key stone problem… key stone problem… next set 24 © 2007 herbert i. gross

Key Stone Problem…Key Stone Problem…

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Set 24© 2007 Herbert I. Gross

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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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…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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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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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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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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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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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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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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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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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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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

key stone problem… key stone problem… next set 24 © 2007 herbert i. gross

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