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Key Stone Problems… Key Stone Problems… next Set 11 © 2007 Herbert I. Gross.

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1 Key Stone Problems… Key Stone Problems… next Set 11 © 2007 Herbert I. Gross

2 You will soon be assigned problems to test whether you have internalized the material in Lesson 11 of our algebra course. The Keystone Illustrations below are prototypes of the problems you'll be doing. Work out the problems on your own. Afterwards, study the detailed solutions we've provided. In particular, notice that several different ways are presented that could be used to solve each problem. Instructions for the Keystone Problems next © 2007 Herbert I. Gross

3 As a teacher/trainer, it is important for you to understand and be able to respond in different ways to the different ways individual students learn. The more ways you are ready to explain a problem, the better the chances are that the students will come to understand. next © 2007 Herbert I. Gross

4 The beauty of logical thought is that we can often solve complicated problems by replacing them by a series of equivalent simpler problems. In particular the goal of this set of exercises is to show how using the rules of the game we can replace a rather complex expression such as… Preface next © 2007 Herbert I. Gross 2 {5 [15 – 4 (3 – c)] – 6} by the simpler, but equivalent, expression 40 c + 18 that is much easier to work with. next

5 We will develop this idea through a step-by-step simplification of the left side of the expression 2 {5 [15 – 4 (3 – c)] – 6}. The two exercises in this set will illustrate our approach. More specifically, the steps that we will use to solve the equation in Question 2 are precisely those which make up the different parts to Question 1. © 2007 Herbert I. Gross

6 next © 2007 Herbert I. Gross 1. Simplify each of the following expressions… a. - 4(3 + - c) b (3 + - c) c. 5[ (3 + - c)] d. 5[ (3 + - c)] e. 2{5[ (3 + - c)] + - 6} 2. Use the result of Problem 1e to solve the equation… 2{5[15 - 4(3 - c)] - 6} = 94 next Keystone Problems for Lesson 10

7 next © 2007 Herbert I. Gross Solution for Problem 1a This exercise asks us to simplify the expression… - 4(3 + - c). By the distributive property… From the arithmetic for signed numbers we know that… If we now replace each term on the right hand side of the first equation by its value in the second equation, we obtain… - 4(3 + - c) = - 4(3) + - 4( - c) - 4(3) = - 12 and - 4( - c) = 4c - 4(3 + - c) = c next

8 In problems such as this, we use the word “simplify” rather reluctantly. The reason is that the meaning of simplify is subjective. That is, what seems ”simpler” to one person might not seem simpler to another. So in the spirit of our game of algebra we will use “simplify” as another word for “paraphrase”. In this particular discussion it will mean to paraphrase each expression into a sum consisting of two terms. © 2007 Herbert I. Gross Note 1a

9 In terms of the above note, someone might prefer to simplify our answer by using the commutative property of addition and rewriting c in the form 4c next © 2007 Herbert I. Gross next And someone else might prefer replacing 4c by the “simpler” expression 4c – 12. Note 1a

10 With respect to the rules of the game, our preference is to write subtraction in the form of addition because our rules are stated in terms of addition. However, once you feel that you’ve internalized the connection between addition and subtraction, you may use expressions such as… next © 2007 Herbert I. Gross Note 1a next 4c and 4c –12 interchangeably.

11 If we were being insistent on playing the game strictly by the rules, we would have to prove such statements as: “The product of two negative numbers is always a positive number”. Such an approach could be tedious and too abstract for our present purposes. Instead, we will assume that the approach we used in Lessons 3, 4, and 5 for developing the arithmetic of signed numbers is sufficiently acceptable to all of our players. next © 2007 Herbert I. Gross Note 1a

12 next © 2007 Herbert I. Gross Solution for Problem 1b This exercise asks us to simplify the expression… (3 + - c) In Problem 1a, we showed that… We now replace - 4(3 + c) by c to obtain… (3 + - c) - 4(3 + - c) = c next c ( )

13 next © 2007 Herbert I. Gross Solution for Problem 1b and since = 3, we may rewrite the above equality as... next (3 + - c) = 3 + 4c (3 + - c) = 15 + ( c) By the associative property of addition we know that… (3 + - c) = c () Thus…

14 One of the most common mistakes made by beginning students is to begin by adding the 15 and - 4 in the expression next © 2007 Herbert I. Gross to obtain the incorrect expression… next Caution (3 + - c) 11(3 + - c) next

15 © 2007 Herbert I. Gross Caution (3) = 3 For example, if we replace c by 0 in the expression (3 + - c) we obtain… On the other hand, if we replace c by 0 in 11(3 + - c), we obtain… 11(3) = 33 The fact that the same input gives different outputs shows that the two expressions can’t be equal. next

