Download Systems of Linear Equations

The Game of Algebra
or
The Other Side of
Arithmetic
Lesson 21 Part 1
by
Herbert I. Gross & Richard A. Medeiros
© 2007 Herbert I. Gross
next
Systems
of
Linear
Equations
next
© 2007 Herbert I. Gross
Introduction
Suppose we are buying items at a cost of $3
each. In terms of function notation, if we let
x denote the number of items and C the cost
of the items in dollars, we see that…
C = f(x) = 3x
As soon as we know the number of items we
want to buy, their cost is determined directly
from the formula C = f(x) = 3x.
By knowing the value of one quantity (in this
case, x), we can compute the value of the
other quantity (in this case, C).
next
© 2007 Herbert I. Gross
However in many “real life” situations
we do not have this luxury.
More often than not, the quantity
we want to measure depends on our
knowing the value of more than one
other quantity.
© 2007 Herbert I. Gross
next
For example, suppose you once again are
ordering items from a catalog. This time,
however, you want to buy quantities of
two items, one of which costs $3 each
and the other of which costs $2 each.
Knowing how many $3 items you bought
gives you no information about the
number of $2 items you bought.
Stated another way, to determine
the total cost we would have to know both
how many $3 items we bought and how
many $2 items we bought.
© 2007 Herbert I. Gross
next
Again in terms of the language of
functions, if we let x denote the
number of items we buy at $3 and let y
denote the number of items we buy at $2,
the total cost in dollars of buying this
number of objects is given by…
C = g(x,y) = 3x + 2y
In this case, we would have to know the
values of both x and y in order to
determine the value of C.
© 2007 Herbert I. Gross
next
The point is that one
important aspect of
mathematics involves the
situation in which we are
dealing with several
variables (the number of
variables is called the
dimension of the problem)
and certain constraints
are forced upon us.
© 2007 Herbert I. Gross
next
Definition
The number of variables that are
necessary to determine the output of a
function is called the dimension of the
domain of the function.
Any restriction that limits our choice
for an input is called a constraint.
next
© 2007 Herbert I. Gross
In this Lesson we shall concentrate
primarily on the case in which the
dimension of the domain is 2.
That is: we will be dealing with two
independent variables, which, unless
otherwise specified, will be denoted by x and
y. We shall look in particular at the situation
in which there are two linear constraints;
that is: constraints of the form…
Ax + By = C
…where A, B, and C are constants.
© 2007 Herbert I. Gross
next
A Bit of Subtlety
In the constraint Ax + By = C, we assume
that not both A and B are 0. For if they were
the expression would become 0 = C; which
is not a constraint but rather a statement
(which is true if C = 0; and false otherwise).
Note that in order for the product AB
to equal 0 either A or B must equal 0.
Hence, the mathematical way of
saying that A and B cannot both be 0
is to write AB ≠ 0.
© 2007 Herbert I. Gross
next
Notes
The form Ax + By = C (where AB ≠ 0) is an
extension of the y = mx + b form.
That is, any equation of the form y = mx + b
can be rewritten in the form Ax + By = C.
Namely, we may rewrite y = mx + b
as follows…
y = mx + b → y – mx = b → -mx + y = b;
which is in the form Ax + By = C where,
A = -m, B = 1 and C = b.
© 2007 Herbert I. Gross
next
On the other hand, if we let
Notes A = 1 and B = 0, the equation
Ax + By = C becomes x = C. Recall that
x = C is the equation of the vertical line that
contains every point whose x-coordinate
is C. On the other hand, no vertical line can
have the y = mx + b form. Namely, any line
that can be written in that form intersects the
y-axis at (0,b).
However, any vertical line, other than the
y-axis, doesn’t intersect the y-axis
anywhere; and if it is the y-axis, all of its
points are y intercepts.
next
© 2007 Herbert I. Gross
Caution
Do not confuse x = C (a vertical line)
with y = C ( a horizontal line). y = C
has one and only one y-intercept,
namely (0,C).
Caution
Even though the equation involves the
letter x, the line x = C is parallel to the
y-axis.
