Post on 02-Jan-2016
Keystone Problems…Keystone
Problems…
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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
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© 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.
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© 2007 Herbert I. Gross
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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
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Answer: 74© 2007 Herbert I. Gross
Solution for Problem 1a
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© 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
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Solution for Problem 1a
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© 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)
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Solution for Problem 1a
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© 2007 Herbert I. Gross
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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
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Solution for Problem 1a
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© 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.
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• 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
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© 2007 Herbert I. Gross
Note 1a
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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
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Answer: 54© 2007 Herbert I. Gross
Solution for Problem 1b
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© 2007 Herbert I. Gross
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
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Solution for Problem 1b
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© 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
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(1)13
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Solution for Problem 1b
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© 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
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) )( (( () ) (2)
Solution for Problem 1b
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© 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)
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(2)
15
(3)
Solution for Problem 1b
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© 2007 Herbert I. Gross
Multiplying next, the expression becomes…
(7 × 8) – (4) + (15) – (13)
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56
Solution for Problem 1b
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© 2007 Herbert I. Gross
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56 – 4 + 15 – 13
Once again we do the addition and subtraction from left to right to obtain …
52
)(
67 54
as our answer.
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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
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Answer: 42© 2007 Herbert I. Gross
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)
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© 2007 Herbert I. Gross
( ) ( ( () ) )
7 × 8 – 4 + 6 ÷ 2 × 5 – 25 (2)
Solution for Problem 1c
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© 2007 Herbert I. Gross
7 × 8 – 4 + 6 ÷ 2 × 5 – 25( ) ( )
56 1552 67 42
as our answer
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• 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.
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© 2007 Herbert I. Gross
Note 1c
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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
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Answer: 28© 2007 Herbert I. Gross
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…
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© 2007 Herbert I. Gross
7 × 8 – 7 × 5 – 3 + 2 × 5 (2)
7 × 8 – (4 + 6 ÷ 2) × 5 – 3 + 2 × 57
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( ( () ) )
Solution for Problem 1d
(7 × 8) – (7 × 5) – 3 + (2 × 5) (2)
We then perform the indicated operations to obtain the sequence of steps…
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© 2007 Herbert I. Gross
56 – 35 – 3 + 10 =(56 – 35) – 3 + 10 =
(21 – 3) + 10 =
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18 + 10 = 28 answer
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(21) – 3 + 10 =
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• 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.
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© 2007 Herbert I. Gross
Note 1d
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Problem #2
For what value of x is it true that…
[(5x -12) ÷ 4] + 2 = 9
Keystone Illustrations for Lesson 2
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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
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© 2007 Herbert I. Gross
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
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© 2007 Herbert I. Gross
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.
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© 2007 Herbert I. Gross
• 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.
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© 2007 Herbert I. Gross
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.
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© 2007 Herbert I. Gross
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…
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© 2007 Herbert I. Gross
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
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• To undo the program [ (5x – 12) ÷ 4] + 2 = 9,
the last step we did is the first step we undo.
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© 2007 Herbert I. Gross
Note 2
Answer isDivide by 5
Add 12 Multiply by 4Subtract 2Start with 9
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Start with xMultiply by 5Subtract 12 Divide by 4
Add 2Answer is 9
88
40 28
9 7
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• 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).
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© 2007 Herbert I. Gross
Final Note