Download y + 2z = 13 + - Adjective Noun Math

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.