1. Keystone Problems… Keystone Problems… next Set 2 © 2007 Herbert I. Gross

2. You will soon be assigned problems to test whether you have internalized the material in Lesson 2 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. next Problem #1a If the PEMDAS agreement is being used what number is named by… 7 × 8 – 4 + 6 ÷ 2 × 5 – 3 + 2 × 5? Keystone Illustrations for Lesson 2 next Answer: 74 © 2007 Herbert I. Gross

5. Solution for Problem 1a next © 2007 Herbert I. Gross There are neither exponents nor grouping symbols. Hence using PEMDAS we perform multiplications and divisions before we perform addition and subtraction. In other words if we insert grouping symbols this expression would become… ( ( ( ) ) ) 7 × 8 – 4 + 6 ÷ 2 × 5 – 3 + 2 × 5 next

6. Solution for Problem 1a next © 2007 Herbert I. Gross In the second set of parentheses above we have both multiplication and division. (7 × 8) – 4 + (6 ÷ 2 × 5) – 3 + (2 × 5) In this case PEMDAS tells us to perform the operations from left to right. Thus we read 6 ÷ 2 × 5 as (6 ÷ 2) × 5, or 3 × 5, or 15. Everything inside the grouping symbols is treated as a single number. (6 ÷ 2 × 5) next

7. Solution for Problem 1a next © 2007 Herbert I. Gross next Therefore, we may rewrite 7 × 8 – 4 + 6 ÷ 2 × 5 – 3 + 2 × 5, as… ( ( ( ) ) ) 56 – 4 + 15 – 3 + 10 7 × 8 – 4 + 6 ÷ 2 × 5 – 3 + 2 × 5 56 10 15or next

8. Solution for Problem 1a next © 2007 Herbert I. Gross 56 – 4 + 15 – 3 + 10 We then do the addition and subtraction from left to right to obtain … ) ( 52 )( 67 ) ( 64 74 is the answer. next

9. If you are comfortable working with signed numbers, the expression 56 – 4 + 15 – 3 + 10 may be rewritten as… 56 + - 4 + 15 + - 3 + 10 And because addition is both commutative and associative we may then add the terms in any order that we wish; for example… (56 + 15 + 10) + ( - 4 + - 3) = 81 + - 7 = 74 next © 2007 Herbert I. Gross Note 1a

10. next Problem #1b If the PEMDAS agreement is being used what number is named by… 7 × 8 – 4 + 6 ÷ 2 × 5 – (3 + 2 × 5)? Keystone Illustrations for Lesson 2 next Answer: 54 © 2007 Herbert I. Gross

11. Solution for Problem 1b next © 2007 Herbert I. Gross In this case, we may work within the parentheses first, and since we multiply before we add, PEMDAS tells us that… 7 × 8 – 4 + 6 ÷ 2 × 5 – ( { 3 + 2 × 5 ) } 10 next

12. Solution for Problem 1b next © 2007 Herbert I. Gross Continuing to work within the parentheses, we then add. Hence we may rewrite the given expression as… 7 × 8 – 4 + 6 ÷ 2 × 5 – ( 3 + 10 ) 13 next (1)(1) 13 next

13. Solution for Problem 1b next © 2007 Herbert I. Gross Another way of saying that we multiply and/or divide before we add and/or subtract is to say that plus and minus signs separate terms, but times and division signs don't. In other words we may read expression (1) as… 7 × 8 – 4 + 6 ÷ 2 × 5 – 13 next ) ) ( ( ( ( ) ) (2)(2)

14. Solution for Problem 1b next © 2007 Herbert I. Gross We know from our solution to Problem 1(a) that (6 ÷ 2 × 5) = 15. Hence expression (2) becomes… (7 × 8) – (4) + (6 ÷ 2 × 5) – (13) next (2)(2) 15 (3)(3)

15. Solution for Problem 1b next © 2007 Herbert I. Gross Multiplying next, the expression becomes… (7 × 8) – (4) + (15) – (13) next 56

16. Solution for Problem 1b next © 2007 Herbert I. Gross next 56 – 4 + 15 – 13 Once again we do the addition and subtraction from left to right to obtain … 52 ) ( 67 54 as our answer. next

17. Problem #1c If the PEMDAS agreement is being used what number is named by… 7 × 8 – 4 + 6 ÷ 2 × 5 – ([3 + 2] × 5)? Keystone Illustrations for Lesson 2 next Answer: 42 © 2007 Herbert I. Gross

18. Solution for Problem 1c This time there are brackets inside the parentheses. So we work within the brackets first. That is ([3 + 2] × 5) = 5 × 5 = 25. Hence the given expression becomes… 7 × 8 – 4 + 6 ÷ 2 × 5 – 25 (1) next © 2007 Herbert I. Gross ( ) ( ( ( ) ) ) 7 × 8 – 4 + 6 ÷ 2 × 5 – 25 (2)

