = d) 8x – 2x = and 8 – 2 = e) 6x – x = and 6 – 1 = f) 8x – 2x + 3x = and 8 – 2 + 3 = g) 7x + 3x – 4x – 2x = and 7 + 3 – 4 – 2 = What do students notice?

EXTrA PrACTiCE for Workbook Question 7: a) 10 – 3 = so 10x – 3x = b) 8 – 5 = so 8x – 5x = c) 7 – 4 = so 7x – 4x = d) 9 – 4 = so 9x – 4x = e) 6 – 2 = so 6x – 2x = f) 5 – 3 = so 5x – 3x = g) 8 – 3 + 1 = so 8x – 3x + x = h) 7 – 4 + 2 – 3 = so 7x – 4x + 2x – 3x =

Write each expression so that there is only one x. a) 6x – 5x b) 6x – 4x c) 6x – 3x d) 6x – 2x e) 6x – x f) 8x – 3x + 2x g) 9x – 2x + 3x h) 7x – 2x + 4xSAMPLE ANSWEr: g) 9x – 2x + 3x = 10x because 9 – 2 + 3 = 10

bonus 9x – 8x + 7x – 6x + 5x – 4x + 3x – 2x + x (Students might notice that each pair of consecutive terms (9x – 8x, 7x – 6x, etc.) makes x, so the answer is x + x + x + x + x = 5x.)


PrOCESS ASSESSMENT

7m1, [PS]

PrOCESS ASSESSMENT

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GoalsStudents will explain the difference between an expression and an equation. Students will identify variables, coefficients, and constant terms in word problems.


Can model and solve equations of the form ax + b = c

PA7-12 Equations and ExpressionsPages 43–44

CUrriCULUM EXPECTATiONSOntario: 7m7, optionalWNCP: 7Pr4, 7Pr7, [C]

VOCAbULAry expression equation coefficient variable constant term

Differences and similarities between equations and expressions. See Questions 1–5. Equations and expressions both contain numbers, symbols, and variables. Equations have equal signs that separate two expressions.

Coefficients and constant terms. Define coefficient and constant term as on the worksheet. Have students complete the table.

Expression Variable(s) Coefficient(s) Constant Term(s)

3x + 5 – 4y + 1 x, y 3, – 4 5, 1

0.2x + 4.2

6 + 2x + y

2 + 3 – 5

2x + 3

Variables, coefficients, and constant terms in word problems. Ask students to identify coefficients and variables in context. EXAMPLE: When she is running, Sara’s heart beats at a rate of 120 beats per minute. Have students write an expression for the number of beats if Sara runs for m minutes. (120m) ASK: What is the coefficient? (120) What information does the coefficient provide? (the number of beats per minute while running) What does the variable represent? (the number of minutes Sara runs for)

Repeat with a word problem that includes a constant term. EXAMPLE: A phone company charges a flat fee of $10 plus $3 per hour of phone conversation. Have students write the expression for the total cost of talking on the phone for h hours. ASK: What is the coefficient? What does the coefficient represent? (the coefficient is 3 and it represents the hourly rate) What does the constant term represent? (the flat fee) What does the variable represent? (the number of hours)

7m7, [C] Workbook Questions 2 and 3

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GoalsStudents will solve problems that can be represented by equations of the form = ,

xb

a where a is not zero.


Can solve equations of the form ax = b and ax + b = c by working backwards

MATEriALS

metre stick rulers pieces of string (see below for lengths)

PA7-13 Dividing by a ConstantPage 45

CUrriCULUM EXPECTATiONSOntario: 7m1, 7m2, 7m6, optionalWNCP: 7Pr7, [r, V]

VOCAbULAry none

review using fractional notation for division. See Question 1.

