Download Lesson 2 Solutions - Adjective Noun Math

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Algebra
Problems…
Solutions
Set 2
© 2007 Herbert I. Gross
By Herbert I. Gross and Richard A. Medeiros
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Problem #1
What number is named by
3 × 10² ?
Answer: 300
© 2007 Herbert I. Gross
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Answer: 300
Solution:
3 × 10²
The PEMDAS agreement tells us
that we do exponents before we
multiply. Since 10² = 100, the
given expression is equivalent to
3 × 100 or 300.
© 2007 Herbert I. Gross
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Note 1
• If we were using a left-to-right agreement
we would first multiply 3 by 10 and then
square the result. There is nothing
illogical about this. However to avoid
misinterpretation, we must all accept the
same agreement; and PEMDAS, by and
large, is the one that is usually chosen.
© 2007 Herbert I. Gross
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Problem #2
What number is named by
(3 × 10)² ?
Answer: 900
© 2007 Herbert I. Gross
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Answer: 900
Solution:
(3 × 10)²
Every thing within parentheses is
one number. In this case, since
3 × 10 = 30, the given expression
is equivalent to 30² or 30 × 30,
which is 900
© 2007 Herbert I. Gross
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Note 2
• If we are using PEMDAS but wish to
multiply 3 by 10 before we square the
result, we must use grouping symbols as
we did in this problem. In other words for
anyone using PEMDAS, 3 × 10² means
3 × 100 or 300.
© 2007 Herbert I. Gross
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Problem #3
What number is named by
3 + 7 × 10² ?
Answer: 703
© 2007 Herbert I. Gross
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Answer: 703
Solution:
3 + 7 × 10²
Using PEMDAS we square before we
multiply, and we multiply before we add.
10² = 100, so 7 × 10² = 700 and therefore
3 + (7 × 10²) is equivalent to
3 + 700 or 703.
© 2007 Herbert I. Gross
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Note 3
• In looking at 3 + 7 × 10², it is tempting to
want to read the expression from left to
right and thus add 3 and 7 before
multiplying by 10². If this is what we had
intended and we were using the PEMDAS
agreement we would have had to write
(3 + 7) × 10²
© 2007 Herbert I. Gross
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Problem #4
Evaluate
5[ 2 (x – 3) + 4] – 6 when x = 7
Answer: 54
© 2007 Herbert I. Gross
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Answer: 54
Solution:
5[ 2 (x – 3) + 4] – 6 when x = 7
x is inside the parentheses. Replacing x by 7,
the number within the parentheses is
7 – 3 or 4.
Thus the bracketed expression becomes
2(4) + 4; and since we multiply before we add,
the expression within the brackets becomes
8 + 4 or 12.
© 2007 Herbert I. Gross
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5[ 2 (x – 3) + 4] – 6 when x = 7
Replacing the bracketed expression by
12, we obtain 5[12] – 6.
And since we multiply before we
subtract, we obtain 60 – 6 or 54 as our
answer.
© 2007 Herbert I. Gross
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Note 4
• If you are uncomfortable working with the
algebraic expression, translate it first into
either a verbal recipe or as a step by step
key stroke sequence using a calculator.
© 2007 Herbert I. Gross
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Note 4
• a sequence of keystrokes for the
algebraic expression would translate as…
x
×2
-3
×5
+4
-6
y
and if we now replace x by 7 we obtain…
7
×2
-3
4
© 2007 Herbert I. Gross
×5
+4
8
12
-6
60
54
54
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Note 4
• In recipe format we have …
© 2007 Herbert I. Gross
Input x
7
Subtract 3
4
Multiply by 2
8
Add 4
12
Multiply by 5
60
Subtract 6
54
answer
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Problem #5
Evaluate
5[ 2 (x – 3) + 4] ÷ 6
when x is = 7
Answer: 10
© 2007 Herbert I. Gross
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Answer: 10
Solution:
5[ 2 (x – 3) + 4] ÷ 6 when x = 7
Up to the last step the solution here is
identical to the solution we had in problem
4. The only difference is that in the last
step we divide 60 by 6 rather than subtract
6 from 60.
