Adapted from the Equivalent Expressions Section of Say It with Symbols: Making Sense Of

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Accelerated Mathematics IUnit 51st Edition

Adapted from the “Equivalent Expressions” section of Say It With Symbols: Making Sense of Symbols in the Connected Mathematics 2 series.

I’ve Got Your Number Learning Task

Equivalent algebraic expressions, also called algebraic identities, give us a way to express results with numbers that always work a certain way. In this task you will explore several “number tricks” that work because of basic algebra rules. It is recommended that you do this task with a partner.

Think of a number and call this number x.

Now think of two other numbers, one that is 2 more than your original number and a second that is 3 more than your original number. No matter what your choice of original number, these two additional numbers are represented by x + 2 and x + 3.

Multiply your x + 2 and x + 3 and record it as Answer 1 ______.

Find the square of your original number, x2 ______,

five times your original number, 5x ______,

and the sum of these two numbers plus six, x2 + 5 x + 6, and record it as Answer 2 ______.

Compare Answers 1 and 2. Are they the same number? They should be. If they are not, look for a mistake in your calculations.

Question: How did the writer of this task know that your Answers 1 and 2 should be the same even though the writer had no way of knowing what number you would choose for x?

Answer: Algebra proves it has to be this way.

To get Answer 1, we multiply the number x + 2 by the number x + 3: (x + 2)(x + 3)

We can use the distributive property several times to write a different but equivalent expression.

First treat (x + 2) as a single number but think of x + 3 as the sum of the numbers x and 3, and apply the distributive property to obtain: / (x + 2)· x + (x + 2) ·3
Now, change your point of view and think of x + 2 as the sum of the numbers x and 2, and apply the distributive property to each of the expressions containing x + 2 as a factor to obtain: / x · x + 2· x + x ·3 + 2·3
Using our agreements about algebra notation, rewrite as: / x 2 + 2 x + 3 x + 6
Add the like terms 2x and 3x by using the distributive property in the other direction: / 2 x + 3 x = (2 + 3) x = 5 x
The final expression equivalent to (x + 2)(x + 3) is: / x 2 + 5 x + 6

The last expression is Answer 2, so we have shown that (x + 2)( x + 3) = x 2 + 5 x + 6 no matter what number you choose for x. Notice that 5 is the sum of 2 and 3 and 6 is the product of 2 and 3.

The above calculations are just an example of the following equivalence of algebraic expressions.

(Pattern 1)

For the remainder of this task, we’ll refer to the above equivalence as Pattern 1. Note that we used a = 2 and b = 3 is our example, but your task right now is to show, geometrically, that a could represent any real numberand so could b.

1. (a) Each of the diagrams below illustrates Pattern 1. Match each diagram with one of the following cases for a and b.

Georgia Department of Education

Kathy Cox, State Superintendent of Schools

July 21, 2008

Unit 5: Page1 of 11

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Accelerated Mathematics IUnit 51st Edition

Case 1: a positive, b positive Figure Case 2: a positive, b negative Figure

Case 3: a negative, b positive Figure

Case 4: a negative,b negative Figure

Georgia Department of Education

Kathy Cox, State Superintendent of Schools

July 21, 2008

Unit 5: Page1 of 11

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Accelerated Mathematics IUnit 51st Edition

(b) Thinking of the case it represents, for each diagram above, find the rectangle whose area is and use a pencil to put diagonal stripes on this rectangle. Then explain how the diagram illustrates the pattern. Note that when a is a negative number, , and when a is a positive number, then ; and similarly for b. Tell two of your explanations to another student and let that student explain the other two to you.

2. Pattern 1 can be used to give an alternate way to multiply two digit numbers that have the same digit in the ten’s place.

For example, (31)(37) can be thought of as (30 + 1)(30 + 7) so we let x = 30, a = 1, and b = 7. Since 302 = 900, 1 + 7 = 8 and 1·7 = 7,

Use Pattern 1 to calculate each of the following products.

(a) (52)(57) =

(b) (16)(13) =

(d) (72)(75) =

3. Look at your result for 2(c). You can get the answer another way by doing the following. Take 4, the common ten’s digit, and multiply it by the next integer, 5, to get 20 as the number of hundreds in the answer. Then multiply the units digits 4 x 6 to get 24 for the last two digits.

4 8

x4 2

2 0 1 6

4 · 5 = 20 8 · 2 = 16

Use this scheme to calculate the products below and verify the answers using Pattern 1.

(a) (34)(36) =(b) (63)(67) =

(e) In each of the products immediately above, look at the pairs of units digits: 8 and 2 in the example, 4 and 6 for part (a), 3 and 7 in part (b), 1 and 9 in part (c), and 5 and 5 in part (d). What sum do each of the pairs have?

(f) Now let’s use Pattern 1 to see why this scheme works for these products with the property you noted in part (e). To represent two-digit numbers with the same ten’s digit, start by using n to represent the ten’s digit. So, n is 4 for the example and 3 for part (a). What is n for parts (b), (c), and (d)?

(g) Next, represent the first two-digit number as 10 n + a and the second one as 10 n + b.
In part (a):
(32)(38) = (30 + 2)(30 + 8) = (10n + a)( 10n + b) for n = 3, a = 2, and b = 8.

List n , a , and b for part (b).

List n , a , and b f for part (c).

List n , a , and b f for part (d)

(h) Use Pattern 1 to multiply (10 n + a)( 10 n + b) where we have numbers like the example and parts (a), (b), (c),and (d). We are assuming that each of n, a, and b is a single digit number. What are we assuming about the sum a + b?