16 © 2007 Herbert I. Gross Caution ( )(3 + - c) That is, by our agreeing to the PEMDAS convention we do all multiplications before we do any additions. Hence, if we had wanted the 15 and the - 4 to be added first, we would have had to write the expression as…

17 In words the expression (3 + - c) tells us… next © 2007 Herbert I. Gross Note 1b *** Start with any number c. *** Replace it by its opposite… - c. *** Add 3… c. *** Multiply by - 4… - 4(3 + - c). *** Add 15… (3 + - c). What we showed in this exercise is that the expression ( c) is equivalent to the expression 3 + 4c. next

18 In words the expression 3 + 4c tells us… © 2007 Herbert I. Gross Note 1b *** Start with c… c. *** Multiply 4… 4c. *** Add 3… 3 + 4c. next

19 © 2007 Herbert I. Gross Note 1b In this sense, what we mean by “simplified” is that the relatively cumbersome set of instructions… next Start with any number c. Replace it by its opposite. Add 3. Multiply by - 4. Add 15 and then write the answer. …can be replaced by the less cumbersome but equivalent set of instructions… Start with any number c. Multiply it by 4. Add 3 and then write the answer.

20 next © 2007 Herbert I. Gross Solution for Problem 1c This exercise asks us to simplify the expression… 5[ (3 + - c)] From the previous exercise we already know that… So by replacing (3 + - c) by 3 + 4c, 5[ (3 + - c)] becomes… 5[3 + 4c] (3 + - c) = 3 + 4c next 3 + 4c

21 next © 2007 Herbert I. Gross By the distributive property, we know that… 5[3 + 4c] = 5(3) + 5(4c) = [4c] next By the associative property of multiplication, we know that… 5[4c] = [5(4)]c = 20c

22 next © 2007 Herbert I. Gross next If we now replace 5[4c] by 20c, we obtain… 5[3 + 4c] = c next In summary, we have thus far shown that… (1) 5[ (3 + - c)] = 5[3 +4c] and (2) 5[3 + 4c] = c 5[ (3 + - c)] = c Thus, by transitivity… 5[ (3 + - c)] c next

23 It’s quite possible that in the above equation you looked at 5[4c] and immediately concluded that it was equal to 20c without even thinking consciously about the associative property of multiplication. Hopefully, such observations will become quite common as the course proceeds. For the time being, we’ll try to be as complete as we feel is necessary in demonstrating how certain results follow inescapably from others just by using the accepted rules of the game. © 2007 Herbert I. Gross Note 1c

24 next © 2007 Herbert I. Gross Solution for Problem 1d This exercise asks us to simplify the expression… 5[ (3 + - c)] In exercise 1c we showed that… Therefore by substitution 5[ (3 + - c)] becomes… 5[ (3 + - c)] ( c) next 5[ (3 + - c)] = c

25 next © 2007 Herbert I. Gross By the commutative property of addition we know that… So we may replace c by 20c + 15 to obtain… Therefore, 5[ (3 + - c)] may be replaced by … next 5[ (3 + - c)] (20c + 15) 5[ (3 + - c)] = c 20c c = 20c + 15

26 next © 2007 Herbert I. Gross Then by the associative property of addition… next 20c In summary… 5[ (3 + - c)] = (20c + 15) and (20c + 15) = 20 c + 9 5[ (3 + - c)] = 20c + 9 next () = 20c + 9 So again by transitivity… 5[ (3 + - c)] c + 9 next

27 © 2007 Herbert I. Gross Solution for Problem 1e This exercise asks us to simplify the expression… 2{5[ (3 + - c)] + - 6} From the previous exercise we know … Hence, we replace 5[ (3 + - c)] in the first expression by its equivalent value in the above expression, to obtain… 2{5[ (3 + - c)] + - 6} = 2{20c + 9} 5[ (3 + - c)] = 20c + 9 next

28 © 2007 Herbert I. Gross By the distributive property we know that… 2{20c + 9} = 2{20c} + 2{9} = 40 c + 18 And if we now replace 2{5[ (3 + - c)] + - 6} In the previous expression with the value in the above expression we obtain... 2{5[ (3 + - c)] + - 6} = 40c + 18 next Solution for Problem 1e

29 In any game there is usually more than one correct strategy for obtaining a given objective. In the game of mathematics, this may be restated as... next © 2007 Herbert I. Gross Enrichment Note Therefore, do not try to memorize our proofs. Rather, concentrate on how we elected to use a particular strategy to achieve each of the given objectives. next “In a well-defined problem there is only one correct answer, but hardly ever only one correct way to derive the answer”.