© 2007 Herbert I. Gross
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Problem #6
For what value of x is it true
that…
5[ 2 (x – 3) + 4] – 6 = 84
Answer: 10
© 2007 Herbert I. Gross
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Answer: 10
Solution:
5[ 2 (x – 3) + 4] – 6 = 84
Reading the left side of the equation we
see that we obtained 84 after we
subtracted 6. So to undo this, we add 6 to
both sides of the equation to obtain…
5[ 2 (x – 3) + 4] = 90
© 2007 Herbert I. Gross
(1)
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5[ 2 (x – 3) + 4] = 90
(1)
Referring to equation (1) we obtained 90
after we multiplied by 5. So to undo this we
divide both sides of equation (1) by 5 to
obtain…
2 (x – 3) + 4 = 18
© 2007 Herbert I. Gross
(2)
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2 (x – 3) + 4 = 18
(2)
Referring to equation (2) we obtained 18
after we added 4 so to undo this we
subtract 4 from both sides of equation (2)
to obtain…
2 (x – 3) = 14
© 2007 Herbert I. Gross
(3)
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2 (x – 3) = 14
(3)
We obtained 14 after we multiplied by 2 so
to undo this we divide both sides of
equation (3) by 2 to obtain…
x–3 =7
(4)
And finally we add 3 to both sides of
equation (4) to obtain… x = 10.
© 2007 Herbert I. Gross
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Note 6
• Remember that you can always check
whether your answer is correct. Namely
let x = 10 as the input and verify that the
output will be 84.
10
×2
-3
7
© 2007 Herbert I. Gross
×5
+4
14
18
-6
90
84
84
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Note 6
10
×2
-3
×5
+4
-6
84
• Once we decide to use the key stroke
model we can solve the problem directly
by writing the “undoing” key stroke
program. Namely…
10
÷2
+3
10
© 2007 Herbert I. Gross
7
÷5
-4
14
18
+6
90
84
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Note 6
• In recipe format we have …
Recipe
10
?
Output is
Input
10
?
10
?
Add 3
Subtract 3
?
7
?
7
Divide by 2
Multiply by 2
14
?
14
?
Subtract 4
Add 4
18
?
18
?
Divide by 5
Multiply by 5
90
?
90
?
Add 6
Subtract 6
84
?
84
?
Input
Output is
84
© 2007 Herbert I. Gross
Undoing Recipe
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Problem #7
For what value of x is it true
that…
5[ 2 (x – 3) + 4] ÷ 6 = 84
Answer: 51.4
© 2007 Herbert I. Gross
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Answer: 51.4
Solution:
5[ 2 (x – 3) + 4] ÷ 6 = 84
Except for the first step in which we begin
the “undoing” by multiplying by 6 rather
than by adding 6, the procedure in this part
is exactly the same as it was in problem 6.
For example, using the
calculator key stroke model we see that…
© 2007 Herbert I. Gross
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5[ 2 (x – 3) + 4] ÷ 6 = 84
For example, using the calculator key
stroke model we see that…
51.4 +3
÷2 - 4 ÷5 ×6
51.4 48.4 96.8 100.8 504
© 2007 Herbert I. Gross
84
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Note 7
• There is a tendency for students to be
suspicious if an answer is anything but a
whole number; especially when all of the
numbers in the computation are whole
numbers. However in the “real world”
answers, more often than not, are not
whole numbers.
© 2007 Herbert I. Gross
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Note 7
• One way of seeing this is to think of the
real numbers as being points on the
number line. If a point was chosen at
random it most likely would fall between
two consecutive whole numbers rather
than on a whole number.
© 2007 Herbert I. Gross
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Problem #8
Rewrite the following recipe
using the PEMDAS agreement
Step 1:Input x
Step 2:Add 4
Step 3: Multiply by 2
Step 4: Add 3
Step 5: Multiply by 5
Step 6: Subtract 23
Step 7: Divide by 2
Step 8: The output is y
Answer: {5[2(x+4)+3]- 23} ÷ 2
© 2007 Herbert I. Gross
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Answer: {5[2(x+4)+3]- 23} ÷ 2
Solution:
We can begin by rewriting the given recipe
step by step in algebraic notation.
