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You will soon be assigned problems to test whether you have internalized the material
in Lesson 24 of our algebra course. The Keystone Illustration below is a
prototype 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 Problem
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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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© 2007 Herbert I. Gross
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In the keystone exercise for Lesson 23 we showed several different, but
equivalent, non-algebraic ways to obtain the answer to the problem…
Three girls shared a sum of money. Cathy received $180 more than Betty.
The total amount Betty and Cathy received was 3 times the amount Alice received.
The total amount Alice and Cathy received was 5 times the amount Betty received.
What was the total sum of the money that was shared by all three girls?
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© 2007 Herbert I. Gross
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Each method led to the result that the three girls had a total of $432.
In the present keystone exercise, we shall revisit the above problem from a purely algebraic point of view. We will find that
the expression “Different strokes for different folks” even applies to using
algebra in order to solve word problems.
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© 2007 Herbert I. Gross
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Different people may approach the same question in different ways and as a result there are often many different but correct
ways to achieve its solution.
In this exercise, we will illustrate this by
presenting 3 different algebraic solutions.
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© 2007 Herbert I. Gross
Method 1Method 1
In the algebraic solution, we will use A to denote the sum of money Alice has;
B, the sum of money Betty has; C, the sum of money Cathy has; and
T, the total sum of money they have. Then…
T = A + B + C
C = B + 180where the
constraints are… B + C = 3A
A + C = 5B
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Algebraic SolutionAlgebraic Solution
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© 2007 Herbert I. Gross
If we replace B + C in
T = A + B + C
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by its value B + C = 3A, we see that…
T = A + (B + C)
T = A + 3A
A = 1/4T
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© 2007 Herbert I. Gross
If we replace A + C in
T = A + B + C
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by its value A + C = 5B, we see that…
T = (A + C) + B
T = 5B + B
B = 1/6T
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© 2007 Herbert I. Gross
If we replace A and B in
T = A + B + C
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by their values A = 1/4T and B = 1/6T, we see that…
T = 1/4T + 1/6T + C
T = 5/12T + C
C = T – 5/12T
C = 7/12T
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© 2007 Herbert I. Gross
Using the results B = 1/6T and C = 7/12T we see that…
C = B + 180
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Thus, the equation tells us that altogether Alice, Betty and Carol have $432.
7/12T = 1/6T + 1807/12T – 1/6T = 180
7/12T – 2/12T = 1805/12T = 180
T = 12/5 × 180
T = 432
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© 2007 Herbert I. Gross
For example…
The equation A = 1/4T tells us that no matter how much money the three girls have; Alice
has 1/4 ( or 3/12) of the total sum.
The equation B = 1/6T tells us that Betty has 1/6 (or 2/12) of the total sum.
The equation C = 7/12T tells us that Cathy has 7/12 of the total sum.
Using Method 1 may have seemed quite lengthy, but it
yielded much more information than simply the correct answer.
Notes on Method 1
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© 2007 Herbert I. Gross
In terms of our “corn bread” model that was discussed in the keystone exercise of
the previous lesson, if we let the corn breaddenote the total sum of money the three
girls have, we may assume that it is divided into 12 equally sized pieces as follows…
B BA A A C C C C C C C
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Thus, independently, of the total amount of money they have, the ratio of their sums,
A:B:C is 3:2:7.
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© 2007 Herbert I. Gross
However, this is far as we can go without additional information.
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The additional information comes in the form that Cathy has $180 more than Betty;
that is C – B = 180. In algebraic terms, this leads to the equation 7/12T – 1/6T, and as we saw above, this led to the equation
T = 432.
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© 2007 Herbert I. Gross
Again in terms of the corn bread…next
A = 3 pieces, B = 2 pieces, and C = 7 pieces.
Therefore, C – B = 7 pieces – 2 pieces = 5 pieces
And if 5 pieces = 180, each piece represents $36.
Hence, the corn bread represents 12 × $36 or $432.