EXTrA PrACTiCE for Question 1: 1. Write these division statements as fractions.

a) 8 ÷ 2 (ANSWEr: 82

) b) 18 ÷ 6 c) 20 ÷ 10

2. Write these fractions as division statements.

a) 204

(ANSWEr: 20 ÷ 4) b) 168

c) 213

Solve for n by using a chart. Have students copy and complete the chart below in their notebooks.

n 0 5 10 15 20 25 30 35 40 45 50

n5

0 1 2

Then have students use the chart to solve for n.

a) n5

4= b) n5

7= c) n5

10= d) n5

8= e) n5

6= f) n5

2= n = n = n = n = n = n =

Solve for n by guessing and checking. EXAMPLE: n/5 = 6. Try n = 10, get 10/5 = 2, which is less than 6, so try a larger number, say n = 20. This gets 20/5 = 4, which is still too small. Try a larger number, say n = 30. This works.

Solve for x by using a model. Show students a piece of string that is 120 cm long and tell students that you want them to measure it using only a metre stick and taking only one measurement. Have students brainstorm possible answers. If necessary, explain that the string is too long to be measured by the metre stick, but half of the string wouldn’t be too long. ASK: How can you measure half of the string with only one measurement? (fold it in half and measure it along the metre stick) How long is half the

Organizing data

PrOCESS EXPECTATiON

Guessing checking and revising

PrOCESS EXPECTATiON

n

Modelling

PrOCESS EXPECTATiON

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string? (60 cm) How long is the whole string? (120 cm) Show this as a series of equations: x/2 = 60 so x = 120.

Now tell students that they have only a ruler instead of a metre stick. How could they measure the string now, again using only one measurement? (fold the string in half twice) ASK: What fraction of the string’s original length is the folded string? (one quarter—to ensure students understand this, have them mark where the folds are and then open the string; they should see four equal parts) Have a volunteer show the corresponding equations on the board: x/4 = 30 so x = 120.

Give each student a piece of string and a ruler. Have students find the length of the string by finding the length of a fraction of the string, as above. Restrict the string lengths to numbers that can be divided by 2, 4, or 8 to make a length less than the length of a standard ruler (30 cm). SAMPLE string lengths: 224 cm (x/8 = 28 cm) 100 cm (x/4 = 25) 52 cm (x/2 = 26 cm) 54 cm (x/2 = 27 cm) 68 cm (x/4 = 17) 176 cm (x/8 = 22 cm)

Now tell students that a string is divided into 3 equal pieces:

ASK: If x is the length of the string, what is the length of each piece? (x/3) Tell students that the length of each piece is 4 m. ASK: What equationcan we write? (x/3 = 4) Show students how to model each piece being 4 m long:

So the whole string is 4 × 3 m long, which means x = 12. Have students draw a model to solve these equations:

a) x2

5= b) x3

5= c) x5

4= d) x3

10= e) x10

3=

Solve for x by working backwards. See Questions 3 and 4.

EXTrA PrACTiCE for Question 3: a) x

210= (x = 10(2) = 20) e) x

67= (x = 7(6) = 42)

b) x7

8= (x = 8(7) = 56) f) 69x

= (x = 6(8) = 48)

c) x9

5= (x = 5(9) = 45) g) x9

6= (x = 6(9) = 54)

d) x7

7= (x = 7(7) = 49) bonus x/12 = 17 (x = 17(12) = 204)

7m6, [V]

PrOCESS ASSESSMENT

Working backwards

PrOCESS EXPECTATiON


PrOCESS ASSESSMENT


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Extensions1. Write as many multiplication and division statements as you can that are equivalent to the given statement. (NOTE: Capital letters can be used as variables to represent numbers just like lower case letters)

a) 12 ÷ 3 = 4 b) AB = C c) XY

Z=

2. Multiply both sides by x and solve the new equation. Explain your strategy and check your answer.

a) 82

x= b) 12

x= 3 c) 20

4x

= d) 248

x=

8 ÷ x = 2 8 ÷ x × x = 2x 8 = 2x x = 8

2 x = 4

7m1, 7m7, [PS, R, C]

PrOCESS ASSESSMENT

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GoalsStudents will solve problems involving equations by translating phrases and sentences into expressions and equations.


Can solve equations

CUrriCULUM EXPECTATiONSOntario: 7m1, 7m2, 7m3, 7m4, 7m5, 7m7, 7m66, 7m69WNCP: 7Pr7, [C, CN, PS, r]

VOCAbULAry expression equation consecutive odd even perimeter area

Associating phrases with operations. Write the phrases from the box at the top of Workbook p. 46 (increased by, product, decreased by, etc.) on the board. Have students decide which operation each phrase makes them think of. Students can then create a chart with the headings Add, Subtract, Multiply, and Divide, and sort each phrase under the correct heading.