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
Step 6:
Step 7:
Step 8:
© 2007 Herbert I. Gross
Input x
x
Add 4
x+4
Multiply by 2
2(x + 4)
Add 3
2(x + 4) + 3
Multiply by 5
5[2(x + 4) + 3]
Subtract 23
5[2(x + 4) + 3]- 23
Divide by 2
{5[2(x + 4) + 3]- 23} ÷ 2
Output is y
y = {5[2(x + 4) + 3]- 23} ÷ 2
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Note 8
• It is just as logical to write 4 + 1 = 3 + 2
as it is to write 3 + 2 = 4 + 1. That is the
equality of two expressions does not
depend on the order in which the
expressions are written. Hence it would
be just as logical to write the answer in
the form. y = {5[2(x + 4) + 3]- 23} ÷ 2
In this respect, it is often customary to
write the output by itself on the left side of
the equal sign.
© 2007 Herbert I. Gross
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Note 8
• We could have used brackets in Step 4
and written [2(x + 4)] + 3 rather than
2(x + 4) + 3 to indicate that we were
adding 3 after we multiplied by 2.
However this is unnecessary because the
PEMDAS agreement tells us that we
multiply (and/or divide) before we add
(and/or subtract). However, there is no
harm in using more grouping symbols
than necessary when in doubt.
© 2007 Herbert I. Gross
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Note 8
• Again because of the PEMDAS
agreement in Step 6, we wrote
5[2(x + 4) + 3]- 23 rather than, say,
{5[2(x + 4) + 3]}- 23
• The grouping symbols can be whatever
you wish to use. For example, we could
have written the answer as…
5{2[x + 4] + 3} - 23 ÷ 2
© 2007 Herbert I. Gross
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Problem #9
For what value of x is it true
that…
{5[ 2 (x +4) +3] – 23} ÷ 2 = 51
Answer: 7
© 2007 Herbert I. Gross
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Answer: 7
Solution:
{5[ 2 (x +4) +3] – 23} ÷ 2 = 51
The left side of the equation is the same
expression that we dealt with in Question 4.
In key stroke format the problem becomes…
x
© 2007 Herbert I. Gross
+4
×2
+3
×5
-23
÷2
51
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Solution:
{5[ 2 (x +4) +3] – 23} ÷ 2 = 51
x
×2
+4
+3
×5
-23
÷2
51
and the “undoing” program is then…
7
÷2
-4
7
© 2007 Herbert I. Gross
11
-3
22
÷5 +23 ×2
25 125 102
51
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Note 9
• In problem 4 we started with the verbal
recipe and converted it into an algebraic
equation. Problem 9 was in essence the
inverse problem. That is we started with
the algebraic equation and converted to
the verbal recipe. If we had been given
problem 9 without having been given
problem 4 first, we might have wanted to
convert it first into the less threatening
verbal recipe format.
© 2007 Herbert I. Gross
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Note 9
• In this case we start with x and because
is inside x the parentheses we next add 4.
• We are then inside the brackets where
we are multiplying by 2 and then adding 3.
Because of PEMDAS we first multiply by 2
and then add 3 to obtain the expression…
2( x + 4) + 3
© 2007 Herbert I. Gross
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Note 9
• The entire expression 2( x + 4) + 3
is being multiplied by 5 so we enclose it in
brackets and write…
5[2( x + 4) + 3]
Next we subtract 23 from 5[2( x + 4) + 3]
and since we multiply before we subtract
there is no need to enclose 5[2( x + 4) + 3]
in grouping symbols. In this way we obtain
the expression…
5[2( x + 4) + 3] – 23
•
© 2007 Herbert I. Gross
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Note 9
• Finally we divide the entire expression
5[2( x + 4) + 3] – 23 by 2. So we enclose
the expression in braces and write…
{5[2( x + 4) + 3] – 23} ÷ 2
• Notice that if we omitted the braces in
the previous step, it would read…
5[2( x + 4) + 3] – 23 ÷ 2
© 2007 Herbert I. Gross
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Note 9
5[2( x + 4) + 3] – 23 ÷ 2
• In this case the PEMDAS agreement
tells us that we divide before we subtract.
So we would first divide 23 by 2 and then
subtract the result from the expression
5[2( x + 4) + 3].
© 2007 Herbert I. Gross