Unit 2 Patterns and Algebra - JUMP Math :: Home for AP...Unit 2 Patterns and Algebra In this unit, students extend and describe patterns, use T-tables to solve problems, and solve

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Teacher’s Guide for Workbook 8.1

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).

Meeting your curriculum All lessons are core material for the WNCP curriculum.

For those following the Ontario curriculum, part of lesson PA8-11 is optional,

namely the part that models equations of the form xa

b= or

xa

b c+ = .

Lessons PA8-13 and PA8-14 are also optional for Ontario students.

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 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).

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C-2 Teacher’s Guide for Workbook 8.1

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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.

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

6 7 8 9 10 11 12 13 14 15 16

+3 +5

46 47 48 49 50 51 52 53 54 55 56

+3 +5

10 + ? = 17 10 + 7 = 17

8 + ? = 10 8 + 2 = 10

6 7 8 9 10 11 12

C-3Patterns and Algebra

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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.


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PA8-1 Extending PatternsPage 92

CUrriCULUM EXPECTATiONSOntario: reviewWNCP: review

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 8-1

EXTrA PrACTiCE for 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 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 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: You slid 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. Then demonstrate turning it into a Möbius strip and show how the


PrOCESS EXPECTATiON

tape

outside

inside

tape


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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

bonus

i) 6, 8.2, 10.4, , , 17

j) 7, , 10, , 13

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, 8m1, 8m7]

PrOCESS ASSESSMENT

Guessing, checking and revising

PrOCESS EXPECTATiON

Using logical reasoning

PrOCESS EXPECTATiON

[R, PS, C, 8m1, 8m7] Workbook Question 4

PrOCESS ASSESSMENT


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PA8-2 Describing PatternsPages 93–94

CUrriCULUM EXPECTATiONSOntario: 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 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 question 6.

EXTrA PrACTiCE for 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.

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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PA8-3 T-tablesPages 95–96

CUrriCULUM EXPECTATiONSOntario: 8m1, 8m6WNCP: [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. 95 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 PrACTiCECount 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 Questions 6 and 7.

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, 8m1, 8m6]

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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4 7 4

5 5 5

6 3 6


PrOCESS EXPECTATiON


PrOCESS EXPECTATiON

Organizing data

PrOCESS EXPECTATiON

Geometry

CONNECTiON

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 it 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)

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Doing a simpler problem first

PrOCESS EXPECTATiON

PA8-4 Patterns (Advanced)Pages 97–98

CUrriCULUM EXPECTATiONSOntario: 8m1, 8m3, 8m5, 8m6, 8m7 WNCP: [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 Question 3.

EXTrA PrACTiCE: Describe the pattern in the numbers along the 2nd diagonal of Pascal’s triangle. bonus 1. 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.

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 6 squares in a


PrOCESS EXPECTATiON

Looking for a pattern, Organizing data

PrOCESS EXPECTATiON


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row. 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 square, 2 squares, 3 squares, and 4 squares, 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 rectangles for each given number of squares:

Numbers of Squares 1 2 3

Number of rectangles

Numbers of Squares 4 5 6

Number of rectangles

Challenge students to predict the expression for the number of rectangles for 6 squares. (6 × 7)/2

ASK: Does your expression give the right answer? (yes, 6 × 7 ÷ 2 = 42 ÷ 2 = 21)

Extensions1. a) Show students the connection between the two ways of finding the

next term in the pattern from the Investigation on Workbook p. 98. For example, to find the sixth term, you can either add 6 to the fifth term, or use the expression: 6 × 7 ÷ 2. To find the expression for the sixth term, add 6 to the expression for the fifth term. Challenge students to find a way to do this. ANSWEr: 5 6 6

2×

+ = 6 5 6

2×

+ since 5 × 6 = 6 × 5

= 6 5 6 2

2 2× ×

+ since 6 = 6 2

2×

= 6 5 6 2

2× + ×

since to add fractions with the same denominator, just add the numerators =

6 (5 2)2

× + by the distributive property

= 6 7

2×

= 6 × 7 ÷ 2

Reflecting on what made the problem easy or hard

PrOCESS EXPECTATiON


PrOCESS EXPECTATiON

Reflecting on the reasonableness of an answer

PrOCESS EXPECTATiON

1 =1 × 2

23 =

2 × 3

26 =

3 × 4

2

15 =5 × 6

210 =

4 × 5

2

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b) Show students another connection between the two ways of finding the next term in the pattern. Notice that, according to the pattern found in part E of the Investigation, the number of rectangles for 6 squares is 1 + 2 + 3 + 4 + 5 + 6. This can be represented visually using dots:

Now double the number of dots as follows.

2 × = + =

The picture shows that 2 × (1 + 2 + 3 + 4 + 5 + 6) = 6 × 7. So, 1 + 2 + 3 + 4 + 5 + 6 = 6 × 7 ÷ 2.

2. Your answers to part E of the Investigation on Workbook p. 98 form a pattern: 1 3 6 10 15 21. Find this pattern in the Pascal’s triangle you drew for Workbook p. 97 Question 3 a). You can find the pattern in two places.

3. Pick one number from each row in the grid below; 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 2 3 4 5

6 7 8 9 10

11 12 13 14 15

16 17 18 19 20

21 22 23 24 25

Visualization

PrOCESS EXPECTATiON


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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 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 sum, to their buddies’ surprise.

4. 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:

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, 8m1, 8m2]

PrOCESS ASSESSMENT

1

3

6

10

[8m1, 8m2, 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

PA8-5 VariablesPages 99–100

CUrriCULUM EXPECTATiONSOntario: 6m63, 7m65, 7m67, 8m1, 8m7, 8m59WNCP: 7PR5, review, [C, PS]

VOCAbULAry variable substitution flat fee hourly rate algebraic expression

introduce variables and algebraic expressions. See Questions 1–5.

A variable represents a changing number. After students do Question 6, 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 7 a) together, then have students write an expression for the cost of renting a boat (at a 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)

Using brackets as another notation for multiplication. See Questions 8 and 9.


PrOCESS EXPECTATiON


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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 Questions 10 and 11.

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 12 and 13.

Substituting for 2 variables. See Question 23.

Extensions1. Write an expression for the cost of a pizza (x) divided among 4 people.

ANSWEr: x ÷ 4

2. 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.

[8m1, 8m7, [R, C] Workbook Question 13

PrOCESS ASSESSMENT

[8m1, R]

PrOCESS ASSESSMENT

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GoalsStudents will substitute fractions and decimals for variables in expressions that include one or more variables.