Translating phrases into expressions. See Questions 1 and 2.

EXTrA PrACTiCE for Questions 1 and 2:Translate each phrase into an expression. a) 5 more than a number (x + 5) b) 5 less than a number (x – 5) c) 5 times a number (5x) d) the product of a number and 5 (5x) e) a number reduced by 5 (x – 5) f) a number divided by 5 (x/5)g) 5 divided into a number (x/5) h) 5 divided by a number (5/x)i) a number divided into 5 (5/x) j) a number decreased by 5 (x – 5)k) a number increased by 5 (x + 5) l) the sum of a number and 5 (x + 5)m) the product of 5 and a number (5x) n) 5 fewer than a number (x – 5) bonus a number multiplied by 3 then increased by 5 (3x + 5)

Translating sentences into equations. Once students can reliably translate phrases into expressions, it is easy to translate sentences into equations: simply replace the word is with an equal sign (=) and replace the two phrases separated by is with the appropriate expressions. Give students some sentences to translate, but use sentences where one of the phrases is just a number. EXAMPLE: Three times a number is 12. bonus Use sentences in which neither phrase is just a number. EXAMPLE:

Three times a number is the number increased by 8. (3x = x + 8)

Have students translate sentences into equations and then solve the equations. Use sentences where only one side of the resulting equation includes a variable. EXAMPLE: 3x + 5 = 20 not 3x + 5 = 2x + 9.

PA7-14 Word ProblemsPages 46–48


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Solving word problems. Show students how to translate word problems into equations. EXAMPLE: Carl has 7 stickers. He has 2 more stickers than John. How many stickers does John have? SOLUTiON: Let n stand for the number of stickers that John has, since that is the unknown that we want to find. Let’s try to find two ways of writing how many stickers Carl has so that we can write an equation of this form: (number of stickers Carl has) = (number of stickers Carl has).

Carl has 2 more stickers than the number John has.So Carl has 2 more stickers than n.So Carl has n + 2 stickers. But Carl has 7 stickers. So n + 2 = 7.

EXTrA PrACTiCE: a) Katie has 10 stickers. She has 3 fewer stickers than Laura. How many

stickers does Laura have? SOLUTiON: Let n be the number of stickers that Laura has, the unknown we are looking for. Katie has 3 fewer stickers than the number Laura has, so Katie has n – 3 stickers. But Katie has 10 stickers. So n – 3 = 10.

b) Katie has 12 stickers. She has 3 times as many stickers as Laura. How many stickers does Laura have?

c) Katie has 12 stickers. She has 3 more stickers than Laura. How many stickers does Laura have?

d) Katie has 12 stickers. She has half as many stickers as Laura. How many stickers does Laura have?

introducing new contexts to word problems. Have students solve the three questions below and then discuss how they are similar and how they are different. a) Bilal has 20 stickers. He has 5 times as many stickers as Ron. How many stickers does Ron have? b) Bilal is 20 years old. He is 5 times older than Ron. How old is Ron? c) Bilal walked 20 km. He walked 5 times farther than Ron. How far did Ron walk?

Challenge students to make up their own contexts for the same numbers.

EXTrA PrACTiCE with word problems: a) Bilal runs 600 m each day. He runs 3 laps each day. How long is

each lap?b) Bilal runs 800 m each day. He runs 40 m more than Ahmed. How far

does Ahmed run?

Review consecutive numbers. See Questions 4 and 5.

Word problems involving consecutive numbers. Challenge students to solve this problem using algebra: The sum of two consecutive numbers is 27. What are the two numbers?

Connecting

PrOCESS EXPECTATiON

Problem solving

PrOCESS EXPECTATiON

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HiNT: Choose one of the unknown numbers to be x, then write an expression for the other number. For example, if you let the smaller number be x, the larger number is x + 1; if you let the larger number be x, the smaller number is x – 1.