© 2007 Herbert I. Gross
next
Notes
In summary, every straight line in the
xy-plane can be represented in the form
Ax + By = C. However, only non-vertical
straight lines can be represented in the
form y = mx + b.
Stated a bit differently, there is a
one-to-one correspondence between
equations of the form Ax + By = C (where
AB ≠ 0) and lines (including vertical lines)
in the xy-plane.
next
© 2007 Herbert I. Gross
Note on Eliminating Fractions
Given an equation such as y = 1/2 x + 2/3,
we may multiply both sides by 6 to get the
equivalent equation 6y = 3x + 4.
This, in turn, may be rewritten in the form
-3x + 6y = 4; which has the form
Ax + By = C (where A = -3, B = 6 and C = 4).
More generally, if m and/or b are common
fractions, the equation y = mx + b can be
rewritten in the form Ax + By = C, where A,
B and C are integers.
next
© 2007 Herbert I. Gross
A constraint will usually reduce the number
of choices we can make in choosing values
for the variables.
That is: with each constraint we usually lose
a certain amount of freedom in our choice
of values for the variables. If, for example,
we start with five independent variables in
the domain of a function (that is: the domain
is 5 dimensional) and after a certain number
of constraints, we are only free to choose
two of the variables at random, we say that
we have two degrees of freedom.
next
© 2007 Herbert I. Gross
Definitions
By the (number of) degrees of freedom in
the domain of a function,
we mean the number of variables that can
be assigned values at random.
The special cases in which the constraints
are either too limiting or not limiting
enough are discussed in Lesson 22.
next
© 2007 Herbert I. Gross
To see what this means in more specific
terms, recall that we have already defined
mathematical expressions in terms of
“computer programs”.
For example, to visualize the mathematical
expression 3x + 5, we may look at the
“program”…
Step 1
Step 2
Step 3
Step 4
© 2007 Herbert I. Gross
Program 1
Command
Algebraic Translation
Pick a number
x
Multiply by 3
3x
Add 5
3x + 5
Write the output
3x + 5
next
Notes
In the language of functions we usually
abbreviate “Program 1” by writing
f(x) = 3x + 5.
In this example, the dimension of the
domain (the set of inputs) is 1 because
there is only one variable, namely: x.
And since there are no restrictions on what
the input can be, we also have
1 degree of freedom.
next
© 2007 Herbert I. Gross
Notice that “1-dimensional”
tells us that there is only
1 variable (x) that is needed to determine
the output. It doesn’t tell us anything about
the number of steps in the program.
Notes
Notice also the connection between
“dimension” as we use it in geometry and
dimension” as defined in this lesson.
Namely, the equation f(x) = 3x + 5 has
a 1-dimensional domain and the domain
in this case is the x-axis which is a line
(1 dimensional).
next
© 2007 Herbert I. Gross
Recall that the graph of f is the set of all
ordered pairs (x,f(x)).
Hence, the
graph of f is
2-dimensional.
In the language
of sets, the
graph of
f(x) = 3x + 5
consists of all
ordered pairs
(x, 3x + 5)
(1,8)
(0,5)
(-1,2)
(-2,-1)
next
© 2007 Herbert I. Gross
f(x) = 3
3x + 5
Pictorially, the
graph is the
straight line
whose
slope is 3,
and whose
y-intercept
is (0,5).
(1,8)
(0,5)
(-1,2)
(-2,-1)
next
© 2007 Herbert I. Gross
If we now impose a constraint
on f(x) = 3x + 5, say, 3x + 5 = 26
(that is, the output in Program 1 has to
be 26), the domain of f is now
0-dimensional.
That is: we no longer have any
degrees of freedom because in order to
obey the constraint
the value of x must be 7.
© 2007 Herbert I. Gross
next
In terms of sets {x: 3x + 5 = 26} consists
of the single number 7.
In other words…
{x: 3x + 5 = 26} = {7}
In terms of the graph of f(x) = 3x + 5,
(7,26) is the only point on the
line whose y-coordinate is 26.
This point is represented by the
intersection of the line y = 3x + 5
and the line y = 26.