Can substitute whole numbers for variables in an expression

PA8-6 Substituting Fractions and Decimals for Variables Pages 101–102

CUrriCULUM EXPECTATiONSOntario: 8m5, 8m6, 8m7, 8m61, 8m62WNCP: essential for 8Pr2, [CN, C]

VOCAbULAry substitution variable fraction decimal

Substituting fractions for variables. Begin with expressions that require only one operation. EXAMPLES:a) 3x, x = 1/2 b) v − 2, v = 11/2 c) w + 3, w = 2/5

Then continue with expressions that require two operations. EXAMPLES:a) 3t + 2, t = 5/3 b) 8m − 1, m = 3/4 c) 5z + 1, z = 3/5

Substituting decimals for variables. Again, begin with expressions that require only one operation and then continue with expressions that require two operations. EXAMPLES:a) 9x, x = 0.2 b) r + 6, r = 0.3 c) w − 5, w = 5.8d) 4u − 2, u = 0.56 e) 5w + 1, w = 0.6 f) 7r + 2, r = 2.34

Looking for equivalent answers. Remind students that the same number can be written as a fraction and a decimal. Tell students that there is an answer from the fraction questions that is the same as an answer from the decimal questions, and ask them to find it. Can they explain why? (5z + 1, z = 3/5 and 5w + 1, w = 0.6 both give the answer 4; this makes sense because the expressions are the same and 3/5 and 0.6 are equivalent numbers)

Substituting whole numbers for two variables. See Workbook Question 4.

EXTrA PrACTiCE: Find the value of each expression for x = 3 and y = 2

a) 5x + 4y (23)

b) 6x − 2y (14)

c) 5x − 6y (3)

bonus 7xy (42)

Equivalent substitutions in two variables. Have students do these substitutions and compare the answers: a) 3x + 4y, x = 5 and y = 2 and 4x + 3y, x = 2 and y = 5b) 2x + 7y, x = 8 and y = 1 and 7x + 2y, x = 1 and y = 8

What do students notice about the answers? (they are the same) Why is that? (because we are multiplying the same numbers; for example, in a), I multiplied 3 by 5 and 4 by 2 and 3(5) + 4(2) is the same as 4(2) + 3(5))

Representing, Connecting

PrOCESS EXPECTATiON

Connecting

PrOCESS EXPECTATiON


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Ask students to decide what they could substitute for the variables in the second expression to obtain the same answer as by doing the first substitution:

a) 2x + 3y, substitute x = 4, y = 5 and 3x + 2y, substitute x = , y =

b) 4x + 5y, substitute x = 1, y = 0 and 5x + 4y, substitute x = , y =

c) x + 4y, substitute x = 3, y = 2 and 4x + y, substitute x = , y =

bonus x + 2y + 3z, substitute x = 4, y = 5, z = 6

and 2x + 3y + z, substitute x = , y = , z =

Have students check their answers by actually making all substitutions.

Substituting fractions and decimals for two variables. Use only whole number coefficients. See Workbook Question 5.

More than two variables. See Workbook Questions 7–10.

EXTrA PrACTiCE

Write an expression for the number of sides (n) in …

a) t triangles

b) s squares

c) t triangles and s squares

d) t triangles, s squares, and p pentagons

Substitute in d) to find how many sides 4 triangles, 2 squares, and

6 pentagons have in total. (50) bonus for Question 9: If the total number of points is 100, find

possible values for m, p, and r. (SAMPLE ANSWEr: m = 5, p = 25, r = 7)

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

2. A triangle has area 12bh . Find the area if b = 8 and h = 3.

3. Find 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)

8m1, [PS]

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

Solving Equations—Guess and CheckPA8-7

Pages 103–105

CUrriCULUM EXPECTATiONSOntario: 6m66; 7m69; 8m1, 8m3, 8m4, 8m7, 8m64 WNCP: 8Pr2, [r, C]

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

review using fractional notation for division. Have students write these division statements as fractions.

a) 8 ÷ 2 (ANSWEr: 82

) b) 18 ÷ 6 c) 20 ÷ 10

Have students 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

5n

0 1 2 Then have students use the chart to solve for n. a) n/5 = 4 b) n/5 = 7 c) n/5 = 9 d) n/5 = 6

Add another row to the chart for n/5 + 4, fill it in, and have students use it to solve similar questions (EXAMPLE: n/5 + 4 = 9). Repeat with rows for 2n and 2n + 5.

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.)

Organizing data

PrOCESS EXPECTATiON

Guessing, checking and revising

PrOCESS EXPECTATiON

Connecting

PrOCESS EXPECTATiON

NOTE: Teachers following the Ontario Curriculum are not required to cover expressions

of the form: 54x

+ as in

Workbook Question 2, and parts c and d of Questions 3 and 4.


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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.

Discuss strategies for finding two equations that mean the same thing. ASK: If two equations mean the same thing, would they have the same answer? (yes, the same number would solve both) Finding equations with the same answer considerably reduces the amount of checking required to find two equations that are just written differently.

Hint for Workbook Question 9: Try splitting the problem into two easier problems: 2x + 1 = 7 and 4y – 1 = 7.

Adding the same number to both sides of an equation doesn’t change the answer. Write this sequence of equations on the board:3x = 12 3x + 1 = 13 3x + 2 = 14 3x + 3 = 15 3x + 4 = 16

Have students solve these equations by guessing and checking. ASK: What do you notice about your answers? (they are all the same) Point out that these equations are all different—they are not just different ways of writing the same thing—but there is a pattern in them. Have students describe the pattern. (the variable is always being multiplied by 3; you add one more to 3x each time and you add one more to the right side each time) Tell students that if two numbers are the same and you add 1 to both numbers, they are still the same. As an example, tell students to suppose that you and a friend both have the same number of dollars. Then you both get one more dollar. ASK: Do we both still have the same number of dollars? (yes) Explain that this is exactly what is going on in the equations. If three times a number is 12, then adding one to three times the same number will get 13, because 12 + 1 is 13, and adding two to three times the same number will get 14, because 12 + 2 is 14.

Choosing the easiest equation to solve. ASK: Which of the five equations was easiest to solve by guessing and checking? (the first one) Why? (the guessing part is easier because to find three times what equals 12, simply divide 12 by 3; the checking part is easier because there is only one operation to do, multiplication instead of multiplication and adding)

Tell students that if they know that two equations are solved by the same number, they can just solve the easier one to get the answer to both! Challenge students to find an easier problem to solve that has the same answer: a) 4x + 1 = 25 b) 3x + 2 = 17 c) 5x + 3 = 13 (ANSWEr of a): If 4x + 1 = 25, then 4x = 24, which is solved by x = 6.)