ANSWEr: If the smaller number is x, the equation becomes 2x + 1 = 27, which gives x = 13, so the two numbers are 13 and 14. If the larger number is x, the equation becomes 2x – 1 = 27, which gives x = 14. Again, the two numbers are 13 and 14.

Now encourage students to solve the same problem using T-tables, as in Question 6a) on Workbook p. 47, and discuss the two methods. (T-tables are a lot more work—algebra saves time and effort!)

In the following problems, there is more than one unknown. Students will see that while you can let the variable represent any of the unknowns, some choices are better than others. In the first two problems, the middle number is the best choice because it makes the equation easier to work with.

1. The sum of three consecutive numbers is 36. What are the three numbers? a) Let the smallest number be x and solve the problem. (x + x + 1 + x + 2 = 3x + 3 = 36 so x = 11) b) Let the middle number be x and solve the problem. (x – 1 + x + x + 1 = 3x = 36 so x = 12) c) Let the greatest number be x and solve the problem. (x – 2 + x – 1 + x = 3x – 3 = 36 so x = 13) d) Did you get the same answer all three ways? (Yes, the numbers are always 11, 12, and 13.) e) Which way was easiest? Explain your choice. (part b because the equation only involved multiplication. This happened because the added and subtracted numbers cancelled)

2. The sum of five consecutive even numbers is 80. What are the five numbers? a) Let the smallest number be x and solve the problem. (5x + 20 = 80, so x = 12; the numbers are 12, 14, 16, 18, and 20) b) Let the middle number be x and solve the problem. (5x = 80, so x = 16; the numbers are 12, 14, 16, 18, and 20) c) Which way was easiest? Explain your choice. (part b was easiest because the equation only involved multiplication, again because of cancelling)

Here is more practice in a new context:

3. Solve the following problem using algebra: Jane has $3 in dimes and quarters. She has 21 coins in all. How many of each coin does she have? Steps for students who need guidance: a) Let the number of quarters be x. b) Write an expression for the number of dimes. (21 – x) c) Write an expression for the value of the quarters, in cents. (25x)d) Write an expression for the value of the dimes, in cents. (10(21 – x))e) Explain why your answer to d) is the same as 10 × 21 – 10x.

Reflecting on other ways to solve a problem

PrOCESS EXPECTATiON

7m1, 7m2, [PS, R]

PrOCESS ASSESSMENT

Selecting tools and strategies

PrOCESS EXPECTATiON


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(This is the distributive law) f) We are given the total value of the coins: $3, or 300¢. Use the expressions in c) and e) to write another expression for the total value of the coins, in cents. Write your answer in the form ax + b. (25x + 10 × 21 – 10x = 15x + 210) g) Write an equation by using the expression for the total value of the coins from f) and the given information. (15x + 210 = 300) Use 300 instead of 3 because the left side is in cents, not dollars) h) Solve your equation. How many quarters does Jane have? How many dimes does she have?(15x = 90, so x = 90/15 = 6; Jane has 6 quarters and 21 – 6 = 15 dimes) i) Verify your answer by totalling the value of the coins from h). (6 quarters = $1.50 and 15 dimes = $1.50, so altogether we have $3 and 6 + 15 = 21 coins)

ExtensionWrite an equation to find the length of the missing side(s).

a) b)

x x

c) d)

x + 4 2x

2m Area = 24 m2 4 A = 16 m2

Perimeter = 48 cm P = 72 cmx x

7m3, 7m4, [R, C] Workbook Questions 7 and 8

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GoalsStudents will understand that variables represent a changing quantity and will represent generalizations arising from number relationships using algebraic equations.


Knows that division by zero is not possible Knows that addition commutes (a + b = b + a)Knows that multiplication commutes (a × b = b × a) and that multiplication distributes over addition and subtraction (e.g., a × (b – c) = a × b – a × c)

PA7-15 investigating EquationsPages 49–50

CUrriCULUM EXPECTATiONSOntario: 6m63, 7m1, essential for 7m19WNCP: 6PR3, essential for 7Pr2, [r]

VOCAbULAry variable

Variables as changing quantities. Show the equations from the top of the worksheet. Emphasize that the variable a is not representing an unknown number, but instead a changing number. The equation 2 × a = a + a is always true, no matter what we substitute for a, as long as we substitute the same number for all three a’s. This is very different from an equation like 3a + 1 = 13, where a represents an unknown number and we have to find the number that makes the equation true.

