Key Stone Problem… Key Stone Problem… next Set 22 © 2007 Herbert I. Gross

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. 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 For what values of x, y, and z is it true that… Keystone Problem for Lesson 22 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 30

next © 2007 Herbert I. Gross Solution Using the method described in Lesson 21, we can eliminate x from all but the top equation of the system… next Namely, since the coefficients of x are 3, 2 and 5, we know that 30 is a common multiple of them. 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 30

next © 2007 Herbert I. Gross next Therefore, we can multiply both sides of the top equation of the system above by 10; Solution 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 30 the middle equation by - 15; - 30x y z = x y z = x y z = and the bottom equation by - 6 to obtain the equivalent system…

next © 2007 Herbert I. Gross next So to eliminate x, we replace the middle equation in the system above by the sum of middle equation and the top equation, Solution - 30x y z = x y z = x y z = and we replace the bottom equation by the sum of the bottom equation and the top equation to obtain… - 30x y z = x y z = x y z = y + - 5z = y + - 2z = - 50 next

© 2007 Herbert I. Gross next If we prefer to work with smaller numbers, we may divide both sides of the top equation in system above by 10, Solution the middle equation by - 5, - 30x y z = y + - 5z = y + - 2z = - 50 next and the bottom equation by - 2 to obtain… - 30x y z = y + - 5z = y + - 2z = x + y + 2z = 13 10y + - z = 19 10y + - z = 25 next

© 2007 Herbert I. Gross Notice that the middle equation and the bottom equation in the system above are contradictions of one another. Solution 3x + y + 2z = 13 10y + - z = 19 10y + - z = 25

next © 2007 Herbert I. Gross Solution 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… 3x + y + 2z = 13 10y + - z = 19 10y + - z = 25 next 10y + - z = y + - z = = - 6

next © 2007 Herbert I. Gross 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… Solution next 3x + y + 2z = 13 10y + - z = = 6 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 30

next © 2007 Herbert I. Gross The bottom equation in the system above often elicits such comments as… next Notes 3x + y + 2z = 13 10y + - z = = 6 “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. next

© 2007 Herbert I. Gross Namely what we have shown is that if there were numbers x, y, and z that satisfied the system… next Notes it would mean that 0 = 6. 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 30 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.

next © 2007 Herbert I. Gross More generally, if event A happening guarantees that event B also has to happen, then if event B doesn’t happen, it means that event A didn’t happen. next Review 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. next

© 2007 Herbert I. Gross However, the above logic is subtle and as a result it is often misused. next Review 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). next

© 2007 Herbert I. Gross Let’s apply the above discussion to solving an algebraic equation. next 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. Notes

next © 2007 Herbert I. Gross Let’s revisit the system… next and see if we can determine why it had no solutions. Notes 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 30 To this end, look what happens when we add the top two equations. 3x + 2y + 2z = 13 2x + 4y + 2z = 15 5x + 5y + 3z = 28 5x + 5y + 3z = 30 next

© 2007 Herbert I. Gross 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. Notes 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 30 In other words, the bottom equation in our system is not compatible with the top two equations. next

© 2007 Herbert I. Gross 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… Discussion 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 28

next © 2007 Herbert I. Gross In other words, our system now can be replaced by the abridged system… 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 28 next 3x + y + 2z = 13 2x + 4y + z = 15 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.

next © 2007 Herbert I. Gross The fact that our system 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.) next 3x + y + 2z = 13 2x + 4y + z = 15 or 3 + y + 2z = y + z = 15 For example, suppose we want the value of x to be 1. If we replace x by 1 in the system we obtain… y + 2z = 10 4y + z = 13

next © 2007 Herbert I. Gross To solve the system, we can first multiply the bottom equation by 2 to obtain… next and if we then subtract the top equation in the system from the bottom equation we see that… y + 2z = 10 4y + z = 13 y + 2z = 10 4y + z = 13 8y + 2z = 26 - y + - 2z = y + 0 = 16 or y = 16 / 7 next 7y next

© 2007 Herbert I. Gross If we now replace y by 16 / 7 in the top equation of system, we see that… y + 2z = 10 4y + z = 13 2z = 10 – 16 / 7 z = 27 / 7 next 16 / 7 2z = 70 / 7 – 16 / 7 2z = 54 / 7

next © 2007 Herbert I. Gross As a check, we replace x by 1, y by 16 / 7, and z by 27 / 7 in the system to verify that… next 3x + y + 2z = 13 2x + 4y + z = 15 3(1) + ( 16 / 7 ) + 2( 27 / 7 ) = 13 2(1) + 4( 16 / 7 ) + ( 27 / 7 ) = 15

next © 2007 Herbert I. Gross More specifically… next 3(1) + ( 16 / 7 ) + 2( 27 / 7 ) = 13 2(1) + 4( 16 / 7 ) + ( 27 / 7 ) = 15 and… / / 7 = / 7 = = / / 7 = / 7 = = 15

next © 2007 Herbert I. Gross Finally, the fact that the truth of the bottom equation in our system 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. 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 28

next © 2007 Herbert I. Gross However, as a double check we see that with these choices for x, y and z. next 5(1) + 5( 16 / 7 ) + 3( 27 / 7 ) = 28 becomes… / / 7 = / 7 = = 28 5x + 5y + 3z = 28

next © 2007 Herbert I. Gross The solution for the system was based on our choosing x to be 1. 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 28 The same procedure would have applied for any value we had chosen for x. next

© 2007 Herbert I. Gross In other words, while there are no values of x, y, and z that satisfy the system 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 = 30 3x + y + 2z = 13 2x + 4y + z = 15 5x + 5y + 3z = 28

next © 2007 Herbert I. Gross 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.

next © 2007 Herbert I. Gross next or 3x 0 + y + 2z = 13 2x 0 + 4y + z = 15 For example, suppose we want the value of x to be x 0. y + 2z = 13 – 3x 0 4y + z = 15 – 2x 0 If we replace x by x 0 in the system we obtain…

next © 2007 Herbert I. Gross next y + 2z = 13 – 3x 0 4y + z = 15 – 2x 0 7y = x 0 next To solve the above system for y, we could multiply the bottom equation by 2 to obtain… y + 2z = x 0 And then subtract the top equation from the bottom equation to obtain… 4y + z = x 0 - 8y + - 2z = x 0 - y + - 2z = x 0 y + 2z = x 0 4y + z = x 0 or 7y = 17 – x 0 next

7y = 17 –x 0 And if we now divide both sides of the above equation by 7, we see that ….. And if we now replace y in the equation y + 2z = 13 – 3x 0 by its above value, we see that ….. y = (17 – x 0 )/7 (17 – x 0 )/7 + 2z = 13 – 3x 0 next

(17 – x 0 )/7 + 2z = 13 – 3x 0 2z = 13 – 3x 0 – (17 – x 0 )/7 14z = 91 – 21x 0 – (17 –x 0 ) 14z = 91 – 21x 0 – 17 + x 0 14z = 74 – 20x 0 7z = 37 – 10x 0 z = (37 – 10x 0 )/7 next

© 2007 Herbert I. Gross 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

…and this agrees with the result we obtained previously. 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 … next y = (17 – x)/7 and z = (37 – 10x)/7

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 Summary

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 Case 2 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… 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.

Although the demonstration is beyond the scope of this course, our summary remains valid for any linear system of equations having “n” unknowns. next Concluding Note More specifically, one and only one of the following situations can occur…

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. next 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. next