Multiplying both sides of an equation by the same number doesn’t change the answer. Write the following sequence of equations on the board: 2x = 6 4x = 12 6x = 18 8x = 24

Have students solve each equation by guessing and checking. ASK: What do you notice about the answers? (they are the same) Why is that the case? PrOMPT: Do you see a pattern in the equations? How can I get the second

Splitting into simpler problems

PrOCESS EXPECTATiON


PrOCESS EXPECTATiON

Selecting tools and strategies, Reflecting on what made the problem easy or hard

PrOCESS EXPECTATiON

Changing into a known problem.

PrOCESS EXPECTATiON

Organizing data

PrOCESS EXPECTATiON

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equation from the first? (multiply both sides by 2) Repeat for the third equation (multiply both sides by 3) and the fourth equation (multiply both sides by 4). Emphasize that if two expressions represent the same number, multiplying both by the same number will still result in the same number. Repeat for the following sequence of equations: 3x + 2 = 8 6x + 4 = 16 9x + 6 = 24 12x + 8 = 32

Then discuss which equation is easiest to solve and why (the first one, because the numbers are smaller and so easier to guess and check). ASK: How can you turn the last equation into the first one? (divide all the numbers by 4) Have students change each of these problems into an easier problem with smaller numbers by finding a common factor of each number in the equation, then solve the easier problem: a) 4x + 2 = 14 b) 6x + 9 = 21 c) 30x + 55 = 115 d) 30x + 24 = 144ANSWErS: a) 2x + 1 = 7 so x = 3 b) 2x + 3 = 7 so x = 2 c) 6x + 11 = 23 so x = 2 d) 5x + 4 = 24 so x = 4

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.

2. 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 just as lowercase letters can.) a) 12 ÷ 3 = 4 b) AB = C c) X Z

Y=

3. a) Multiply both sides by 3 to write an equivalent multiplication statement: i) 24 ÷ 3 = 8 ii) 30 ÷ 3 = 10 iii) 15 ÷ 3 = 5 SAMPLE ANSWEr: 24 ÷ 3 × 3 = 8 × 3 so 24 = 8 × 3.

b) Multiply both sides by x to write an equivalent multiplication statement: i) 24 ÷ x = 8 ii) 24 ÷ x = 6 iii) 24 ÷ x = 2

c) Rewrite each fraction as a division statement, then multiply by x to write an equivalent multiplication statement. i) 8/x = 2 ii) 12/x = 3 iii) 20/x = 4 iv) 24/x = 8 SAMPLE ANSWEr: i) 8 ÷ x = 2, so 8 ÷ x × x = 2 × x, so 8 = 2x

d) Solve the equations from c). SAMPLE ANSWEr: i) 8 = 2x, so 8 ÷ 2 = 2x ÷ 2, so 4 = x or x = 4


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GoalsStudents will “undo” many operations to get back to the number they started with, and will “undo” one operation 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

PA8-8 Solving Equations—Preserving EqualityPages 106–107

CUrriCULUM EXPECTATiONSOntario: 7m69, 8m1, 8m6, 8m7, 8m64 WNCP: 8Pr2, [r, C]

VOCAbULAry variable

[7m1, R]

PrOCESS ASSESSMENT

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. 106 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 16Add 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. Each sequence should include

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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), unless both partners are comfortable with decimals. 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:

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.107 (top).

Writing in words what was done to the variable. See Questions 8 and 9.

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 10.

Preserving equality. Give students equations with one variable and 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 = 5 ANSWErS: 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. bonus for Question 12: Solve for x by doing the same thing to both

sides of the equation, then check by substitution. a) 5x = 23 b) 4.2 + x = 9 c) x/5 = 3.1 d) x − 5.34 = 7

[8m6, R] Workbook Question 11

PrOCESS ASSESSMENT

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

back of Card ? = 7


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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

PA8-9 Solving Equations—Two OperationsPages 108–110

CUrriCULUM EXPECTATiONSOntario: 7m69, 8m1, 8m2, 8m5, 8m6, 8m7, 8m59, 8m64 WNCP: 8Pr2, [r, PS, C, CN]

VOCAbULAry variable substitute solve for

NOTE: Students following the Ontario curriculum are not required to solve equations of the form x b c

a+ = .

Thus, Workbook p. 109 Question 2 and Workbook p. 110 Questions 3 c), d), g), h); 4 a), c); and 6 are optional. Students in Ontario can answer Question 7 based on Question 5 instead of based on Question 6.

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

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 = 48

Working backwards

PrOCESS EXPECTATiON

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ANSWEr 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! Solving equations of the form

x b ca

+ = . Undo each operation in

reverse order as before. Now students must subtract b from both sides and then multiply by a, since the left side was obtained by dividing by a and then adding b.

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 ANSWErS: i) 3 bags 5(3) = 15¢ ii) 5 bags 5(5) = 25¢ iii) 10 bags 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?


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8m1, 8m6, [PS, C]

PrOCESS ASSESSMENT

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.)

Challenge students to make up their own word problem based on a real-life situation in which algebra can be used.

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)

Real World

CONNECTiON


PrOCESS ASSESSMENT

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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

PA8-10 Solving Equations—AdvancedPage 111

CUrriCULUM EXPECTATiONSOntario: 8m1, 8m2WNCP: [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 Question 5.

EXTrA PrACTiCE for 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 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

[8m1, PS]

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 and a(x + b) = c. Students will learn to solve a(x + b) = c by turning it into ax + ab = c.


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

MATEriALS

bLM The balance Model (p C-56)

PA8-11 Modelling Equations—AdvancedPages 112–113

CUrriCULUM EXPECTATiONSOntario: 7m69; 8m1, 8m2, 8m6, 8m7, 8m61, 8m64WNCP: 7PR3, 7PR6, 7PR7; 8Pr2, [V, PS, C]

VOCAbULAry distributive property

Using mass to model expressions. 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.

Have students write an expression for the mass of the whole set. (x + 2) Then draw 3 copies of the set, one on top of each other, and repeat the question. (the mass is 3x + 6 because there are now 3 triangles (x’s) and 6 circles) ASK: How many times more is the mass than it was before? (3 times) How do you know? (you just copied the set 3 times) Write on the board: 3(x + 2) = 3x + 6.