More equations in one variable. Show the following equations: 8 + 1 – 1 = 8 8 + 2 – 2 = 8 8 + 3 – 3 = 88 + 17 – 17 = 8 8 + 134 – 134 = 8

ASK: What numbers are changing? Have volunteers replace these numbers with a variable, such as a: 8 + a – a = 8. Explain that it doesn’t matter what we add and subtract—as long as we add and subtract the same number, we will always end up with 8.

Have students try to come up with more examples of equations in one variable that are always true, no matter what you substitute for the variable. EXAMPLES: a + 4 – 4 = a 3 × a = a + a + a a × 1 = a a × 0 = 0 a + 0 = a

Equations that are almost always true. Write the following equations on the board: 1 ÷ 1 = 1 2 ÷ 2 = 1 3 ÷ 3 = 1 4 ÷ 4 = 1 5 ÷ 5 = 1

Have a volunteer replace the changing number with a variable: a ÷ a = 1. ASK: Can we substitute any value for a and make the equation true? As a prompt, ASK: Is there a number that you are not allowed to divide by? (0) Explain that 0 ÷ 0 has no answer, so the equation a ÷ a = 1 is true as long as a doesn’t equal 0.

Equations in two variables. Remind students that a + 4 – 4 = a is true for any value of a, as long as we substitute the same number for both a’s. Now, write these equations:


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a + 1 – 1 = a a + 2 – 2 = a a + 3 – 3 = aa + 4 – 4 = a a + 5 – 5 = a

Have students create more equations that follow the same pattern. ASK: What number is changing in these equations? Have a volunteer come and replace the changing number with a variable. Then write on the board: a + a – a = a. ASK: Does this equation show the pattern here? Why not? Emphasize that the equation a + a – a = a tells you what happens when you substitute the same number for all four variables; for example, 5 + 5 – 5 = 5 and 6 + 6 – 6 = 6. But for our pattern, we need two different variables, so we need the more general a + b – b = a, which tells you that as long as both occurrences of a are replaced with the same number and both occurrences of b are replaced by the same number, the equation is true. The first equation, a + a – a = a, is a special case of the second equation (the case where a = b).

The statements “a and b are both not zero” and “a and b are not both zero”. Assign different values, including 0, to a and b and have students decide when these statements are true. EXAMPLE: a = 3 and b = 0 a = 0 and b = 0 a = 5 and b = 2 a = 0 and b = 7

The statement “a and b are both not zero” is only true for a = 5 and b = 2; “a and b are not both zero” is true for the first, third and fourth examples.Have students finish this sentence: The equation (a ÷ b) × (b ÷ a) = 1 is true provided that . ANSWEr: a and b are both not zero.

Substituting for the variables. Have students substitute a = 3 and b = 5 into various equations. a) a + b − b = a (3 + 5 − 5 = 3)b) a + b = b + a (3 + 5 = 5 + 3)c) a × b = b × a (3 × 5 = 5 × 3)d) 2(b − a) = 2b − 2a (2(5 − 3) = 2(5) − 2(3))e) 5(a + b) = 5a + 5b (5(3 + 5) = 5(3) + 5(5))

Verifying equations. Review with students how to verify that an equation is true for given values of the variable: calculate both sides and make sure they both equal the same number. See Workbook p.50 Question 6.