30 next © 2007 Herbert I. Gross Solution for Problem 2 In this exercise we are asked to use the result of Problem 1e to solve the equation… 2 {5[15 – 4(3 – c)] – 6} = 94 We begin our solution by observing that by the “add the opposite” rule, the equation… 2{5[ (3 + - c)] + - 6} = 94 next 2 {5[15 – 4(3 – c)] – 6} = 94 …can be rewritten in the equivalent form…

31 © 2007 Herbert I. Gross Solution for Problem 2 The left side of our equation is precisely the expression we simplified in Exercise 1e. In other words the fact that… …means that we may replace the equation 2{5[ (3 + - c)] + - 6} = 94 by the equivalent but simpler equation… next 40c + 18 = 94 2{5[ (3 + - c)] + - 6} = 40c + 18

32 next © 2007 Herbert I. Gross Solution for Problem 2 To solve equation 40c + 18 = 94, we first subtract 18 from both sides to obtain the equivalent equation… Finally, we divide both sides of the above equation by 40 to obtain… next 40c = 76 c = 76 / 40 = 19 / 10 or in decimal form… = 1.9

33 We could have omitted Exercise 1 in its entirety and just asked you to find the value of c for which... © 2007 Herbert I. Gross Note 2 In that case one of the ways we could have used to answer the question would have been to rewrite the equation in the form… 2 {5[15 – 4(3 – c)] – 6} = 94 2{5[ (3 + - c)] + - 6} = 94 next

34 Notice that the left side of 2{5[ (3 + - c)] + - 6} = 94 is the expression that we simplified in Exercise 1e. © 2007 Herbert I. Gross Note 2 Then since grouping symbols are removed starting from the innermost set first, we would have begun the simplifying process by starting with the removal of the parentheses; and this is precisely what we did in Exercises 1a, 1b, 1c, 1d, and 1e. next More visually… 2{ 5 [ (3 + - c)] } - 4(3 + - c)[ (3 + - c)] 5 [ (3 + - c)] 5[ (3 + - c)] next 1a, 2 {5[ (3 + - c) + - 6} 1b, 1c, 1d, 1e.

35 In summary, we used the rules in Lessons 9 and 10 to show that the more complicated expression… next © 2007 Herbert I. Gross Summary …could be replaced by the equivalent, but much simpler expression… next 2{5[ (3 + - c)] + - 6} 40c + 18

36 next © 2007 Herbert I. Gross At least until you become more comfortable with complicated algebraic expressions; it is a good idea to rewrite every subtraction problem in terms of the add-the-opposite rule. That is, wherever we see a minus sign we replace it by a plus sign and change the sign of the number immediately after the minus sign. Feeling Comfortable

37 next © 2007 Herbert I. Gross *** we first locate the minus signs… *** Then we replace each minus sign by a plus sign… next By way of Review 2 {5[15 – 4(3 – c)] – 6} = 94 2 {5[15 – 4(3 – c)] – 6} = 94 …and we replace each number immediately after the changed minus sign by its opposite… 2 {5[15 + 4(3 + c)] + 6} = next +++

38 © 2007 Herbert I. Gross *** Notice that the steps we used in solving the equation… …had nothing to do with the fact that the right side was 94. That is, we demonstrated that 2{5[15 – 4(3 – c)] – 6} = 40c + 18 without any reference to the right hand side of the above equation. By way of Review 2 {5[15 – 4(3 – c)] – 6} = 94 next

39 © 2007 Herbert I. Gross Recall that to evaluate the expression… 2 {5[15 – 4(3 – c)] – 6} …for any given value of c, we start with c and remove the grouping symbols by working our way from the inside to the outside. The above comments may seem more impressive if you look at them in terms of “programs”. next

40 © 2007 Herbert I. Gross next Program #1 (1) Start with c (2) Subtract it from 3 (3) Multiply the result by 4 (4) Subtract this result from 15 (5) Multiply the result by 5 (6) Subtract 6 (7) Multiply by 2 (8) Write the answer Following this procedure we see that the expression 2 { 5 [ 15 – 4 ( 3 – c ) ] – 6 } is equivalent to Program #1, where… c( 3 – c )4 ( 3 – c )15 – 4 ( 3 – c )5 [ 15 – 4 ( 3 – c ) ]5 [ 15 – 4 ( 3 – c ) ] – 6 2 { 5 [ 15 – 4 ( 3 – c ) ] – 6 } next

41 © 2007 Herbert I. Gross next Program #2 (1) Start with c (2) Multiply by 40 (3) Add 18 (4) Write the answer On the other hand, the expression 40 c + 18 is equivalent to Program #2, where… c next

42 © 2007 Herbert I. Gross If we had not used the rules of the game, how likely is it that we could have guessed that Programs #1 and #2, as different as they look, are equivalent? It is in this sense that the “game” is important. That is, we use our rules to demonstrate that certain things which may not seem obvious are indeed true. Because it contains fewer steps, Program #2 is preferable to Program #1. In other words, the fewer steps we use, the less chance there is that we will make a computational mistake. This is true in other things as well. For example, the fewer “moving parts” a machine has, the easier it is to maintain the machine.

