Download Lesson15 Solutions - Adjective Noun Math

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Algebra
Problems…
Solutions
Set 15
© 2007 Herbert I. Gross
By Herbert I. Gross and Richard A. Medeiros
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Problem #1
For what value of y is it true that
7(y + 4) – 3(y – 5) = 6y + 35 ?
Answer: y = 4
© 2007 Herbert I. Gross
Answer: y = 4
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Solution:
The right-hand side of the equation already
has the mx + b (actually, the my + b) form.
So we begin by using our “rules of the
game” to transform the left-hand side of the
equation…
7(y + 4) – 3(y – 5) = 6y + 35
and rewrite it by replacing each subtraction
by the “add the opposite” rule. That is…
7(y + 4) +– - 3(y +– - 5)
© 2007 Herbert I. Gross
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Solution:
By using the distributive property, we may
rewrite 7(y + 4) + -3(y + -5) as…
7(y) + 7(4) + -3(y) + -3(-5)
or…
7y + 28 + -3y + 15
or…
7y + - 3y + 28 + 15
or…
(7y + - 3y) + (28 + 15)
or…
4y + 43
© 2007 Herbert I. Gross
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Solution:
We may now replace the left-hand side of
7(y + 4) – 3(y – 5) = 6y + 35 by its value
4y + 43 to obtain…
4y + 43 = 6y + 35
Next, we might subtract 4y from both sides
in the above equation to obtain…
43 = 2y + 35
We may then subtract 35 from both sides
above to obtain…
8 = 2y,
from which we see that…
© 2007 Herbert I. Gross
y=4
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Notes on #1
Technically speaking we have not as yet
proved that y = 4 is a solution of the
equation 7(y + 4) – 3(y – 5) = 6y + 35.
Rather what we have shown is that if our
equation does have a solution, it must be
y = 4.
As a check that y = 4 is a solution;
we should get a true statement,
if we replace y by 4 in our equation
© 2007 Herbert I. Gross
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Notes on #1
Replacing y by 4 in the equation below,
we see that…
4
(4)
4
7( y + 4 ) + -3( y + -5 ) = 6 y + 35
or…
7(8) + -3(-1) = 24 + 35
or…
56 + 3 = 59
and…
© 2007 Herbert I. Gross
59 = 59
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Notes on #1
Notice that in working with the equation
7(y + 4) – 3(y – 5) = 6y + 35, we began by
rewriting the left-hand side into the form
my + b. We did this because we
already knew how to solve equations that
are in the form my + b = ny + c.
The point is that in situations that involve
logical thought, we often try to reduce
unsolved problems to a series of one or
more previously solved problems.
© 2007 Herbert I. Gross
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Notes on #1
Problem # 1is an good example of how we
develop new techniques in mathematics,
one step at a time.
In contrast, “cramming” rarely works in
learning to do mathematics. Namely, since
each step follows logically from a previous
step, it is important to be aware of the
significance of each step as it occurs.
© 2007 Herbert I. Gross
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Notes on #1
In summary…
A major objective in the “game” of
algebra is to use the “rules of the game”
to paraphrase equations we haven’t yet
learned to solve into equations that we
already know how to solve.
© 2007 Herbert I. Gross
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Problem #2
For what value of p is it true that
2 [3p – 2(4 – p)] = 7p + 32 ?
Answer: p = 16
© 2007 Herbert I. Gross
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Answer: p = 16
Solution
Again the basic strategy will be to
paraphrase the left side of the equation
into the mp + b form. As in the previous
problem, one way to begin is by replacing
each subtraction by the “add the opposite”
rule. That is, we rewrite the left side of the
equation…
2 [3p – 2(4 – p)] = 7p + 32
as…
2 [3p + -2(4 + -p)] = 7p + 32
© 2007 Herbert I. Gross
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Solution:
Recalling that we simplify an algebraic
expression by starting with the innermost
set of grouping symbols (in this case, the
parentheses), we next use the distributive
property to rewrite
2 [3p + -2(4 + -p)] as…
2 [3p + -2(4) + -2(-p)]
or…
© 2007 Herbert I. Gross
2 [3p + -8 + 2p]
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Solution:
Then by using the commutative and
associative properties of addition, the
expression 2 [3p + -8 + 2p] within the
brackets can be rewritten as…
2 [5p + -8]
…which by the distributive property can
be rewritten as…
10p + -16 (or 10p – 16)
© 2007 Herbert I. Gross
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Solution:
We may now replace the left side of the
equation 2 [3p + -2(4 + -p)] = 7p + 32
by its value 10p + -16 to obtain the
equivalent equation…
-16 = 7p + 32
2 [3p + -2(4
10p++-p)]
Subtracting 7p from both sides of the
above equation, and then adding 16
to both sides of the new equation, we see
that…
3p = 48 or p =16
© 2007 Herbert I. Gross
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Notes on #2
In terms of a
“program”,
3p – 2
2 [3p
4–p
2(4
p)]
means…
Program #1
Start with p.