© 2007 Herbert I. Gross
next
Viewing the constraint 3x + 5 = 26 in terms
of “Program 1”, we now have the following
situation…
Program 1
Command
Algebraic Translation
Step 1
Pick a number
x
Step 2
Multiply by 3
3x
Step 3
Add 5
3x + 5
Constraint Output must be 26
3x + 5 = 26
In other words, up through Step 3, our input
can be any number we want.
However, once the constraint 3x + 5 = 26 is
imposed, the only input we can choose is 7.
© 2007 Herbert I. Gross
next
Although the language may have changed,
what we have done so far in this lesson is
to review what we were already doing in
Lessons 1 and 2. In this Lesson we shall
be doing the same kind of thing but only in
cases where the dimension of the domain
of f is greater than 1.
In fact, we shall concentrate on linear
equations in 2 unknowns. The most general
form of a linear equation in 2 unknowns is…
Ax + By = C
…where A, B and C represent constants.
© 2007 Herbert I. Gross
next
Important Aside
One of the ways in which linear equations
differ from non-linear equations is in
terms of the relationship between
degrees of freedom and the
number of elements in the solution set.
For example, if a linear equation in x has
no degrees of freedom, then the solution
set cannot have more than a single
member.
© 2007 Herbert I. Gross
next
For example, the solution set of the
equation 3x + 5 = 26 has only the
one member, x = 7
However, look at an equation such as
x2 = 49. This equation is not linear
(because the x term is squared) and it has
no degrees of freedom. That is, in order to
solve the equation, we are not free to
choose values of x at random.
Yet the solution set, S, of this equation has
two members, 7 and -7.
© 2007 Herbert I. Gross
next
In fact, an equation can have no degrees
of freedom; yet its solution set can have
many members.
For example, consider the equation…
(x – 1)(x – 2)(x – 3)(x – 4) = 0
The only way a product of numbers can
equal 0 is if one of its factors is equal to 0.
© 2007 Herbert I. Gross
next
Hence, the equation
(x – 1)(x – 2)(x – 3)(x – 4) = 0
can be true only if either…
(x – 1) = 0; in which case, x = 1
(x – 2) = 0; in which case, x = 2
(x – 3) = 0; in which case, x = 3
(x – 4) = 0; in which case, x = 4
© 2007 Herbert I. Gross
next
In other words, the equation
(x – 1)(x – 2)(x – 3)(x – 4) = 0
has no degrees of freedom (because we
are not free to choose a value of x at
random and have it be a solution). Yet its
solution set, S, contains four members.
That is…
S = {x: (x – 1)(x – 2)(x – 3)(x – 4) = 0}
= {1, 2, 3, 4}
© 2007 Herbert I. Gross
next
Let’s now suppose that instead of a
1-dimensional domain we had a
2-dimensional domain; that is, a function of
the type f(x,y) = 3x + 2y.
In terms of a program, the situation might
look something like this…
Step 1
Step 2
Step 3
Step 4
Program 2
Command
Algebraic Translation
Pick a number (Input).
x
Multiply by 3.
3x
Pick another number.
y
Multiply it by 2.
2y
Step 5
Add the number in Step 4
to the number in Step 2.
© 2007 Herbert I. Gross
3x +2y
next
Step 1
Step 2
Step 3
Step 4
Program 2
Command
Algebraic Translation
Pick a number (Input).
x
Multiply by 3.
3x
Pick another number.
y
Multiply it by 2.
2y
Step 5
Add the number in Step 4
to the number in Step 2.
3x +2y
“Program 2” is 2-dimensional because there
are two independent inputs (x and y), and
it also has 2 degrees of freedom because
there are as yet no constraints on the
choices for the two variables.
© 2007 Herbert I. Gross
next
Another
way
to
describe
Notes
“Program 2” is by saying
that it has two independent variables.
When we say that x and y are independent,
we mean that in choosing the value of one
of the variables, we do not, in any way,
restrict the value of the other variable.
