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next Key Stone Problem… Set 22 © 2007 Herbert I. Gross next Instructions for the Keystone Problem You will soon be assigned problems to test whether you have internalized the material in Lesson 22 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. © 2007 Herbert I. Gross next 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. © 2007 Herbert I. Gross next Keystone Problem for Lesson 22 For what values of x, y, and z is it true that… 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 30 © 2007 Herbert I. Gross next Solution Using the method described in Lesson 21, we can eliminate x from all but the top equation of the system… 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 30 Namely, since the coefficients of x are 3, 2 and 5, we know that 30 is a common multiple of them. © 2007 Herbert I. Gross next Solution 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 30 Therefore, we can multiply both sides of the top equation of the system above by 10; the middle equation by -15; and the bottom equation by -6 to obtain the equivalent system… © 2007 Herbert I. Gross -30x + -10y + -20z = -130 -30x + -60y + -15z = -225 -30x + -30y + -18z = -180 next Solution -30x + -10y + -20z = -130 -30x + -60y + -15z = -225 -30x + -30y + -18z = -180 So to eliminate x, we replace the middle equation in the system above by the sum of middle equation and the top equation, and we replace the bottom equation by the sum of the bottom equation and the top equation to obtain… © 2007 Herbert I. Gross -30x + -10y + -20z = -130 -30x -5z = -225 - 95 + --60y 50y ++-15z -30x -2z = -180 -50 + -30y 20y + -18z next -30x Solution + -10y + -20z = -130 -50y + -5z = - 95 -20y + -2z = -50 If we prefer to work with smaller numbers, we may divide both sides of the top equation in system above by 10, the middle equation by -5, and the bottom equation by -2 to obtain… -30x + 3x -10y + y+ +-20z 2z == -130 13 -z -50y 10y+ + -5z = - 19 95 -z = -25 -20y 10y+ + -2z = 50 © 2007 Herbert I. Gross next Solution 3x + y + 2z = 13 10y + -z = 19 10y + -z = 25 Notice that the middle equation and the bottom equation in the system above are contradictions of one another. © 2007 Herbert I. Gross next 3x + y + 2z = 13 10y + -z = 19 Solution 10y + -z = 25 If we didn’t notice this, our next step would be to replace the bottom equation in our system by the bottom equation minus the middle equation to conclude that… -10y © 2007 Herbert I. Gross + -z = -19 10y + -z = -25 -0 = -6 next Solution 3x + y + 2z = 13 10y + -z = 19 -0 = 6 Since the bottom equation in the above system is a false statement, it tells us that there are no numbers x, y, and z that are solutions to our system… 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 30 © 2007 Herbert I. Gross next Notes 3x + y + 2z = 13 10y + -z = 19 -0 = 6 The bottom equation in the system above often elicits such comments as… “How in the world is it possible for 0 to equal 6?” The answer, of course, is that it can’t, and that’s exactly the point. © 2007 Herbert I. Gross next Notes Namely what we have shown is that if there were numbers x, y, and z that satisfied the system… 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 30 it would mean that 0 = 6. Since 0 = 6 is a false statement, we have to conclude that there are no values of x, y and z that satisfy our system. © 2007 Herbert I. Gross next More generally, if event A Review happening guarantees that event B also has to happen, then if event B doesn’t happen, it means that event A didn’t happen. As a non-mathematical example, suppose a person says… “Whenever the roads are icy, I don’t drive my car.” So if he’s telling the truth, and if we see the person driving his car, we know that the roads are not icy. © 2007 Herbert I. Gross next Review However, the above logic is subtle and as a result it is often misused. For example, the statement, “If I get 100 on the final exam, the professor will give me an A”, means the same thing as, “ If the professor didn’t give me an A, I didn’t get 100 on the final exam”. However, it doesn’t mean that if I didn’t get 100 on the final exam the professor didn’t give me an A! (For example, you might have got 95 on the final exam and still got an A is the course). © 2007 Herbert I. Gross next Notes Let’s apply the above discussion to solving an algebraic equation. When we say that the solution of the equation x + 3 = 5 is x = 2, what we have really proven is that is if x ≠ 2, then x + 3 ≠ 5. If x = 2, we still have to check that x + 3 = 5. © 2007 Herbert I. Gross Let’s revisit the system… next Notes 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 30 and see if we can determine why it had no solutions. To this end, look what happens when we add the top two equations. 3x + 2y + 2z = 13 2x + 4y + 2z = 15 © 2007 Herbert I. Gross 5x + 5y + 3z = 28 next 3x + y + 2z = 13 Notes 2x + 4y + z = 15 5x + 5y + 3z = 30 That is, if x, y, and z satisfy the top two equations in our system, it means that 5x + 5y + 3z has to equal 28; thus making the bottom equation a false statement. In other words, the bottom equation in our system is not compatible with the top two equations. © 2007 Herbert I. Gross next Discussion A follow-up discussion about the above system might be, “But what would happen if 5x + 5y + 3z = 28?” That is, suppose we wanted to find values of x, y, and z that satisfied the system… 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 28 © 2007 Herbert I. Gross next 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 28 To solve the system above, we begin by “ignoring” the bottom equation, since it tells us nothing that we didn’t already know from the top two equations. In other words, our system now can be replaced by the abridged system… 3x + y + 2z = 13 2x + 4y + z = 15 © 2007 Herbert I. Gross next 3x + y + 2z = 13 The fact that our system 2x + 4y + z = 15 has 3 variables but only 2 conditions indicates that we can pick any