Subtract it from 4.
Multiply the result by 2.
Subtract this result from 3 times p.
Multiply the result by 2.
© 2007 Herbert I. Gross
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What the
expression
10
p – 16 told
10p
us was that
this was
equivalent to
the simpler
program…
© 2007 Herbert I. Gross
Notes on #2
Program #2
Start with p.
Multiply it by 10.
Subtract 16.
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Notes on #2
In terms of a chart…
p
1
2
3
4
5
6
4– p
3
2
1
0
-1
-2
© 2007 Herbert I. Gross
2(4 – p)
6
4
2
0
-2
-4
3p 3p – 2(4– p)
-3
3
6
2
9
7
12
12
15
17
18
22
2[3p –2(4 –p)] 10p 10p – 16
-6
-6
10
4
20
4
14
30
14
24
40
24
34
50
34
44
60
44
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Notes on #2
In terms of a “program”, the right-hand side
p + 32
of the equation 2 [3p + -2(4 + -p)] = 7
7p
becomes…
Program #3
Start with p.
Multiply it by 7.
Add 32.
© 2007 Herbert I. Gross
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Notes on #2
And what we showed in solving this
problem is that the only time
Program #1 (or, equivalently, Program #2)
and
Program #3 yield the same output
for a given input is
when p = 16
© 2007 Herbert I. Gross
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Notes on #2
As a check, we see that…
p 10p 10p – 16
© 2007 Herbert I. Gross
7p
7p + 32
1
10
-6
7
39
2
20
4
14
46
…
…
…
…
…
16 160
144
112
144
17 170
154
119
151
18 180
164
126
158
etc.
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Notes on #2
In essence, p = 16 is the equilibrium point.
p 10p 10p – 16
7p
7p + 32
15 150
134
105
137
16 160
144
112
144
17 170
154
154
119
151
As the chart indicates; if p is less than 16,
10p – 16 is less than 7p + 32.
However, when p is greater than 16,
10p – 16 is greater than 7p + 32.
© 2007 Herbert I. Gross
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Problem #3
For what value of q is it true that
1/ (q
7
+ 3) + 6 = 1/2(q + 5) ?
Answer: q = 16
© 2007 Herbert I. Gross
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Answer: q = 11
Solution
Since 14 is a common multiple of 7 and 2,
we can paraphrase…
1/ (q
7
+ 3) + 6 = 1/2(q + 5)
into an equation involving only whole
numbers by multiplying both sides of the
equation by 14. Doing this we obtain…
14 [1/7 (q + 3) + 6] = 14[1/2(q + 5)]
© 2007 Herbert I. Gross
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Solution:
Since the left side of the equation
14 [1/7(q + 3) + 6] = 14[1/2(q + 5)]
has the form a(b + c), where
a = 14, b = 1/7(q + 3), and c = 6;
we may use the distributive property to
rewrite our equation as…
14 [1/7(q + 3)] + 14(6) = 14[1/2(q + 5)]
© 2007 Herbert I. Gross
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Solution:
Since 14(1/7) = 2, 14 × 6 = 84 and 14 (1/2) = 7, it
is not difficult to see from the properties of
multiplication that we may rewrite
the equation…
14 [1/7(q + 3)] + 14(6) = 14[1/2(q + 5)] as…
2(q + 3) + 84 = 7(q + 5)
or…
2q + 6 + 84 = 7q + 35
or…
2q + 90 = 7q + 35
or…
90 = 5q + 35
or…
55 = 5q
or…
q = 11
© 2007 Herbert I. Gross
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Notes on #3
Some students find it easier to first use the
distributive property and then multiply both
sides by the least common denominator.
That is, we could have solved the equation
1/ (q + 3) + 6 = 1/ (q + 5) by first using the
7
2
distributive property to rewrite it as…
q/
© 2007 Herbert I. Gross
3/ + 6 = q / + 5
+
7
7
2
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Notes on #3
q/
7
+ 3/7 + 6 = q/2 + 5
We could then have multiplied each term
on both sides of the above equation by 14
to obtain…
2q + 6 + 84 = 7q + 35
…whereupon we could have then solved the
equation as we did in our previous solution.
© 2007 Herbert I. Gross
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Notes on #3
Remember that you can (and should) check
your answer by replacing it in the original
equation and seeing whether it becomes
a true statement. Replacing q by 11 in
equation 1/7(q + 3) + 6 = 1/2(q + 5) gives us…
1/ ( 11
7 q
+ 3 ) + 6 = 1/2( 11
q + 5)
1/ (
7
14 ) + 6 = 1/2( 16 )
2 +6=8
© 2007 Herbert I. Gross
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Notes on #3
In simplifying q/7 + 3/7 + 6 = q/2 + 5, it was not
necessary to multiply by the least common
multiple of 7 and 2. Any other common
multiple would have worked as well.
