key stone problem… key stone problem… next set 22 © 2007 herbert i. gross

Key Stone Problem…Key Stone Problem…

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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

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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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For what values of x, y, and z is it true that…

Keystone Problem for Lesson 22Keystone Problem for Lesson 22

3x + y + 2z = 13

2x + 4y + z = 15

5x + 5y + 3z = 30

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SolutionSolution

Using the method described in Lesson 21, we can eliminate x from all but the top equation of the system…

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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

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Therefore, we can multiply both sides of the top equation of the system above by 10;

SolutionSolution 3x + y + 2z = 13

2x + 4y + z = 15

5x + 5y + 3z = 30

the middle equation by -15;

-30x + -10y + -20z = -130-30x + -60y + -15z = -225-30x + -30y + -18z = -180

and the bottom equation by -6 to obtain the equivalent system…

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So to eliminate x, we replace the middle equation in the system above by the sum of middle equation and the top equation,

SolutionSolution -30x + -10y + -20z = -130

-30x + -60y + -15z = -225

-30x + -30y + -18z = -180

and we replace the bottom equation by the sum of the bottom equation and the topequation to obtain…

-30x + -10y + -20z = -130-30x + -60y + -15z = -225-30x + -30y + -18z = -180

-50y + -5z = - 95-20y + -2z = -50

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If we prefer to work with smaller numbers, we may divide both sides of the top equation in system above by 10,

SolutionSolution

the middle equation by -5,

-30x + -10y + -20z = -130-50y + -5z = - 95-20y + -2z = -50

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and the bottom equation by -2 to obtain…

-30x + -10y + -20z = -130-50y + -5z = - 95

-20y + -2z = -50

3x + y + 2z = 13 10y + -z = 19

10y + -z = 25

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Notice that the middle equation and the bottom equation in the system above

are contradictions of one another.

SolutionSolution

3x + y + 2z = 13 10y + -z = 19 10y + -z = 25

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SolutionSolution

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

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10y + -z = -25

-10y + -z = -19

-10y + -z = -19

-0 = - 6

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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…

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3x + y + 2z = 13 10y + -z = 19

-0 = 6

3x + y + 2z = 13

2x + 4y + z = 15

5x + 5y + 3z = 30

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The bottom equation in the system above often elicits such comments as…

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Notes 3x + y + 2z = 13

10y + -z = 19 -0 = 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.

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Namely what we have shown is that if there were numbers x, y, and z that satisfied the

system…

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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.

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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.

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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.

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However, the above logic is subtle and as a result it

is often misused.

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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).

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Let’s apply the above discussionto solving an algebraic equation.

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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

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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 = 28

5x + 5y + 3z = 30

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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.

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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

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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

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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.

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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.)

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3x + y + 2z = 13

2x + 4y + z = 15

or 3 + y + 2z = 13

2 + 4y + 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

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To solve the system, we can first multiply the bottom equation by 2 to obtain…

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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 = -10

7y + 0 = 16 or y = 16/7

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7y

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If we now replace y by 16/7 in the top equation of system, we see that…

y + 2z = 10

y + 2z = 10

4y + z = 13

2z = 10 – 16/7

z = 27/7

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16/7

2z = 70/7 – 16/7

2z = 54/7

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As a check, we replace x by 1, y by 16/7, and z by 27/7 in the system to verify that…

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3x + y + 2z = 13

2x + 4y + z = 15

3(1) + (16/7) + 2(27/7) = 13

2(1) + 4(16/7) + (27/7) = 15

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More specifically…

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3(1) + (16/7) + 2(27/7) = 13

2(1) + 4(16/7) + (27/7) = 15and…

3 + 16/7 + 54/7 = 13

3 + 70/7 = 13

3 + 10 = 13

2 + 64/7 + 27/7 = 15

2 + 91/7 = 15

2 + 13 = 15

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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

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However, as a double check we see that with these choices for x, y and z.

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5(1) + 5(16/7) + 3(27/7) = 28

becomes…

5 + 80/7 + 81/7 = 28

5 + 161/7 = 28

5 + 23 = 28

5x + 5y + 3z = 28

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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.

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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

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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.

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or

3x0 + y + 2z = 13

2x0 + 4y + z = 15

For example, suppose we want the value of x to be x0.

y + 2z = 13 – 3x0

4y + z = 15 – 2x0

If we replace x by x0 in the system we obtain…

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y + 2z = 13 – 3x0

4y + z = 15 – 2x0

7y = 17 + -x0

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To solve the above system for y, we could multiply the bottom equation by 2 to obtain…

y + 2z = 13 + -3x0

And then subtract the top equation from the bottom equation to obtain…

4y + z = 15 + -2x0 -8y + -2z = -30 + -4x0

-y + -2z = -13 + +3x0

y + 2z = 13 + -3x0

4y + z = 15 + -2x0

or

7y = 17 – x0

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7y = 17 –x0

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 – 3x0 by its above value,we see that …..

y = (17 – x0)/7

(17 – x0)/7 + 2z = 13 – 3x0

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(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

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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…

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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.

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SummarySummary

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Case 1Case 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.

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Case 2Case 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.

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Case 3Case 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.

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Concluding NoteConcluding 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.

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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.

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key stone problem… key stone problem… next set 22 © 2007 herbert i. gross

Documents

y z

values of x

coefficients of x

gross slide

sum of middle equation

gross solution

gross notice

equivalent system slide