Have students draw circles and triangles for each expression. a) x + 3 b) 3x + 1 c) 2x + 5 d) 5x + 3 e) 4x + 4f) x + 1 g) 2(x + 1) h) 3(x + 1) i) 4(x + 1) j) 5(x + 1) ASK: Which two are the same? (e and i) Why does that happen? (because in both cases there are 4 of each shape)

Have students draw a picture to write each expression without brackets. a) 2(x + 3) b) 3(x + 2) c) 4(x + 3) d) 3(x + 5) e) 5(x + 3)f) 3(x + 4) g) 2(x + 5) h) 4(x + 2) i) 2(x + 6) j) 5(x + 2) bonus k) 3(2x + 1) l) 5(3x + 2) m) 4(3x + 2)

Have students draw a picture to write each of the first five expressions without brackets, and then look for a pattern to do part f. a) 2(x + 1) b) 2(x + 2) c) 2(x + 3) d) 2(x + 4) e) 2(x + 5) f) 2(x + a)

SAMPLE ANSWEr: c) so 2(x + 3) = 2x + 6 Then have students use the pattern to write these expressions without brackets, without drawing a picture: g) 2(x + 11) h) 2(x + 7) i) 2(x + 8) j) 2(x + 100) bonus 2(x + 1 000 000 000)

Modelling

PrOCESS EXPECTATiON

Looking for a pattern, Generalizing from examples

PrOCESS EXPECTATiON

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

NOTE: The content of Workbook p. 112 is optional for students in Ontario, but is core material in the WNCP curriculum.


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Changing into a known problem

PrOCESS EXPECTATiON

Substitute x = 5 in the first expression above (part a) and in the corresponding expression without brackets: 2(x + 1) = 2(5 + 1) = 2(6) = 12 and 2x + 2 = 2(5) + 2 = 10 + 2 = 12Have students do the same for x = 7. Remind students that this is the distributive property.

Have students use the distributive property to rewrite each of these expressions: ANSWErS:a) 2(1 + 5) 2(1) + 2(5) = 2 + 10 b) 2(2 + 5) 2(2) + 2(5) = 4 + 10 c) 2(3 + 5) 2(3) + 2(5) = 6 + 10d) 2(4 + 5) 2(4) + 2(5) = 8 + 10 e) 2(5 + 5) 2(5) + 2(5) = 10 + 10 f) 2(x + 5) 2(x) + 2(5) = 2x + 10

Have students use the distributive property to write these expressions without brackets.

a) 3(x + 4) = 3x + b) 2(x + 7) = 2x +

c) 5(x + 4) = 5x + d) 4(x + 6) =

e) 9(x + 8) = f) a(x + b) = bonus 3(2x + 4) =

review working backwards to solve equations of the form ax + b = c. For example, to solve 3x + 4 = 19, note that to get the left-hand side, we multiply by 3 and add 4 and we ended with 19, so to undo those operations, we have to subtract 4 (19 − 4 = 15) and then divide by 3 (15 ÷ 3 = 5) so x = 5:

3x + 4 = 19 3x + 4 − 4 = 19 − 4 3x = 15 3x ÷ 3 = 15 ÷ 3 x = 5Indeed, 3(5) + 4 = 19.

Solve equations of the form a(x + b) = c by writing the expression on the left without brackets. It is now exactly the type of problem that students already know how to do. For example: 5(x + 2) = 20 becomes 5x + 10 = 20. Solve this as before:

5(x + 2) = 20 5x + 10 = 205x + 10 − 10 = 20 − 10 5x = 10 5x ÷ 5 = 10 ÷ 5 x = 2

EXTrA PrACTiCE Solve: a) 2(x + 7) = 20 b) 3(x + 5) = 27 c) 4(x + 5) = 40

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bonus Solve 2(3x + 5) = 52 ANSWEr: 6x + 10 = 52 so 6x = 42 and x = 7.

Using mass to model equations. Have students write the equation for various scale models, as on Workbook p. 113 Question 5.

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 mass of the triangle (the unknown).

a) b)

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)


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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 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.

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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 1. 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

2. 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

g)

A B

3. Which part of Extension 2 shows that 2x + 1 = 7 is solved by x = 3? (part e) Which part shows that 2(x + 3) = 10 is showed by x = 2? (part g)

4. On bLM The balance Model, students use the same method as on Workbook p. 113 to model equations of the form a(x + b) = c.

Modelling equations of the form x ba

= . A rectangle holds x dots. Have

students solve the equation by using the picture.

[8m6, V] Workbook Question 9

PrOCESS ASSESSMENT


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a)

Note that, according to the model, the dots are divided into 3 equal groups (x ÷ 3 or x/3) and the number in each group is 5 (x/3 = 5).

b) so x =

c) so x =

Have students draw the same number of dots in each part to solve the equation.

a) so x =

b) so x =

c) so x =

Have students draw the model themselves to solve the equation.

a) so x = 12 b) so x =

c) so x = d) so x =

e) so x = f) so x =

Modelling equations of the form x b ca

+ = . A rectangle holds x dots.

Show students how to use this model to solve 3x

+ 5 = 7.

so 53x

= so x =

26x

=

38x

=

54x

=

65x

=

37x

=

34x

= 52x

=

25x

= 23x

=

72x

= 54x

=

Modelling

PrOCESS EXPECTATiON

Modelling

PrOCESS EXPECTATiON

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Solution: Step 1. Draw a rectangle divided into 3 parts to model x/3:

Step 2. Add 5 circles to model x/3 + 5:

Step 3. Add more circles, all in one section, to make a total of 7, to model x/3 + 5 = 7:

+ 5

Step 4. Use the picture to find x/3. In this case x/3 = 2.

Step 5. Add the same number of circles to each remaining part of the rectangle to find x.

So x = 6.

a)

b)

Draw a model to solve the equation.

a) 53x

= b) 4 72x

+ = c) 2 55x

+ =

d) 6 114x

+ =e) 4 7

3x

+ =f) 5 8

8x

+ =

3x

3 84x

+ =

34x

+

+ =4 92x

42x

+

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

7m6, [V]

PrOCESS ASSESSMENT

So 2x

=

So x =

So 4x

=

So x =


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

a)

A B

Cb)

A B

C

bonus A square weighs x kg, a triangle weighs y kg and a circle weighs 1 kg. Write an equation for each of the three scales in parts a and b.

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.

8m1, [PS]

PrOCESS ASSESSMENT

8m1, 8m6, 8m7, [PS, V, C]

PrOCESS ASSESSMENT

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GoalsStudents will identify false statements of the form “all [of these] are [like this]” by using counter-examples. Students will also be able to write the reverse (mathematically, the converse) of a statement.