The main reason for choosing the least
common multiple is to keep the arithmetic as
simple as possible. However, if it is not
obvious what the least common multiple is, it
is often easier (especially with the help of a
calculator) to use any common denominator
and then solve the resulting equation.
© 2007 Herbert I. Gross
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Problem #4
A car leaves from point O moving at a
constant speed of 50 miles per hour.
One hour later, a second car, traveling in
the same direction as the first car, leaves
point O and travels at a constant speed of
60miles per hour.
How many miles must the second car
travel before it catches up to the first car?
© 2007 Herbert I. Gross
Answer: 300 (miles)
Answer: 300 miles
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Solution:
We may let t stand for the number of hours
the second car has to travel to overtake the
first car, and we may let d denote the
number of miles the second car is from the
point O. Since the speed of the second car
is 60 miles per hour, the relationship for the
second car between d (distance) and t (time)
for the second car is given by…
d = 60t
© 2007 Herbert I. Gross
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Solution:
Since the first car started an hour earlier, it
has traveled one hour longer than the
second car, and since t denotes the time the
second car traveled, t + 1 denotes the time
the first car traveled.
Therefore, the relationship between
d and t for the first car is given by…
d = 50(t + 1)
© 2007 Herbert I. Gross
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Solution:
Since the two cars would have traveled the
same distance d from O when they meet;
we may equate the value of d in the
equation d = 60t to the value of d in the
equation d = 50(t + 1) to obtain…
60t = 50(t + 1)
…which, by the distributive property
may be rewritten as…
60t = 50t + 50
© 2007 Herbert I. Gross
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Solution:
We then subtract 50t from both sides of the
equation below to give us…
60t = 50t + 50
50t 50t
10t =
50
t=5
And replacing t by 5 in either equation
d = 60t or equation d = 50(t + 1) we see
that d = 300.
© 2007 Herbert I. Gross
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Notes on #4
As a check, notice that at the instant the
second car catches up to the first car, both
cars have traveled the same number of miles.
Since the first car started 1 hour before
the second car; if the second car traveled
for 5 hours, the first car must have
traveled for 6 hours. At 50 miles per hour
for 6 hours, we see that the second car
also travels 300 miles.
© 2007 Herbert I. Gross
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Notes on #4
From a “real world” point of view, the
“weakness” of this exercise is that it is
very unlikely that a car would travel at a
constant speed for 5 or 6 hours.
A more realistic version would be to say
that the first car had an average speed of
50 miles per hour and that the second car
had an average speed of
60 miles per hour.
© 2007 Herbert I. Gross
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Notes on #4
Note that, to find the average speed we
would have to know how far the car
traveled and how long that took.
That is, we define the average speed
to be the quotient we get when we
divide the distance by the time.
If the speed happens to be constant, then
the average speed is the same as the
constant speed.
© 2007 Herbert I. Gross
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Notes on #4
A common strategy for solving algebra
problems is to let x (or any other letter)
stand for the quantity we are asked to find.
Hence, as a first step, we read the problem
and see what it asks for.
Since this problem asks for how far the second
car traveled in order to catch up to the first car;
we might let d stand for the distance from the
point O to the point P at which the second car
catches up to the first car.
© 2007 Herbert I. Gross
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Notes on #4
From the relationship…
distance = speed × time
…we see that it is also true that…
distance
time =
speed
Therefore, the time it took the first car to
get to the point P is given by d/50 and the
time it took the second car to get to the
point P is d/60.
© 2007 Herbert I. Gross
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Notes on #4
Since the first car traveled for one hour
more than the second car, the equation
that relates their times is given by…
The number of
hours that the first
car traveled
© 2007 Herbert I. Gross
=
The number of
hours that the
second car traveled
+1
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Notes on #4
Since we know that the number of hours
the first car travels is d/50 and the number
of hours the second car travels is d/60, the
equation…
The number of
hours that the first
car traveled
=
The number of
hours that the
second car traveled
becomes…
d/
50
© 2007 Herbert I. Gross
= d/60 + 1
+1
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Notes on #4
Aside: Since 60 is more than 50, d/60 is less
than d/50, and that’s why in the equation
d/
d/
=
50
60 + 1, we had to add 1 to the
right-hand side.
Since 300 is the least common multiple of 50
and 60, we may multiply both sides of our
equation by 300 to obtain…
300(d/50) = 300(d/60 + 1)
6d = 5d + 300
© 2007 Herbert I. Gross
d = 300
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Notes on #4
In general, we seek the simplest solution to
any problem we face, and we look for a more
complicated solution only when the simpler
one gets to be either inadequate or too
cumbersome.