value we wish for one of the variables and that will determine the values of the remaining two variables. (That is, the original system has one degree of freedom.) For example, suppose we want the value of x to be 1. If we replace x by 1 in the system we obtain… 3 + y + 2z = 13 2 + 4y + z = 15 © 2007 Herbert I. Gross or y + 2z = 10 4y + z = 13 next y + 2z = 10 4y + z = 13 To solve the system, we can first multiply the bottom equation by 2 to obtain… and if we then subtract the top equation in the system from the bottom equation we see that… -y + -2z -10 = 10 4y ++ 2z z = 13 8y 26 7y + 0 = 16 or © 2007 Herbert I. Gross y = 16/7 next y + 2z = 10 4y + z = 13 If we now replace y by 16/7 in the top equation of system, we see that… 16/y 7 + 2z = 10 2z = 10 – 16/7 2z = 70/7 – 16/7 2z = 54/7 © 2007 Herbert I. Gross z = 27/7 next 3x + y + 2z = 13 2x + 4y + z = 15 As a check, we replace x by 1, y by 16/7, and z by 27/7 in the system to verify that… 3(1) + (16/7) + 2(27/7) = 13 2(1) + 4(16/7) + (27/7) = 15 © 2007 Herbert I. Gross next More specifically… 3(1) + (16/7) + 2(27/7) = 13 3 + 16/7 + 54/7 = 13 3 + 70/7 = 13 3 + 10 = 13 and… 2(1) + 4(16/7) + (27/7) = 15 2 + 64/7 + 27/7 = 15 2 + 91/7 = 15 2 + 13 = 15 © 2007 Herbert I. Gross next Finally, the fact that the truth of the bottom equation in our system 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 28 follows inescapably from the truth of the top two equations means that we do not have to check to see whether the choices for x, y, and z that satisfy the top two equations also satisfy the bottom equation. © 2007 Herbert I. Gross next However, as a double check we see that with these choices for x, y and z. 5x + 5y + 3z = 28 becomes… 5(1) + 5(16/7) + 3(27/7) = 28 5 + 80/7 + 81/7 = 28 5 + 161/7 = 28 5 + 23 = 28 © 2007 Herbert I. Gross next The solution for the system 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 28 was based on our choosing x to be 1. The same procedure would have applied for any value we had chosen for x. © 2007 Herbert I. Gross next In other words, while there are no values of x, y, and z that satisfy the system 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 30 there are infinitely many sets of values for x, y and z that satisfy the system 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 28 © 2007 Herbert I. Gross next Namely we may choose a value at random for x, (or for that matter either y or z), and once we do this, the first two equations in system 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 28 become a system of 2 equations in 2 unknowns from which we can determine the values of the other two variables. © 2007 Herbert I. Gross next For example, suppose we want the value of x to be x0. If we replace x by x0 in the system we obtain… 3x0 + y + 2z = 13 2x0 + 4y + z = 15 or y + 2z = 13 – 3x0 4y + z = 15 – 2x0 © 2007 Herbert I. Gross next y + 2z = 13 – 3x0 4y + z = 15 – 2x0 or y + 2z = 13 + -3x0 4y + z = 15 + -2x0 To solve the above system for y, we could multiply the bottom equation by 2 to obtain… And then subtract the top equation from the bottom equation to obtain… -y y++-2z 2z==-13 13 ++ +-3x0 -8y 4y++-2zz == -30 15 7y © 2007 Herbert I. Gross + -4x 2x0 = 17 + -x0 7y = 17 – x0 next 7y = 17 –x0 And if we now divide both sides of the above equation by 7, we see that ….. y = (17 – x0)/7 And if we now replace y in the equation y + 2z = 13 – 3x0 by its above value, we see that ….. (17 – x0)/7 + 2z = 13 – 3x0 next (17 – x0)/7 + 2z = 13 – 3x0 2z = 13 – 3x0 – (17 – x0)/7 14z = 91 – 21x0 – (17 –x0) 14z = 91 – 21x0 – 17 + x0 14z = 74 – 20x0 7z = 37 – 10x0 z = (37 – 10x0)/7 next In summary, once we choose a value of x at random, the system of equations…. 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 28 will have a solution if and only if… y = (17 – x)/7 and z = (37 – 10x)/7 © 2007 Herbert I. Gross next y = (17 – x)/7 and z = (37 – 10x)/7 As a “plausibility” check, if we replace x by 1, we see from the above equations that if x = 1, then y = 16/7 and z = 27/7… …and this agrees with the result we obtained previously. next Summary To generalize what we have demonstrated in Lesson 21 and 22, it turns out that given a system of three linear equations in three unknowns, one and only one of the following conditions can be true. next Case 1 There will be one and only one set of values for x, y, and z that satisfy each of the 3 equations. This situation (which occurred in Lesson 21) will occur whenever the three equations are neither redundant nor contradictory. next Cases 2 and 3 will occur when the system has one or more degrees of freedom (as was the case in this lesson). More specifically… Case 2 The system has no solution. That is, there are no values of x, y, and z that satisfy the given system of equations. This will occur when at least one of the equations contradicts the given information in the other equations. next Case 3 There are infinitely many sets of values for x, y, and z that satisfy the given system of equations. This situation will occur whenever the truth of at least one of the three equations follows inescapably from the truth of the other equations. In other words in this situation, the solution set of the system has one or more degrees of freedom. next Concluding Note Although the demonstration is beyond the scope of this course, our summary remains valid for any linear system of equations having “n” unknowns. More specifically, one and only one of the following situations can occur… next The solution set has one and only one member. That is, there is only one set of values for the variables that satisfies each of the given equations. The solution set is empty. That is, there is no set of values for the variables that satisfies each of the given equations. In other words the constraints are contradictory. The solution set has infinitely many members. That is, we may choose one or more variables at random whereupon the remaining variables are uniquely determined.