Can identify vowels and consonants Can identify even and odd numbers Can identify prime numbers Can identify perfect squares

MATEriALS

bLM Who is breaking the Law? (p C-57)bLM Sudoku (pp C-58–C-63)bLM Always, Sometimes, Never (p C-64)bLM Define a Number (pp C-65–C-66)

PA8-12 Counter-ExamplesPages 114-116

CUrriCULUM EXPECTATiONSOntario: 8m2, 8m7WNCP: [r, C]

VOCAbULAry counter-example reverse true false

False statements in context. Make a statement about the class that is obviously false. EXAMPLE: All girls (or boys) in the class wear glasses. When students name a girl (or boy) that proves the statement wrong (and allow more than one counter-example) have them explain their choice. Tell students that an example that proves a statement false is called a counter-example. Then, if using the statement “All girls wear glasses,” name a boy who does not wear glasses and ask if he is a counter-example. SAy: He doesn’t wear glasses, so why isn’t he a counter-example? Emphasize that the statement is about girls, and he’s not a girl, so the statement doesn’t say anything at all about him. If the statement was “All boys wear glasses,” then he would be a counter-example. Now name a girl who does wear glasses. ASK: Is she a counter-example? Why not? (because she does wear glasses, so she is an example) Emphasize that though there might be many girls in the class who wear glasses, the statement says that all girls wear glasses, so we just need to name one girl who doesn’t wear glasses to prove the statement false.

Repeat for various statements where the students themselves are counter-examples. Students can raise their hands if they themselves are a counter-example. EXAMPLES: All girls are wearing earrings. All boys are wearing running shoes. All girls are wearing skirts. All girls like soccer. All boys like math. All students are girls. All students are boys. All people who like soccer also like baseball. All people who like pizza also like spinach.

ACTiViTy

Make a false statement (you could use any statement given above). Have everyone who is a counter-example stand up. Continue with


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other statements until everyone is standing up. Then have students individually make up and write a statement that applies to some but not all students, and challenge the class to get everyone standing up in exactly 5 turns. Repeat the game until everyone has had a chance to say their sentence.

bLM Who is breaking the Law?

False statements about shapes. See Workbook Question 1. Write the following statement and picture on the board:

All circles are shaded.

ASK: What is this statement about? (circles) Underline all the circles. ASK: Are all circles shaded? (no) Have a student circle the counter-example. Erase the underlining and the circling, and repeat with new statements. EXAMPLES: • All squares are big. • All squares are shaded. • All big squares are shaded. • All small circles are shaded. • All shaded shapes are circles. • All white shapes are small. • All shaded shapes are big. • All white shapes are squares. • All big shapes are squares. • All big shapes are shaded. • All small shapes are white. • All small shapes are squares.

False statements about words. Write the following sentence on the board: All words start with the letter b. Ask in turn if each of these is a counter-example to the statement and have students explain why or why not: • bat (no, because it does start with b) • cat (yes, it is a word that does not start with b) • boat (no, because it does start with b) • bxcv (no, because it does start with b OR no, it’s not a word) • xcvb (no, because it’s not a word, and the statement only talks about words, so something that’s not a word cannot be a counter-example)

ASK: To which of these statements is “Bob” a counter-example:• All names have three letters. • All names have 4 letters. • All boys’ names start with D. • All names are boys’ names. • All names read the same backwards as forwards. ANSWEr: All names have 4 letters; All boys’ names start with D.

[8m7, 8m2, R, C]

PrOCESS ASSESSMENT

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Have students find a counter-example to the three statements for which “Bob” is not a counter-example. Challenge students to find one example that works as a counter-example to all three statements at the same time. SAMPLE ANSWEr: Sara.

bonus Explain why there cannot be a counter-example to all five statements. ANSWEr: To be a counter-example to the third statement, the name would have to be a boys’ name, but to be a counter-example to the fourth statement, the name would have to not be a boys’ name.

Deciding whether a statement is true or false by checking examples. Create and display copies of the following cards (large enough for students to see across the room).

Point out that each card has a number and a letter. Write on the board: All cards with an odd number have a vowel. Ask students to look for and identify any cards that are counter-examples. Encourage students to reread the statement if necessary; emphasize the value of rereading statements carefully. After students have had a chance to individually write down the card(s) they think are counter-examples, ASK: What type of cards is the statement saying something about? (cards with odd numbers) Can a card with an even number be a counter-example? (no) Why not? (the sentence is only saying something about cards with odd numbers) SAy: I have a card that is a counter-example. What can you tell me about that card? What type of number is on it? (odd) What type of letter is on it? (a consonant)

Determine whether each statement in Workbook Question 5 is true or false for this set of cards. A statement will be true if it is true for every card (you need to check all examples and not find a counter-example) and false if you find at least one counter-example. Emphasize how much more work it can be to prove a statement true than false because you need to check all examples. For these cards, sample answers are: a) False: A,7 b) False: A, 6 c) False: A, 6 d) False: e, 1 e) True f) False: j, 5 g) False: A, 6 h) True i) False: A, 7 j) True

bonus All cards with small letters have prime numbers. (False: e, 1) All cards with perfect squares have consonants. (False: e,1 or U, 9)

Discuss any relationships between answers. Are there any answers you could have predicted based on previous answers. For example, can you tell that statement j is true from looking at any other true statement? (yes; statement j is true because statement h is true) If you did statement c first, which other statement could you tell is false without checking any examples? (statement b)

A b 1 e j7 6 5 1 5

K U H W A11 9 4 8 6


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The reverse of a statement means to “go the other way.” Write these two false statements on the board: All girls are teenagers. All teenagers are girls. Discuss how the two statements are related. How are they the same and different? Do they mean the same thing? Note that one of them is talking about girls and saying they are all teenagers and the other is talking about teenagers and saying they are all girls. You can think of “all teenagers are girls” as the reverse of “all girls are teenagers.”

Write on the board: All carrots are vegetables. ASK: Is this a true statement? (yes) Now try going the other way—what is the reverse of this statement? (All vegetables are carrots.) Is the reverse true? (no) What is a counter-example? (broccoli, for example, because it is a vegetable that is not a carrot)

Have students write the reverse of each true statement below and decide if the reverse is true too. Only assign the first two statements if students are familiar with all terms involved. Use blanks and underlining at first, to scaffold students’ answers, but gradually remove these aids.

a) All triangles are polygons. Reverse: All are .Counter-example: . b) All squares are shapes with 4 equal sides.Reverse: All are .Counter-example: . c) All numbers with ones digit 0 are even numbers.d) All people are mammals. e) All girls are people. f) All dogs are animals with tails. g) All humans are animals with legs. bonus All spiders are organisms with 8 legs.

Challenge students to make up a) a true statement whose reverse is false, b) a false statement whose reverse is true, c) a true statement whose reverse is true, d) a false statement whose reverse is false.

To help students practise making logical deductions, go through the book Anno’s Hat Tricks by Akihiro Nozaki and Mitsumasa Anno with the class. Work through the book over several days, a few pages at a time. The book is suitable for grades 5–12 and students will let you know when the logic becomes too tough.

