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

Keystone Problems…Keystone

Problems…

next

Set 2© 2007 Herbert I. Gross

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

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


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

Solution for Problem 1a

next


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


next


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


next


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 15 or

next


next


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.

nextnextnextnext

• 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


Note 1a

next

Problem #1b


7 × 8 – 4 + 6 ÷ 2 × 5 – (3 + 2 × 5)?


next


Solution for Problem 1b

next


In this case, we may work within the parentheses first, and since we multiplybefore we add, PEMDAS tells us that…

7 × 8 – 4 + 6 ÷ 2 × 5 – ({ 3 + 2 × 5) }

10

nextnext


next


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)13

next


next


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)


next


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)

15

(3)


next


Multiplying next, the expression becomes…

(7 × 8) – (4) + (15) – (13)

next

56


next


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.

nextnext

next

Problem #1c


7 × 8 – 4 + 6 ÷ 2 × 5 – ([3 + 2] × 5)?


next


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


( ) ( ( () ) )

7 × 8 – 4 + 6 ÷ 2 × 5 – 25 (2)

Solution for Problem 1c

next


7 × 8 – 4 + 6 ÷ 2 × 5 – 25( ) ( )

56 1552 67 42

as our answer

nextnextnextnext

• 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


Note 1c

next

Problem #1d


7 × 8 – (4 + 6 ÷ 2) × 5 – 3 + 2 × 5?


next


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


7 × 8 – 7 × 5 – 3 + 2 × 5 (2)

7 × 8 – (4 + 6 ÷ 2) × 5 – 3 + 2 × 57

nextnextnext

( ( () ) )

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


56 – 35 – 3 + 10 =(56 – 35) – 3 + 10 =

(21 – 3) + 10 =

nextnextnext

18 + 10 = 28 answer

next

(21) – 3 + 10 =

next

• 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


Note 1d

next

Problem #2

For what value of x is it true that…

[(5x -12) ÷ 4] + 2 = 9


next

Answer: x = 8© 2007 Herbert I. Gross

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



[(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



(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


• 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


Note 2

• 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


Note 2

• 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


Note 2

Start with x xMultiply by 5 5xSubtract 12 5x – 12 Divide by 4 (5x – 12) ÷ 4

Add 2Answer is 9

[(5x – 12) ÷ 4] + 2

next

• To undo the program [ (5x – 12) ÷ 4] + 2 = 9,

the last step we did is the first step we undo.

next


Note 2

Answer isDivide by 5

Add 12 Multiply by 4Subtract 2Start with 9

next

Start with xMultiply by 5Subtract 12 Divide by 4

Add 2Answer is 9

88

40 28

9 7

next

• 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


Final Note

Date post:	02-Jan-2016
Category:	Documents
Upload:	leonard-lawrence
View:	216 times
Download:	0 times

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

Documents