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© 2007 Herbert I. Gross
Hence…
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AA
108108
BB
7272
CC
252252
B + 180B + 180
252252
A + CA + C
360360
5B5B
360360
B + CB + C
324324
3A3A
324324
A + B + CA + B + C
432432
A = 3 pieces × 36 = 108
B = 2 pieces × 36 = 72
C = 7 pieces × 36 = 252
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C = B + 180
B + C = 3A
A + C = 5B
CC
252252
5B5B
360360
B + CB + C
324324
Check…nextnextnext
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© 2007 Herbert I. Gross
Method 2Method 2
From the given information we have the following system of linear equations…
C = B + 180
B + C = 3A
A + C = 5B
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A System of Three Linear EquationsA System of Three Linear Equations
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© 2007 Herbert I. Gross
In more traditional format, we may rewrite the system in a way that all the variables appear on the left hand side of the equations. That is…
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C = B + 180
B + C = 3A
A + C = 5B
C – B - = 180
C + B + -3A = 0
C + -5B + A = 0
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© 2007 Herbert I. Gross
We may then multiply both sides of the bottom
two equations in our system by -1 to obtain the
equivalent system…
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C – B - = 180
C + B + -3A = 0
C + -5B + A = 0
C + -B -- = 180
C + B + -3A = 0
C + -5B + A = 0
-C + -B + 3A = 0
-C + 5B + 3-A = 0
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© 2007 Herbert I. Gross
Next we may replace the second equation in the system by the second plus the first…
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C + -B = 180-C + -B + 3A = 0
C + -B = 180-C + -B + 3A = 0-C + 5B + -A = 0
C + -B = 180-C + -B + 3A = 0-C + 5B + -A = 0
- 2B + 3A = 180- 2B + 3A = 180
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© 2007 Herbert I. Gross
…and replace the third equation by the third plus
the first to obtain the equivalent system…
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C + -B = 180-C + 5B + -A = 0
C + -B = 180 -2B + 3A = 180
-C + 5B + -A = 0
C + -B = 180 -2B + 3A = 180-C + 5B + -A = 0
4B + -A = 1804B + -A = 180
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© 2007 Herbert I. Gross
We may then multiply both sides of the middle equation in the system
by 2 to obtain the equivalent system…
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C + -B = 180
4B + -A = 180
-2B + 3A = 180-4B + 6A = 360
-2B + 3A = 180
C + -B = 180
4B + -A = 180
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© 2007 Herbert I. Gross
Next we replace the third equation in the
system by the third plus the second to obtain the
equivalent system…
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-4B + 6A = 360 4B + -A = 180
C + -B = 180 -4B + 6A = 360
4B + -A = 180
C + -B = 180 -4B + 6A = 360
4B + -A = 180
5A = 5405A = 540
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© 2007 Herbert I. Gross
The bottom equation in the system tells us that
A = 108 . That is, Alice has $108.
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C + -B = 180 -4B + 6A = 360
5A = 540
-4B + 6A = 360Once we know that A = 108, we
may replace A by 108 in the middle equation of the
system above to obtain…
-4B + 6(108) = 360-4B + 648 = 360
-4B = 360 – 648 -4B = -288
B = 72
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© 2007 Herbert I. Gross
And if we then replace B by 72 in the top equation of the system, we see that…
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C + -B = 180 -4B + 6A = 360
5A = 540
C + -B = 180
Therefore, the total amount of money they have (A + B + C) is…
$108 + $72 + $252 = $432
C + -72 = 180C = 180 + 72
C = 252
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© 2007 Herbert I. Gross
Method 3Method 3
Starting with the three constraints…
C = B + 180
B + C = 3A
A + C = 5B
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SubstitutionSubstitution
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© 2007 Herbert I. Gross
We may replace C in the equations B + C = 3A and A + C = 5B by its value in the equation C = B + 180 to obtain…
B + C = 3A
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B + (B + 180) = 3A
2B + 180 = 3A
2B + -3A = -180
and…
A + C = 5B
A + (B + 180) = 5B
A + 180 = 4B
180 = 4B – A
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4B + -A = 180
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© 2007 Herbert I. Gross
Equations 2B + -3A = -180 and 180 = 4B + -Aconstitute the linear system…
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2B + -3A = -1804B + -A = -180
To eliminate B, multiply both sides of the top equation in our system by -2 to obtain
the equivalent system…
2B + -3A = -1804B + -A = 180
-4B + 6A = 360
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© 2007 Herbert I. Gross
And if we now replace the bottom equation in the system by the sum of the two equations, we obtain the equivalent system…
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-4B + 6A = 3604B + -A = 180
4B + -A = 180
-4B + 6A = 360
2B + -3A = -1804B + -A = -180
5A = 5405A = 540
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© 2007 Herbert I. Gross
Since 5A = 540, A = 108.
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-4B + 6A = 360
5A = 540
-4B + 6A = 360
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We can then replace A by 108 in the top equation above to find that B = 72…
-4B + 6(108) = 360-4B + 648 = 360
-4B = 360 – 648-4B = -288
B = 72
In the solution of the keystone problem for both this lesson and Lesson 23, we have shown several ways, both algebraic and
otherwise, for approaching the solution to a “word problem”.
The nice thing about trial and error is that you do not need to have a great
mathematics background in order to be able to use this method.
SummarySummarynextnext
The nice thing about the algebraic method is that it is often more efficient than trial and error, and unlike with trial and error,
the algebraic solution can tell us how many numbers are in the solution set of
the equation.
Try to internalize both methods so that you have confidence when you set out to
solve any word problem.
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SummarySummary