Key Stone Problem… Key Stone Problem… next 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 next © 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. next © 2007 Herbert I. Gross

next © 2007 Herbert I. Gross next 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?

next © 2007 Herbert I. Gross next 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.

© 2007 Herbert I. Gross next 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.

© 2007 Herbert I. Gross Method 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 where the constraints are… B + C = 3A A + C = 5B next Algebraic Solution

next © 2007 Herbert I. Gross If we replace B + C in T = A + B + C next by its value B + C = 3A, we see that… T = A + (B + C) T = A + 3A A = 1 / 4 T

next © 2007 Herbert I. Gross If we replace A + C in T = A + B + C next by its value A + C = 5B, we see that… T = (A + C) + B T = 5B + B B = 1 / 6 T

next © 2007 Herbert I. Gross If we replace A and B in T = A + B + C next by their values A = 1 / 4 T and B = 1 / 6 T, we see that… T = 1 / 4 T + 1 / 6 T + C T = 5 / 12 T + C C = T – 5 / 12 T C = 7 / 12 T

next © 2007 Herbert I. Gross Using the results B = 1 / 6 T and C = 7 / 12 T we see that… C = B next Thus, the equation tells us that altogether Alice, Betty and Carol have $ / 12 T = 1 / 6 T / 12 T – 1 / 6 T = / 12 T – 2 / 12 T = / 12 T = 180 T = 12 / 5 × 180 T = 432

next © 2007 Herbert I. Gross For example… The equation A = 1 / 4 T 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 / 6 T tells us that Betty has 1 / 6 (or 2 / 12 ) of the total sum. The equation C = 7 / 12 T 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 next

© 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 bread denote the total sum of money the three girls have, we may assume that it is divided into 12 equally sized pieces as follows… BBAAACCCCCCC next Thus, independently, of the total amount of money they have, the ratio of their sums, A:B:C is 3:2:7.

© 2007 Herbert I. Gross However, this is far as we can go without additional information. next 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 / 12 T – 1 / 6 T, and as we saw above, this led to the equation T = 432.

© 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.

© 2007 Herbert I. Gross Hence… next A 108 B 72 C 252 B A + C 360 5B 360 B + C 324 3A 324 A + B + C 432 A = 3 pieces × 36 = 108 B = 2 pieces × 36 = 72 C = 7 pieces × 36 = 252 next C = B B + C = 3A A + C = 5B C 252 5B 360 B + C 324 Check… next

© 2007 Herbert I. Gross Method 2 From the given information we have the following system of linear equations… C = B B + C = 3A A + C = 5B next A System of Three Linear Equations

next © 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… next C = B B + C = 3A A + C = 5B C – B - = 180 C + B + - 3A = 0 C + - 5B + A = 0

next © 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… next 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 A = 0

next © 2007 Herbert I. Gross Next we may replace the second equation in the system by the second plus the first… next C + - B = C + - B + 3A = 0 C + - B = C + - B + 3A = 0 - C + 5B + - A = 0 C + - B = C + - B + 3A = 0 - C + 5B + - A = 0 - 2B + 3A = 180 next

© 2007 Herbert I. Gross …and replace the third equation by the third plus the first to obtain the equivalent system… next C + - B = C + 5B + - A = 0 C + - B = B + 3A = C + 5B + - A = 0 C + - B = B + 3A = C + 5B + - A = 0 4B + - A = 180 next

© 2007 Herbert I. Gross We may then multiply both sides of the middle equation in the system by 2 to obtain the equivalent system… next C + - B = 180 4B + - A = B + 3A = B + 6A = B + 3A = 180 C + - B = 180 4B + - A = 180

next © 2007 Herbert I. Gross Next we replace the third equation in the system by the third plus the second to obtain the equivalent system… next - 4B + 6A = 360 4B + - A = 180 C + - B = B + 6A = 360 4B + - A = 180 C + - B = B + 6A = 360 4B + - A = 180 5A = 540 next

© 2007 Herbert I. Gross The bottom equation in the system tells us that A = 108. That is, Alice has $108. next C + - B = B + 6A = 360 5A = B + 6A = 360 Once we know that A = 108, we may replace A by 108 in the middle equation of the system above to obtain… - 4B + 6(108) = B = B = 360 – B = B = 72

next © 2007 Herbert I. Gross And if we then replace B by 72 in the top equation of the system, we see that… next C + - B = B + 6A = 360 5A = 540 C + - B = 180 Therefore, the total amount of money they have (A + B + C) is… $108 + $72 + $252 = $432 C = 180 C = C = 252 next

© 2007 Herbert I. Gross Method 3 Starting with the three constraints… C = B B + C = 3A A + C = 5B next Substitution

© 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 to obtain… B + C = 3A next B + (B + 180) = 3A 2B = 3A 2B + - 3A = and… A + C = 5B A + (B + 180) = 5B A = 4B 180 = 4B – A next 4B + - A = 180

next © 2007 Herbert I. Gross Equations 2B + - 3A = and 180 = 4B + - A constitute the linear system… next 2B + - 3A = B + - A = To eliminate B, multiply both sides of the top equation in our system by - 2 to obtain the equivalent system… 2B + - 3A = B + - A = B + 6A = 360

next © 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… next - 4B + 6A = 360 4B + - A = B + 6A = 360 2B + - 3A = B + - A = A = 540 next

© 2007 Herbert I. Gross Since 5A = 540, A = 108. next - 4B + 6A = 360 5A = B + 6A = 360 next We can then replace A by 108 in the top equation above to find that B = 72… - 4B + 6(108) = B = B = 360 – B = 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. Summary next

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. Summary