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Key Stone Problem… Key Stone Problem… next Set 21 © 2007 Herbert I. Gross.

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Presentation on theme: "Key Stone Problem… Key Stone Problem… next Set 21 © 2007 Herbert I. Gross."— Presentation transcript:

1 Key Stone Problem… Key Stone Problem… next Set 21 © 2007 Herbert I. Gross

2 You will soon be assigned problems to test whether you have internalized the material in Lesson 21 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

3 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

4 next © 2007 Herbert I. Gross Two numbers are defined implicitly by the information that (1) the sum of twice the first number plus the second number is 13, and (2) the sum of 3 times the first number plus twice the second number is 21. If we use x to denote the first number and y to denote the second number, in explicit form, find the value of x and y. next Keystone Problem for Lesson 21 next

5 © 2007 Herbert I. Gross Let’s begin by translating the two constraints into algebraic equations. Solution next If x denotes the first number, then 2x denotes twice the first number; and if y denotes the second number then 2x + y denotes the sum of twice the first number plus the second number. Since we are told that this sum is 13, our first algebraic equation is… 2x + y = 13

6 next © 2007 Herbert I. Gross Solution Again letting x represent the first number and y represent the second number; 3x represents 3 times the first number, 2y represents twice the second number and 3x + 2y represents the sum of 3 times the first number plus twice the second number. Since we are told that this sum is 21, our second algebraic equation is… 3x + 2y = 21

7 next © 2007 Herbert I. Gross Solution Since x and y must satisfy both equations they must be the solution of the linear system… 3x + 2y = 21 2x + y = 13 To solve our linear system we use our rules to eliminate either x or y. If we want to eliminate x, one way to do it is to multiply the top equation by - 3, - 6x + - 4y = x + - 3y = - 39 and the bottom equation by 2 to obtain the equivalent system… next

8 © 2007 Herbert I. Gross Solution If we now add the top and bottom equations in our system the terms involving x “cancel”, and we see that y = 3. If we then replace y by 3 in the equation 2x + y = 13, we see that 2x + 3 = 13, which in turn means that 2x = 10 or x = 5. next - 6x + - 4y = x + - 3y = y = 3

9 next © 2007 Herbert I. Gross Solution As a final check, in the equation 3x + 2y = 21, we replace x by 5 and y by 3, next 3x + 2y = 21 2x + y = 13 and verify that it is a true statement. 3 x + 2 y = 21 (5)(3) = 21

10 next © 2007 Herbert I. Gross Notes on our Solution A more formal (but admittedly a much longer) way to present the above solution is in terms of continuing the process of replacing one system of equations by an equivalent system of equations. For example, in our solution we started with the system… next 3x + 2y = 21 2x + y = 13

11 next © 2007 Herbert I. Gross Notes on our Solution We then multiplied both sides of the top equation by - 3 and the bottom equation by 2 to obtain the equivalent system… 3x + 2y = 21 2x + y = x + - 4y = x + - 3y = - 39

12 next © 2007 Herbert I. Gross Notes on our Solution We could continue in this manner by replacing the bottom equation in the system above by the sum of the bottom equation and the top equation to obtain the equivalent system of equations… - 6x + - 4y = x + - 3y = y = x + - 3y = - 39

13 next © 2007 Herbert I. Gross Notes on our Solution …and if we then divide both sides of the top equation above by - 3 we obtain the equivalent system… - y = x + - 3y = - 39y = 3 2x + y = 13

14 next © 2007 Herbert I. Gross Notes on our Solution While both systems look quite different they are equivalent in the sense that they possess the same solution set. y = 3 2x + y = 13 That is, the ordered pair (x,y) is a solution of the first system if and only if it is also a solution of second system. However, the second system is easier to solve than the first system because the bottom equation tells us explicitly what number y has to be. 3x + 2y = 21 2x + y = 13 next

15 © 2007 Herbert I. Gross Notes on our Solution We can continue this process by replacing our system by an equivalent system in which the top equation contains only x as the variable. More specifically, we can multiply both sides of the bottom equation above by - 1 to obtain the equivalent system… y = 3 2x + y = 13 - y = - 3 2x + y = 13

16 next © 2007 Herbert I. Gross Notes on our Solution Next we can replace the top equation in our system by the sum of the top and bottom equation - y = - 3 2x + y = 13 - y = - 3 2x = 10 2x + 0 = 10 to obtain the equivalent system… next

17 © 2007 Herbert I. Gross Notes on our Solution Finally, we may divide both sides of the top equation in our system by 2 y = 3 x = 5 - y = - 3 2x = 10 and the bottom equation by - 1 to obtain the equivalent system of equations…

18 next © 2007 Herbert I. Gross y = 3 x = 5 There is a tendency to think of the system 3x + 2y = 21 2x + y = 13 as being the solution of the system In actuality, it is just an equivalent system of equations that happens to be “trivial” to solve. next

19 © 2007 Herbert I. Gross It is similar to asking a question of the type, “What was the color of Paul Revere’s white horse?” Namely, it’s still a question but one that already contains its own answer. To continue the above analogy, suppose we were told that the color of Paul Revere’s horse was the color of freshly fallen snow. next This would implicitly tell us that in explicit terms the color of the horse was white.