EXTrA PrACTiCEbLM Sudoku (Warm-Up, Introduction, Another Strategy, Advanced) introduces students to these popular puzzles, which require substantial logical thinking to solve. bLM Always, Sometimes, Never and bLM Define a Number provide practice with logical thinking. bLM Define a Number should be photocopied single-sided so that students don’t need to continually turn over the sheet.

Literature

CONNECTiON

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Extensions1. Alana thinks that if n is a prime number, then 6n − 1 will also be prime.

Find a counter-example to prove her wrong. Hint: Check the prime numbers in order: n = 2, n = 3, n = 5, and so on.ANSWEr: the first counter-example is n = 11, since 6(2) − 1 = 11 is prime, 6(3) − 1 = 17 is prime, 6(5) − 1 = 29 is prime, 6(7) − 1 = 41 is prime, but 6(11) − 1 = 65 = 5 × 13 is not prime.

2. Take the cards you made above and fold them so that only one side is visible to students. Take these four cards and make visible the part shown here as shaded:

ASK: Which card or cards would you need to turn over to decide whether each statement is true: A All cards with odd numbers have a vowel. B All cards with vowels have an odd number. C All cards with capital letters have an odd number. D All cards with odd numbers have a capital letter. ANSWErS: A—Check j, because if that card has an odd number, the statement is false. Also check 7, because if that card has a consonant, the statement is false. B—Check e and 8. C—Check 8. D—Check 7, e, and j.

What do you notice about your answers? PrOMPT: How are statements A and B related? (they are the reverse of each other) How are statements C and D related? (they are the reverse of each other) You had to check j and 7 for A—which cards did you have to check for its reverse? (the other two cards, e and 8) You had to check 8 for statement C—which cards did you have to check for its reverse? (the other three cards, 7, e, and j)

Have students make a conjecture about reverse statements in general and then check it. ASK: To check the reverse of a statement, which cards do you have to check? (Exactly the cards you didn’t have to check for the original statement.) Students could make up their own statements and write the reverse to check the conjecture, or students could check with a different set of four cards.

j 5 A 7 e 1W 8

Organizing data

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GoalsStudents will solve equations of the form a(x + b) = c directly by working backwards.


Can solve equations of the form ax + b = c by working backwardsCan solve equations of the form a(x + b) = c by changing it into ax + ab = cKnows that solving the same equation two ways should get the same answer Can apply the order of operations

PA8-13 Exploring Preservation of EqualityPages 117–188

CUrriCULUM EXPECTATiONSOntario: 8m2, 8m5, 8m7, optionalWNCP: 8Pr2, [CN, r, C]

VOCAbULAry working backwards counter-example

Writing phrases from expressions. Tell students that you are starting with 10. Write several expressions and have students tell you what you are doing to the 10. EXAMPLES:a) 10 + 3 (adding 3) b) 3 × 10 (multiplying by 3) c) 4(10) (multiplying by 4) d) 10 ÷ 5 (dividing by 5) e) 10 − 7 (subtracting 7) f) 15 − 10 (subtracting from 15)Replace 10 with a variable and repeat the line of questioning (What am I doing to x?). EXAMPLES: a) 7x (multiplying by 7) b) x + 7 c) x − 7 d) 7 − x e) x ÷ 7 f) 7 + x

Write expressions from phrases. Tell students to write the expression for x:a) add 4 to x b) subtract 4 from x c) subtract x from 4 d) divide x by 5 e) divide 5 by x f) multiply x by 3

Write phrases from expressions combining more than one operation. See Workbook Question 1. Emphasize that operations in brackets are performed first, then multiplication and division from left to right, then addition and subtraction from left to right. For example, in 3(x + 5) we are being told to add 5 to x first, and then multiply by 3, whereas in 3x + 5, we are being told to multiply by 3 first and then add 5. ASK: What operation do you do first in 5 + 3x? (still multiply by 3, because multiplication always comes before addition. Emphasize that 5 + 3x is really just 5 + x + x + x. Also point out that when there is a fraction sign, everything in the numerator can be treated as though it is inside brackets and everything in the denominator can be treated as though it is inside brackets. So, for example:

5 ( 5) (3 4)3 4x x+

= + ÷ ××

Thus, in Workbook Question 1 f), you add 1 first then divide by 2, whereas in Question 1 e), you divide by 2, then add 1.

Write expressions from phrases combining more than one operation. See Workbook Question 2.

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Expressions that mean the same thing. ASK: Which of these sequences of operations is the same as not doing anything to x?a) Multiply x by 3. Then subtract 3.b) Add 3 to x, then subtract 3. c) Subtract x from 3, then add 3. d) Add 3 to x, then divide by 3. e) Multiply x by 3, then divide by 3. ANSWEr: b), e). Have students write the expressions represented by each sequence of operations above. ANSWErS: a) 3x − 3 b) x + 3 − 3 c) 3 − x + 3 d) (x + 3) ÷ 3 e) 3x ÷ 3

Emphasize that because b) and e) are the same as not doing anything to x, we can write this as x + 3 − 3 = x and 3x ÷ 3 = x. These are true for all x, because no matter what you start with, adding 3 and subtracting 3 gets you back where you started; same with multiplying by 3, then dividing by 3.

Checking guesses by substitution. ASK: Which of these sequences of operations is the same as adding 2 to x?a) Add 2. Multiply by 3. Divide by 3. b) Multiply by 3. Add 2. Divide by 3. Have students write expressions for both. a) 3(x + 2) ÷ 3 b) (3x + 2) ÷ 3

Have students predict which of these is true for all x: 3(x + 2) ÷ 3 = x + 2 or (3x + 2) ÷ 3 = x + 2. Then write the following table on the board for students to copy and complete in their notebooks:

x 3(x + 2) ÷ 3 3(x + 2) ÷ 3 x + 2

0

1

2

3

bonus Check for x = 1/2. Have students find a counter-example for the equation that is not true. Emphasize that even though you are doing the same three operations to x each time, the order you do the operations in matters.

Tell students that substitution is a good way to check their prediction, but they must check more than just one number! For example, have students check that, for x = 1, 2x + 5 = 5x + 2. ASK: Are these expressions equal for all x? (no) Have students find a counter-example. (In this case, any other value for x will be a counter-example).

EXTrA PrACTiCE for Question 6: Investigate: Which two expressions are equal for all x:2(x + 3) 2(x − 3) 2x − 3 2x + 32x − 6 6x − 2 3x − 6

a) Substitute x = 3 into each expression above.