Write your result in the form 100 k + ab where k is an expression containing the variable n and numbers.

Write the expression for k .

Explain why k represents the product of two consecutive integers.

(j) Create three other multiplication exercises that can be done with this scheme and exchange your exercises with a classmate. When you both are done, check each other’s answers.

(i)(ii) (iii)

(k) Explain the quick way to calculate each of the following:

(15)(15) .

(45)(45)

(85)(85)

Why do each of these products fit the pattern of this question?

(l) Summarize in your own words what you’ve learned in the parts of question 3.

4. The parts of question 3 explored one special case of Pattern 1. Now, let’s consider another special case to see what happens when a and b in Pattern 1 are the same number. Start by considering the square below created by adding 4 to the length of each side of a square with side length x.

(a) What is the area of the square with side length x?

(b) The square with side length x + 4 has greater area. Use Pattern 1 to calculate its total area. When you use Pattern 1, what are a, b, a + b?

(c) How much greater is the area of the square with side length x + 4? Use the figure to show this additional area. Where is the square with area 16 square units?

(d) How would your answers to parts (b) and (c) change if the larger square had been created to have side length x + y, that is, if both a and b are both the same number y?

(e) At the right, draw a figure to illustrate the area of a square with side length x + y assuming that x and y are positive numbers. Use your figure to explain the pattern below.

(Pattern 2 - the Square of a Sum)

5. This pattern gives a rule for squaring a sum. Use it to calculate each of the following by making convenient choices for x and y.

(a) 3022 =

(b) 542 =

(d) 2.12=

(e) Look back at question 3, (k). Why will the rule for squaring a sum also work on those exercises? Can the method of Question 3 be used for the square of any sum?

6. We can extend the ideas of questions 4 and 5 to cubes.

(a) What is the volume of a cube with side length 4?

(b) What is the volume of a cube with side length x?

(c) Now determine the volume of a cube with side length x + 4. First, use the rule for squaring a sum to find the area of the base of the cube.

Now use the distributive property several times to multiply the area of the base by the height, x + 4. Simplify your answer.

(d) Repeat parts (b) and (c) for a cube with side length x + y. Write your result as a rule for the cube of a sum.

area of the base of the cube

area of base multiplied by the height, x + y:

– (Pattern 3 – the Cube of a Sum)

(e) Making convenient choices for x and y, use Pattern 3 to find the following cubes.

113 =

233 =

1013 =

Use the rule for cubing a sum to cube 2 = 1 + 1. Do you get the same number as (2)(2)(2)?

(f) Use the cube of the sum pattern to simplify the following expressions.

(t + 5)3 =

(w + 2) 3 =

7. (a) Let y represent any positive number. Go back to Pattern 1 and substitute – y for a and for b to get the following rule for squaring a difference.

(Pattern 4 - The Square of a Difference)

(b) In the diagram below find the square with side x, the square with side y, and two different rectangles with area xy. Now, use the diagram to give a geometric explanation of the rule for the Square of a Difference.

(c) By making a convenient choices for x and y, use the Square of a Difference pattern to find the following squares. Note that 99 = 100 – 1, 38 = 40 – 2, and 17 = 20 – 3.

992

382

172

8. (a) Find a rule for the cube of a difference.

(Pattern 5 – The Cube of a Difference)

(b) Check your rule for the Cube of a Difference by using it to calculate:

the cube of 1 using 1 = 2 – 1 and the cube of 2 using 2 = 5 – 3.

9. Now let’s consider what happens in Pattern 1 if a and b are opposite real numbers. Use Pattern 1 to calculate each of the following. Substitute other variables for x as necessary.

(a) Calculate (x + 8)( x – 8). Remember that x – 8 can also be expressed as x + (– 8).

(x + 8)( x – 8)

(b) Calculate (x – 6)( x + 6) . Remember that x – 6 can also be expressed as x + (– 6).

(x – 6)( x + 6)

(z + 12)( z – 12)

(d) Calculate (w + 3)( w – 3) . Remember that w – 3 can also be expressed as w + (– 3).

(w + 3)( w – 3)

(e) Calculate (t – 7)( t + 7) . Remember that t – 7 can also be expressed as t + (– 7).

(t – 7)( t + 7)

(f) Substitute y for a and – y for b in Pattern 1 to find an pattern for the product (x + y)(x – y).

(x + y)(x – y)(Pattern 6)

10. Make appropriate choices for x and y to use Pattern 6 to calculate each of the following.

(a) (101)(99)

(b) (22)(18)

Extension

(d) (63)(57)

(e) (6.3)(5.7)

(f)

11.

(a) In Question 10, you computed several products of the form (x + y)(x – y) verifying that the product is always of the form x2 – y2. Thus, if we choose values for x and y so that x = y, then the product (x + y)(x – y) will equal 0. If x = y, what is x – y?

(b) Is there any other way to choose numbers to substitute x and y so that the product

(x + y)(x – y) will equal 0? If so, what is x + y?

(d) Consider Pattern 2 for the Square of a Sum:. Is there a way to choose numbers to substitute for x and y so that the product xy equals 0?

(e) Is it ever possible that (x + y)2 could equal x2 + y2? Explain your answer.

(f) Could (x – y)2 ever equal x2 + y2? Could (x – y)2 ever equal x2 – y2? Explain your answer.

Georgia Department of Education

Kathy Cox, State Superintendent of Schools

July 21, 2008

Unit 5: Page1 of 11