43 next © 2007 Herbert I. Gross For example, we used Program #2 to prove that if the output of Program #1 was 94 then the input had to be 1.9. To check that our solution was correct, we would replace c by 1.9 in Program #1 and show that we got 94 as the output. Moreover, if we have to evaluate a program for many different values of the input, the shorter program has the potential to save us significant amounts of time. next

44 © 2007 Herbert I. Gross next Program #1 (1) Start with C (2) Subtract it from 3 (3) Multiply the result by 4 (4) Subtract this result from 15 (5) Multiply the result by 5 (6) Subtract 6 (7) Multiply by 2 (8) Write the answer That is…

45 next © 2007 Herbert I. Gross next Program #2 (1) Start with c (2) Multiply by 40 (3) Add 18 (4) Write the answer If now we wanted to find the output in Program #1 when the input was, say, 6.5, we could go through the same tedious process we used above, or we could replace Program #1 by the equivalent Program #2 and see that…

46 next © 2007 Herbert I. Gross The important thing isn't just that Program #2 is simpler than Program #1; but rather that it is equivalent to Program #1. That is, if we replace c by any given value using Program #2, we will find the same answer more simply than we would have found it by using the more cumbersome Program #1. Important Point

47 next © 2007 Herbert I. Gross Keep in mind that while we simplify expressions by removing the grouping symbols from the inside to the outside; we simplify equations by removing the grouping symbols from the outside to the inside. The point is that when a complicated expression is part of an equation, we may not have to simplify the complicated expression first. For example, another way to solve the equation 2{5[ (3 + - c)] + - 6} = 94 without first doing the steps in Exercise 1 is by working on both sides of the equation at the same time. next

48 © 2007 Herbert I. Gross More specifically, in solving the equation 2{5[ (3 + - c)] + - 6} = 94 by the undoing method, we notice that c is inside the parentheses; the parentheses are inside the brackets; and the brackets are inside the braces. So the last step we did to obtain 94 was to multiply by 2. Hence, we divide both sides of the equation by 2 to obtain… 5[ (3 + - c)] = 472{5[ (3 + - c)] + - 6} = 94 next

49 There is a tendency for beginning students to begin the solution of 2{5[ (3 + - c)] + - 6} = 94 by adding 6 to both sides of the equation. However, notice that since - 6 is inside the braces, and the braces are being multiplied by 2, it behaves as 12. More specifically, PEMDAS tells us that we multiply by 2 after we add - 6. © 2007 Herbert I. Gross Caution

50 To undo subtracting 6, we add 6 to both sides of 5[15 – 4(3 – c)] – 6 = 47 to obtain… next © 2007 Herbert I. Gross To undo multiplying the left side of 5[15 – 4(3 – c)] = 53 by 5, we divide both sides of by 5 to obtain… 5[15 – 4(3 – c)] = – 4(3 – c) = 10.6 next

51 Since it's often easier to deal with undoing addition than with undoing subtraction, we may prefer to replace each subtraction by the add-the-opposite rule and rewrite equation 15 – 4(3 – c) = 10.6 in the form… © 2007 Herbert I. Gross Since the last step we did in the above equation to obtain 10.6 was to add 15, we now subtract 15 from both sides of (3 + - c) = 10.6 to obtain… (3 + - c) = (3 + - c) = next

52 Since the last step in obtaining - 4(3 + - c) = - 4.4, was to multiply by - 4, we divide both sides of the equation by - 4 to obtain… © 2007 Herbert I. Gross Next we subtract 3 from both sides of equation c = 1.1 to obtain… c = c = next Since the opposite of c (that is, - c) is - 1.9, c itself must be equal to 1.9… c = 1.9 next

53 © 2007 Herbert I. Gross We may solve Question 2 by either of the previous two methods. However, the method we used in Question 1 is the only one we can use if we are asked to simplify an expression rather than to solve an equation. next Important Point

54 As a final note observe the subtlety involved in the language of subtraction. For example, it makes no difference whether we say “Pick a number and add 7” (in the language of algebra, x + 7) or “Pick a number and add it to 7” (in the language of algebra, 7 + x) because addition obeys the commutative rule. next © 2007 Herbert I. Gross Final Note next

55 © 2007 Herbert I. Gross However, it makes a big difference whether we say… “Pick a number and subtract 7 (in the language of algebra, x – 7) or “Pick a number and subtract it from 7” (in the language of algebra 7 – x) …because subtraction doesn't obey the commutative property.

56 next © 2007 Herbert I. Gross This is one reason we prefer to rewrite subtraction in terms of “add the opposite”. That is, while 15 – 7 ≠ 7 – 15; = In fact, as we’ll discuss in more detail in the Exercise Set, x – 7 and 7 – x are opposites of one another. For example, 15 – 7 = 8 while 7 – 15 = - 8. next


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