Making and investigating conjectures

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Connecting

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b) When x = 3, which expressions are equal? (2 (x − 3) = 2x − 6 and 2x − 3 = 3x − 6)c) Substitute x = 5 into each expression above. d) When x = 5, which expressions are equal? (2(x − 3) = 2x − 6) e) Which two expressions are equal for x = 3 and x = 5? (2 (x − 3) and 2x − 6)f) Explain why the two expressions from part e are the same for all x? SAMPLE ANSWEr: 2(x − 3) = x − 3 + x − 3 = x + x − 3 − 3 = 2x − 6.

2. Circle the two expressions that are equal for all x.

a) 3x + 5 3 + 5x 5 + 3x 8x 5 − 3x

b) (3)3x 3

3x + 3(x − 3) x 3x − 3

c) x − 4 × 4 x ÷ 4 × 4 x

4 4x+

x + 4 ÷ 4

d) 2(x + 3) − 2 2(x + 3) ÷ 2 x + 3 2x + 6 x + 6

Choose one part and check your answers to that part by substituting the same value for x into each expression.

Solving equations of the form a(x + b) = c by working backwards. Write on the board: 3(x + 2) = 24 ASK: What two operations are you doing? (multiply by 3 and add 2) Which operation do you do first? (add 2) How do you know? (it is in brackets) Tell students that you add 2 then multiply by 3 and end up with 24. Show this on the board, as in the margin. ASK: How can I undo those operations to find what I started with? Remind students that the operations need to be undone in reverse order, so if you did multiplying by 3 last, you have to undo it first.

Write the equation again: 3(x + 2) = 24Undo multiplying by 3 by dividing by 3. 3(x + 2) ÷ 3 = 24 ÷ 3Write the new equation x + 2 = 8Undo adding 2 by subtracting 2. x + 2 − 2 = 8 − 2 Write the new equation. x = 6

Tell students that you solved for x! Encourage students to check if this is correct by substituting x = 6 into the original expression. Do students get 24? ANSWEr: 3(6 + 2) = 3(8) = 24. Yes, it works.

ExtensionSolve for x.

a) 3 72

x +=

b) 2( 3) 45x +

= c) 6( 1) 3

5 1x +

=−

ANSWErS: a) x = 11 b) x = 7 c) x = 1

[CN]

PrOCESS ASSESSMENT

Working backwards

PrOCESS EXPECTATiON

8m2, [R] Workbook Question 7

PrOCESS ASSESSMENT

Start with x. xAdd 2. x + 2Multiply by 3. 3(x + 2) Get 24. 3(x + 2) = 24

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GoalsStudents will identify and correct an error in a given incorrect solution of a linear equation.


Can solve equations of the form:

ax = b, , 0x b aa

= ≠ , ax + b = c, + = ≠, 0x b c aa

, a(x + b) = c. Can substitute values for the variable in equations Can translate operations into phrases

CUrriCULUM EXPECTATiONSOntario: 8m3, 8m7, optionalWNCP: 8Pr2, [r, C]

VOCAbULAry none

A shortcut for solving equations. Tell students that sometimes people don’t write out every step when they solve equations. For example, the solution: 3x + 12 = 33 3(x + 4) = 33 Rewrite the left side. 3(x + 4) ÷ 3 = 33 ÷ 3 Divide both sides by 3. x + 4 = 11 Rewrite both sides. x + 4 − 4 = 11 − 4 Subtract 4 from both sides. x = 7 Rewrite both sides.

Can be shortened to: 3x + 12 = 33 3(x + 4) = 33 Rewrite the left side. x + 4 = 11 Divide both sides by 3. x = 7 Subtract 4 from both sides.

Instead of rewriting both sides, we can do that step mentally.

ASK: How would you rewrite this solution: 3x + 12 = 333x + 12 − 12 = 33 − 12 Subtract 12 from both sides 3x = 21 Rewrite both sides 3x ÷ 3 = 21 ÷ 3 Divide both sides by 3. x = 7 Rewrite both sides

ANSWEr:3x + 12 = 33 3x = 21 Subtract 12 from both sides x = 7 Divide both sides by 3.

Emphasize that if students can subtract 12 from both sides and rewrite them mentally, then students can skip writing that step. However, they might be more likely to make mistakes when they don’t write down a step, so they have to be more careful.

PA8-14 Correcting MistakesPage 119


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Have students write the missing steps in each of these solutions: a) 3x + 5 = 23 b) 2(x + 5) = 24 c) 4x − 3 = 25 d) 5x − 2 = 18 3x = 18 x + 5 = 12 4x = 28 5x = 20 x = 6 x = 7 x = 7 x = 4SAMPLE ANSWEr:a) 3x + 5 = 23 3x + 5 − 5 = 23 − 5 Subtract 5 from both sides 3x = 18 Rewrite both sides. 3x ÷ 3 = 18 ÷ 3 Divide both sides by 3. x = 6 Rewrite both sides.

Tell students that when solving problems, they will never be expected to skip steps—in fact, they should always write out all steps so that you can check their understanding. They should, however, know how to read solutions that have skipped steps.

identifying mistakes by using substitution. Tell students that you saw two different solutions to the same equation. Write these on the board. Solution 1 Solution 2 2x + 6 = 18 2x + 6 = 182x + 6 − 6 = 18 − 6 2(x + 3) = 18 2x = 12 2(x + 3) ÷ 2 = 18 ÷ 2 2x ÷ 2 = 12 ÷ 2 x + 3 = 16 x = 6 x + 3 − 3 = 16 − 3 x = 13

ASK: These students are both trying to find the x that satisfies 2x + 6 = 18, so they should get the same answer. Did they? (no) How can we decide who is right? (we could check each step of both) Is there a way to know if the answer is right before checking so that we can know ahead of time that we are looking for a mistake? (yes, substitute the answer into the original expression and check if you get 18)

Then have students substitute both of the answers into the expression and determine which answer is correct. (2(6) + 6 = 12 + 6 = 18, but 2(13) + 6 = 26 + 6 = 32, so 6 is correct and 13 is incorrect) Have students individually describe each step of Solution 1.

Then go through Solution 2 together to find the mistake. The first step is correct, since it is true that 2x + 6 = 2(x + 3) for all x. Have students show why this is true individually. Students could show this in different ways, for example by drawing a picture with triangles and circles, or by using algebra: 2(x + 3) = x + 3 + x + 3 = x + x + 3 + 3 = 2x + 6

The second step is correct because they just divided both sides by 2, and since both sides were equal before dividing by 2, they will be equal after dividing by 2. The left-hand side of the fourth line is also correct; it is the right-hand side that is incorrect, since 18 ÷ 2 = 9, not 16 (maybe the person misread their “÷” as “−” by mistake). Now show students these solutions to the same problem.