19. Solution for Problem 1c next © 2007 Herbert I. Gross 7 × 8 – 4 + 6 ÷ 2 × 5 – 25 ( ) ( ) 56 15 52 67 42 as our answer next

20. Remember that when grouping symbols are nested within other grouping symbols we remove the grouping symbols from the inside to the outside. For example in the expression 3{4[5(x + 7)] + 2x}, the x + 7 is first multiplied by the 5, which is then multiplied by the 4, and then multiplied by the 3. However, the 2x term is multiplied only by 3. next © 2007 Herbert I. Gross Note 1c

21. next Problem #1d If the PEMDAS agreement is being used what number is named by… 7 × 8 – (4 + 6 ÷ 2) × 5 – 3 + 2 × 5? Keystone Illustrations for Lesson 2 next Answer: 28 © 2007 Herbert I. Gross

22. Solution for Problem 1d 7 × 8 – (4 + 6 ÷ 2) × 5 – 3 + 2 × 5 In this situation we first work within the parentheses to obtain that (4 + 6 ÷ 2) = 4 + (6 ÷ 2) = 4 + 3 = 7 Therefore… next © 2007 Herbert I. Gross 7 × 8 – 7 × 5 – 3 + 2 × 5 (2) 7 × 8 – (4 + 6 ÷ 2) × 5 – 3 + 2 × 57 next ( ( ( ) ) )

23. Solution for Problem 1d (7 × 8) – (7 × 5) – 3 + (2 × 5) (2) We then perform the indicated operations to obtain the sequence of steps… next © 2007 Herbert I. Gross 56 – 35 – 3 + 10 = (56 – 35) – 3 + 10 = (21 – 3) + 10 = next 18 + 10 = 28 answer (21) – 3 + 10 =

24. Notice that without the grouping symbols all four parts of Problem 1 would look like… 7 × 8 – 4 + 6 ÷ 2 × 5 – 3 + 2 × 5. In that context, PEMDAS is not so much a matter of logic as it is a way of ensuring that everyone reads an otherwise ambiguous expression the same way. If we did not accept an agreement such as PEMDAS we would often be required to use a large number of grouping symbols. next © 2007 Herbert I. Gross Note 1d

25. next Problem #2 For what value of x is it true that… [(5x -12) ÷ 4] + 2 = 9 Keystone Illustrations for Lesson 2 next Answer: x = 8 © 2007 Herbert I. Gross

26. Solution for Problem 2 [(5x -12) ÷ 4] + 2 = 9 The way the given equation is written tells us that the last operation prior to obtaining 9 as the output was to add 2. Hence to solve for x, our first step is to subtract (that is, “unadd”) 2 from both sides of the above equation to obtain… [(5x -12) ÷ 4] = 7 next © 2007 Herbert I. Gross

27. Solution for Problem 2 [(5x -12) ÷ 4] = 7 The last operation we did in obtaining 7 was to divide by 4. Hence we multiply both sides by 4 to obtain … (5x -12) = 28 next © 2007 Herbert I. Gross

28. Solution for Problem 2 (5x -12) = 28 The last operation we did to obtain 28 was to subtract 12; hence we undo that operation by adding 12 to both sides of the above equation to obtain… 5x = 40 Whereupon, dividing both sides by 5 we obtain x = 8. next © 2007 Herbert I. Gross

29. When we are evaluating an expression we remove the grouping symbols, starting on the inside and working our way out. On the other hand in solving an algebraic equation we remove the grouping symbols by starting on the outside and working our way inside. next © 2007 Herbert I. Gross Note 2

30. Using PEMDAS notice that we could have written (5x -12) ÷ 4 + 2 = 9 rather than [(5x -12) ÷ 4] + 2 = 9. However using the extra grouping symbols ensures that even folks who are not using PEMDAS will interpret the equation in the same way. next © 2007 Herbert I. Gross Note 2

31. If you are still uncomfortable with the algebra, it often helps to translate an equation into a verbal program. In this way the equation [ (5x – 12) ÷ 4] + 2 = 9 becomes… next © 2007 Herbert I. Gross Note 2 Start with xx Multiply by 55x5x Subtract 125x – 12 Divide by 4(5x – 12) ÷ 4 Add 2 Answer is9 [(5x – 12) ÷ 4] + 2 next

32. To undo the program [ (5x – 12) ÷ 4] + 2 = 9, the last step we did is the first step we undo. next © 2007 Herbert I. Gross Note 2 Answer is Divide by 5 Add 12 Multiply by 4 Subtract 2 Start with 9 next Start with x Multiply by 5 Subtract 12 Divide by 4 Add 2 Answer is 9 8 8 40 28 9 7 next

33. Notice that in itself PEMDAS is neither arithmetic nor algebra. It is simply an agreement for determining the order of operations when ambiguities exist. In Exercise 1, we illustrated PEMDAS by a problem that involved direct computation (arithmetic), and in Exercise 2 we illustrated it by a problem that involved indirect computation (algebra). next © 2007 Herbert I. Gross Final Note