20 next © 2007 Herbert I. Gross In this context the system… 3x + 2y = 21 2x + y = 13 …tells us implicitly what the system… tells us explicitly. y = 3 x = 5

21 next © 2007 Herbert I. Gross The above discussion becomes more visual if we think in terms of the geometric graphs of the two equations in the system… 3x + 2y = 21 2x + y = 13

22 next © 2007 Herbert I. Gross next Namely, the geometric graph of the set consisting of all ordered pairs (x,y) for which 2x + y = 13 is the line L 1 (which is determined by the two intercepts (0,13) and (6.5,0)). In a similar way the graph of the set of ordered pairs (x,y) for which 3x + 2y = 21 is the line L 2 (which is determined by the two intercepts (0,10.5) and (7,0)).

23 © 2007 Herbert I. Gross The graph of 2x + y = 13 is (0,13) (6.5,0) next L1L1 When x = 0, y = 13 When y = 0, x = 6.5

24 © 2007 Herbert I. Gross The graph of 3x – 2y = 21 is (0,10.5) (7,0) next When x = 0, y = 10.5 When y = 0, x = 7 L2L2

25 © 2007 Herbert I. Gross (0,13) (6.5,0) next L1L1 (0,10.5) (7,0) L2L2 Clearly these two lines meet at one and only one point, and we have shown that this point is (5,3). (5,3) 2x + y =10 2x + 3y =13 next

26 © 2007 Herbert I. Gross In our graphical solution, we already knew from the algebraic solution that the point of intersection of the two lines was (5,3). However, notice from the graph that if we didn’t already know this fact, it would have been difficult to pinpoint exactly where the two lines meet. next Note For example, by looking at the graph, it is difficult to distinguish between such points as (5,3) and (4.98,3.01).

27 next © 2007 Herbert I. Gross Thus, whenever possible, especially when an exact answer is required, we should strive for an algebraic solution. next Note However, some problems are too complex to allow for an exact algebraic solution. In those cases, we use graphical methods to approximate the exact answer, and then use trial-and-error to refine our approximation.

28 next © 2007 Herbert I. Gross next The following systems… each represents a pair of lines whose point of intersection is (5,3). - 6x + - 4y = x + - 3y = y = x + - 3y = - 39 y = 3 2x + y = 13 - y = - 3 2x + y = 13 - y = - 3 2x = 10 y = 3 x = 5 3x + 2y = 21 2x + y = 13

29 next © 2007 Herbert I. Gross next In particular, the above system represents the intersection of the vertical line x = 5 and the horizontal line y = 3. (5,3) x = 5 y = 3 next y = 3 x = 5

30 next © 2007 Herbert I. Gross next A nice feature about mathematical proofs is that there is usually only one correct answer to a question but never just one correct way to find the answer. For example, we chose to eliminate x in the system… …in order to first find the value of y. It would have been just as logical to first eliminate y and instead find the value of x. 3x + 2y = 21 2x + y = 13 This would have led to a different set of equivalent systems, but each system would have as its solution set {(5,3)}. next

31 © 2007 Herbert I. Gross In fact we could have rewritten 2x + y = 13 in the form y = 13 – 2x and then replaced y by 13 – 2x in the equation 3x + 2y = 21 to obtain the equation… which is linear in x. 3x + 2(13 – 2x) = 21 next Solving this equation leads to the steps… 3x + 2(13 – 2x) = 21 3x + 26 – 4x = 21 3x– 4x + 26 = 21 - x = - 5 x = 5

32 next © 2007 Herbert I. Gross Rewriting the linear equations in the form y = mx + b, often referred to as the method of substitution, does not lend itself very well to situations that deal with linear systems of equations that have more than two unknowns. next Note In fact, even in the case of two unknowns, the arithmetic can become a bit “messy”. For example, 3x + 7y = 9 would become y = -3 / 7 x + 9 / 7.

33 next © 2007 Herbert I. Gross next Systems of equations often arise when we try to solve real world” problems algebraically. For example, the system 3x + 2y = 21 2x + y = 13 2 boxes of chocolate chip cookies and 1 box of mint chip cookies cost $13. However, 3 boxes of chocolate chip cookies and 2 boxes of mint chip cookies cost $21. How much does each box of mint chip cookies cost? next might arise if we were trying to solve the following problem…

34 © 2007 Herbert I. Gross Sometimes using the generic letters x and y causes us to confuse which letter represents the cost of a box of chocolate chip cookies and which letter represents the cost of a box of mint chip cookies. In this context, it might be easier to keep track of the two items if we were to use, say, c to represent the cost of a box of chocolate chip cookies and m represent the cost of a box of mint chip cookies. In that way our system… 3x + 2y = 21 2x + y = 13 next 3c + 2m = 21 2c + m = 13 becomes…

35 next © 2007 Herbert I. Gross When we solve this system and find that c = 5 and m = 3, we immediately know that the cost of each box of mint chip cookies (that is: m) is $3. As a final note, keep in mind that the “short cut” solution we gave earlier for this exercise is the one that is most often used in solving systems of equations. next Note

36 next © 2007 Herbert I. Gross However, it’s important to understand the sequence by which the implicit system is translated, step by step, into the explicit system (such as how we transformed… Note into 3x + 2y = 21 2x + y = 13 y = 3 x = 5 by a sequence of equivalent systems of equations).


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