Reflecting on the reasonableness of an answer

PrOCESS EXPECTATiON

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Solution 3 Solution 4 2x + 6 = 18 2x + 6 = 18 2(x + 3) = 18 2(x + 6) = 182(x + 3) ÷ 2 = 18 ÷ 2 2(x + 6) ÷ 2 = 18 ÷ 2 x + 3 = 9 x + 6 = 9 x + 3 − 3 = 9 − 3 x + 6 − 6 = 9 − 6 x = 6 x = 3

Solution 5 Solution 6 2x + 6 = 18 2x + 6 = 182x + 6 − 6 = 18 + 6 2(x + 3) = 18 2x = 24 x + 3 = 18 2x ÷ 2 = 24 ÷ 2 x + 3 − 3 = 18 − 3 x = 12 x = 15

ASK: Which solution is correct? (Solution 3) How do you know? (it gets the right answer) Have students go through each incorrect solution line by line to find the mistake.

ANSWErS: Solution 4: wrote 2x + 6 as 2(x + 6)—this is not true since 2(x + 6) = 2x + 12, not 2x + 6.

Solution 5: subtracted 6 from the left side and added 6 to the right side, so the two sides are no longer equal.

Solution 6: if 2 times (x + 3) is 18, then x + 3 isn’t equal to 18; someone forgot to divide 18 by 2.

Have students identify and correct any mistakes below.

a) 5(x + 2) = 3 b) 5(x + 2) = 30 5x + 2 = 30 5(x + 2) ÷ 5 = 30 5x + 2 − 2 = 30 − 2 x + 2 = 30 5x = 28 x + 2 − 2 = 30 − 2 5x ÷ 5 = 28 ÷ 5 x = 28 x = 5.6

c) 2x + 8 = 22 d) 2x + 8 = 22 2x + 8 − 8 = 22 − 8 2(x + 4) = 22 2x = 14 x + 4 = 11 2x ÷ 2 = 14 + 2 x + 4 − 4 = 11 − 4 x = 16 x = 7

Solving problems, checking the answer using substitution, and identifying and correcting their own mistakes. Have students use the distributive law (see Method 2 from Workbook p. 119 Question 1) to solve these equations and say what they do at each step. Students should check their answer by substituting it into the original expression. If incorrect, have students find their own mistake! a) 3x + 9 = 15 b) 4x + 20 = 36 c) 2x + 8 = 20 d) 3x + 12 = 27 e) 2x + 4 = 11 f) 3x − 6 = 14 g) 5x + 10 = 27 h) 5x − 15 = 21 ANSWErS: a) x = 2 b) x = 4 c) x = 6 d) x = 5 e) x = 3.5 f) x = 6 2/3 g) x = 3.4 h) x = 7.2

[R], 8m3 Workbook Question 2

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


Can solve equations

Word ProblemsPA8-15

Pages 120-122

CUrriCULUM EXPECTATiONSOntario: 7m66, 7m69; 8m1, 8m2, 8m3, 8m4, 8m5, 8m7, 8m61, 8m64 WNCP: 7PR7; 8Pr2, [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. 120 (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)

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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.

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? d) One side of a rectangle is 20 cm. The other side is 5 times shorter. how long is the other side?

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?

Connecting

PrOCESS EXPECTATiON


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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?

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 5 a) on Workbook p. 121, 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 problems using algebra: Jane has $3 in dimes and quarters. She has 21 coins in all. How many of each coin does she have?

Reflecting on other ways to solve a problem

PrOCESS EXPECTATiON

[8m1, 8m2, PS, R]

PrOCESS ASSESSMENT

Selecting tools and strategies

PrOCESS EXPECTATiON

Problem-Solving

PrOCESS EXPECTATiON

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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. (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)

4. Solve the problem in Question 3 above using tables. Did you get the same answer both ways? ANSWEr:

Quarters Dimes needed to make 300¢

total # of coins

0 30 30

1 ---- ----

2 25 27

3 ---- ----

4 20 24

5 ---- ----

6 15 21 Extensions1. Write 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 Area = 16 m2

Perimeter = 48 cm Perimeter = 72 cmx x

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

PrOCESS ASSESSMENT

Web link for more word problems

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2. Use bLM Solving Equations with Variables on both Sides (pp C-67–C-68) to teach students to solve equations with the variable on both sides. Here are some problems that require solving equations with a variable on both sides of the equation.

a) Solve for x:

i) 5 26 6x x

× = + ii) 5 28 8x x

× = + iii) 15 107 7x x

× = +

ANSWErS: i) 5 12

6 6x x+

=so 5x = 12 + x, or 4x = 12, so x = 3

ii) x = 4 iii) x = 5

b) Problem: Sara has twice as many brothers as sisters. Her brother Tom has the same number of brothers as sisters. How many children are in the family?

Let x be the number of sisters that Sara has.

i) Write an expression for • the number of girls in the family (x + 1)• the number of boys in the family (2x)• the number of children in the family (3x + 1)• the number of brothers that Tom has (2x − 1, since Tom isn’t his own brother) • the number of sisters that Tom has (x + 1, since the number of sisters Tom has is just the number of girls in the family)

ii) Write an equation that shows that Tom has the same number of brothers as sisters. (2x − 1 = x + 1)

iii) Solve the equation you found in b) for x. (2x − x − 1 = x − x + 1, so x − 1 = 1 and x = 2)

iv) Substitute your answer to iii) into your expression for the number of children in the family. How many children are in the family? (3(2) + 1 = 7; Sara has 2 sisters and 4 brothers, Tom has 3 sisters and 3 brothers)

c) Problem: Nomi is twice as old as Bilal. In two years, Ahmed will be 3 times as old as Bilal is now. Nomi is 4 years younger than Ahmed. Find their ages.

Let x be Bilal’s age.

i) Write an expression for: • Nomi’s age (2x) • Ahmed’s age in two years (3x) • Ahmed’s age right now (3x − 2)

ii) Nomi is 4 years younger than Ahmed. Use this and your answers to a) to write an equation. = (2x = 3x − 2 − 4)

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iii) Solve your equation for x.

( 2x = 3x − 6 3x − 6 = 2x 3x − 2x − 6 = 2x − 2x x − 6 = 0, so x = 6)

iv) Substitute your answer for x into the expressions for Nomi’s age and Ahmed’s age. How old is each child? (Since x = 6, Bilal is 6; Nomi’s age is 2(6) = 12; Ahmed’s age is 3(6) − 2 = 16.)

v) Check your answer. (Nomi is indeed twice as old as Bilal and 4 years younger than Ahmed. Also, in two years, Ahmed will be 18 which is 3 times as old as Bilal is now.)

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Unit 2 Patterns and Algebra - JUMP Math :: Home for AP...Unit 2 Patterns and Algebra In this unit, students extend and describe patterns, use T-tables to solve problems, and solve