In this sense, we prefer to use algebra for
solving a word problem only in the event
that we cannot see an easier way to solve
the problem.
© 2007 Herbert I. Gross
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Notes on #4
Very often so-called “word problems” in
algebra can be done by using arithmetic. In
fact, we usually elect to use algebra only
when it is not easy for us to see what the
arithmetic method is.
For example, one non-algebraic way to do
this problem is to notice that the first car
has a 50 miles head start on the second car.
© 2007 Herbert I. Gross
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Notes on #4
That is, the first car has a 1 hour head start,
and during that hour, it travels 50 miles.
The second car gains 10 miles on the first
car each hour it travels.
In other words, each hour after the first
hour, the first car travels 50 miles and the
second car travels 60 miles.
© 2007 Herbert I. Gross
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Notes on #4
Since the second car gains 10 miles on the
first car for each hour that the second car
travels, it will take the second car 5 hours
to make up the 50 mile head start the first
car has.
And at a constant speed of 60 miles per
hour, the second car travels 300 miles in
those 5 hours.
© 2007 Herbert I. Gross
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Notes on #4
As we’ve mentioned before it is
sometimes helpful to make
a chart and look for patterns.
For example, in the present situation,
suppose we denote by “Car 1” the car
that started first, and we denote the
second car by “Car 2”.
© 2007 Herbert I. Gross
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Notes on #4
Distance
Traveled
The
chart
might
then
look
like…
After 1 hour
After 2 hour
After 3 hour
After 4 hour
After 5 hour
After 6 hour
After 7 hour
© 2007 Herbert I. Gross
Car 1
50 miles
100 miles
150 miles
200 miles
250 miles
300 miles
350 miles
Car 2
0 miles
60 miles
120 miles
180 miles
240 miles
300 miles
360 miles
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Notes on #4
Notice the need to keep track of
what our variable represents.
Distance
Traveled
Car 1
Car 2
After 5 hour
250 miles
240 miles
After 6 hour
300 miles
300 miles
Namely in solving the problem, we found
that the two cars were at the same place
when t = 5. Yet the chart seems to indicate
that it happens when t = 6?
How can t equal both 5 and 6?
© 2007 Herbert I. Gross
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Notes on #4
The answer is that there is no
contradiction, because in our case, the
time was measured relative to Car 2, while
in the chart the time is measured relative
to Car 1.
In more mathematically precise terminology,
we might have let t1 represent the time (t) the
first car traveled, and t2 represent the time
(t + 1) the second car traveled.
© 2007 Herbert I. Gross
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Notes on #4
In this particular example, it didn’t take
long for the chart to help us arrive at the
correct answer. Notice, however, that even
if the problem had been more complex the
chart showed us rather quickly that every
hour the distance between the two cars
decreased by 10 miles.
© 2007 Herbert I. Gross
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Problem #5
The relationship between the Fahrenheit (F)
temperature scale and the Celsius (C)
temperature scale is given by the linear
relationship…
C = 5/9(F – 32)
At what temperature will both scales
yields the same reading?
Answer:
© 2007 Herbert I. Gross
-40°C
(or equivalently -40°F)
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Answer: -40°C (or equivalently -40°F)
Solution
When the scales have the same reading,
it means that C = F.
Hence, we may replace C by F in the formula
C = 5/9(F – 32) to obtain…
F = 5/9(F – 32)
To eliminate the fraction in the equation
above, we may multiply both sides of the
equation by 9 to obtain…
© 2007 Herbert I. Gross
9F = 5(F – 32)
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Solution
We may then use the distributive property to
obtain…
9F = 5(F – 32)
9F = 5F – 160
…and if we next subtract 5F from both
sides of the above equation we obtain…
4F = -160
F = -40
© 2007 Herbert I. Gross
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Notes on #5
In F = 5/9(F – 32) we replaced C by F in
formula C = 5/9(F – 32).
It would have been just as logical to
replace F by C in the formula
C = 5/9(F – 32).
In this case, the equation would
have been…
C = 5/9 (C – 32)
© 2007 Herbert I. Gross
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Notes on #5
Notice that equations F = 5/9(F – 32) and
C = 5/9(C – 32) are exactly the same,
except for the name of the variable.
However, since either equation tells us the
temperature at which the two scales would
have the same reading…
(that is, when C = F);
it doesn’t matter which equation we use.
© 2007 Herbert I. Gross
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Notes on #5
What we’ve shown in this problem is that
-40° means exactly the same thing on either
scale (-40°C or -40°F).
In general, it does make a difference
whether the temperature reading is in
Celsius degrees or Fahrenheit degrees.
© 2007 Herbert I. Gross