EXTrA PrACTiCE for Question 6. Verify that each equation is true for a = 4 and b = 7.a) a × b = b × a b) 3(a + b) = 3a + 3b c) (a ÷ 2) × (b × 2) = a × b d) a + 3 + b – 3 = a + b e) (b + 3) – (a + 3) = b – a

Equations in two variables that are sometimes true. Tell students that two of these equations are true for all values of a and b. (1) a × b + b = (a + 1) × b (2) a × b + b = a × (b + 1) (3) a × b + a = a × (b + 1)

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Have students predict which two equations are true for all a and b. Then have students substitute values to determine which of the equations are true for: i) a = 3 and b = 4 ii) a = 2 and b = 6 iii) a = 4 and b = 4

Have students use their answers to decide which two equations are always true. (Equations 1 and 3) Which equation is only sometimes true? (Equation 2)

Explain that because Equation 2 worked for a = 4 and b = 4, you wonder if it will always be true when a = b. Have students check directly for a = 10 and b = 10, and then choose and check their own (equal) values for a and b.

Then show students another way to check if the equation is true when a = b: Substitute a for b into both sides. If you do this in Equation 2, a × b + b becomes a × a + a, and a × (b + 1) becomes a × (a + 1), so the equation becomes a × a + a = a × (a + 1), an equation in one variable that is true for any value of a. EXAMPLE: 7 × 7 + 7 = 7 × 8

Equations in three variables. Show students the following equations in two variables. a + b + 2 = 2 + a + ba + b + 3 = 3 + a + ba + b + 4 = 4 + a + b

Explain that all these equations are true, no matter what you substitute for a and b. Have students replace the changing number with a different variable (i.e., not a or b). (EXAMPLE: a + b + c = c + a + b) Explain that this is an equation in three variables. Have students verify these equations in three variables for a = 2, b = 5, and c = 4:a) 3(a + b + c) = 3a + 3b + 3c b) 3(a + b – c) = 3a + 3b – 3c c) a × (b + c) = a × b + a × cd) a × (b – c) = a × b – a × ce) (a + b) × c = a × c + b × cf) (c – a) × b = c × b – a × b

Extensions1. If 4 3 = (4 × 3) + (4 + 3)

and 2 5 = (2 × 5) + (2 + 5)

calculate 7 9

ANSWEr: 7 9 = (7 × 9) + (7 + 9) = 63 + 16 = 79

2. The numbers greater than 1 are arranged in the following array. The columns are numbered 1 to 5.

7m1, 7m2, 7m3, [PS, R]

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(1) (2) (3) (4) (5)

2 3 4 5

9 8 7 6

10 11 12 13

17 16 15 14

18 19 20 21

25 24 23 22

a) Describe the pattern in each column. Which columns were easier to describe? Why? ANSWEr: (1) Start at 9 and add 8 each time. (2) Start at 2. Add 6 then 2, and repeat.

This column could also be separated into the odd terms and the even terms:

Odd terms: Start at 2 and add 8 each time. Even terms: Start at 8 and add 8 each time. (3) Start at 3 and add 4 each time. (4) Start at 4. Add 2 then 6, and repeat. OR: Odd terms: Start at 4 and add 8 each time. Even terms: Start at 6 and add 8 each time. (5) Start at 5 and add 8 each time. The patterns in columns (1), (3), and (5) are easy to describe because you only have to add the same number each time, whereas in columns (2) and (4), you have to add different numbers. b) In which column does the number 584 appear? ANSWEr: Look at the pattern in the whole array. The numbers are placed in sequence across the first row (in columns 2, 3, 4, 5) and then backwards across the second row (in columns 4, 3, 2, 1). The pattern repeats after each 8 terms. Since the (a + 8)th term will always be in the same column as the ath term, all the terms that are divisible by 8 (8, 16, 24, …) will be in the same column. Since 584 ÷ 8 = 73 (no remainders), 584 is in the same column as 8, 16, 24, etc., which is Column (2).

c) A number leaves a remainder of 4 when divided by 8. In which column does it appear? ANSWEr: Column (4) has all the numbers that have a remainder of 4 when divided by 8: 4, 12, 20, etc.

d) In which row does the number 584 appear? ANSWEr: There are 4 numbers in each row, so every 4th number is in a row one higher. Notice that 584 is divisible by 4 and that the numbers 4, 8, 12, 16,… are in rows 1, 2, 3, 4, …, and, in general, 4n is in row n. So since 584 ÷ 4 = 146, we must have that 584 is in row 146.

unit 2 patterns and algebra - jump math for ap book 7-1... · unit 2 patterns and algebra in this...

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