For example, in Step 3 when the “program”
says “Pick another number”, we are free to
choose any number (including the same
number we picked in Step 1) regardless of
what number we chose in Step 1.
next
© 2007 Herbert I. Gross
If we call the numbers x
Notes and y, the two numbers we
are determining must come from the set of
all ordered number pairs (x,y).
Since there are two variables (or
geometrically speaking, since the graph of
this set is the xy-plane), we refer to the set
as being 2-dimensional.
That is, the algebraic definition of the
xy-plane is…
{(x,y): x and y are any numbers}
© 2007 Herbert I. Gross
next
Notes
Since we now have a 2-dimensional
problem, we can expect to need
two constraints in order to determine what
the two numbers are.
In general (but as we shall see in the next
Lesson there are exceptions), each
constraint reduces the number of
degrees of freedom in our equation by 1.
© 2007 Herbert I. Gross
next
For example, suppose we add the single
constraint in “Program 2” that
the sum in Step 5 has to be 18. That is…
Step 1
Step 2
Step 3
Step 4
Program 2
Command
Algebraic Translation
Pick a number (Input).
x
Multiply by 3.
3x
Pick another number.
y
Multiply it by 2.
2y
Step 5
Add the number in Step 4
to the number in Step 2.
3x +2y
Constraint
The sum is 18.
3x + 2y = 18
© 2007 Herbert I. Gross
next
Constraint
The sum is 18.
3x + 2y = 18
In other words, through Step 5, we are free
to choose x and y in any way that we wish.
However, the constraint 3x + 2y = 18
means that we can no longer choose
x and y at random.
That is: x and y are no longer
independent variables because once we
choose a value for either x or y, the value
of the other is determined.
© 2007 Herbert I. Gross
next
For example, suppose we elect to let x = 2.
If we replace x by 2 in the 2-dimensional
equation…
3x + 2y = 18;
3(2) + 2y = 18;
we obtain the 1 dimensional equation…
6 + 2y = 18,
for which the solution is y = 6. In other
words, while we aren’t forced to let x = 2,
once we make this choice, y must be 6. next
© 2007 Herbert I. Gross
More generally, we can solve the equation
3x + 2y = 18 for y to obtain that 2y = 18 – 3x,
or, y = 1/2(18 – 3x) = 9 – 3/2x.
In other words, to belong to the solution set
of 3x + 2y = 18, the ordered number pair
must have the form (x,9 – 3/2x)
In this form, we see that the solution set
is now 1-dimensional because its
members are determined by a single
variable (x).
© 2007 Herbert I. Gross
next
Notice that while there are infinitely many
number pairs (x,y) that will satisfy the
equation 3x + 2y = 18, the likelihood that an
ordered pair of numbers (x,y) chosen at
random will satisfy the equation is
very small.
In terms of a graph, this is the set of all
points (x,y) that are on the line whose
equation is 3x + 2y = 18; or, in the mx + b
form, y = -3/2x + 9 (that is, the line whose
slope is -3/2 and whose y-intercept is (0,9)).
© 2007 Herbert I. Gross
next
So even though a point (x,y) chosen at
random is very likely not to be on this line
there are, nevertheless, infinitely many
points (x,y) that are on the line.
Thus, even with the constraint
3x + 2y = 18,
we still have one degree of freedom.
What we’ll be doing in this Lesson and the
next is to see what happens to a linear
expression in two variables when we
impose two linear constraints.
next
© 2007 Herbert I. Gross
For example, let’s add another step
And a second constraint to Program #2.
Step 1
Step 2
Step 3
Step 4
Program 2
Command
Algebraic Translation
Pick a number (Input).
x
Multiply by 3.
3x
Pick another number.
y
Multiply it by 2.
2y
Step 5
Add the number in Step 4
to the number in Step 2.
3x +2y
Constraint
The sum is 18.
3x + 2y = 18
Step 6
Subtract the number in
Step 4 from the number
in Step 2.
3x – 2y
Constraint
The difference is 6.
3x – 2y = 6
© 2007 Herbert I. Gross
next
To indicate that both equations have to be
satisfied simultaneously (that is, by the
same ordered number pair), we usually
connect the equations by means of a
brace. That is, rather than write, for
example, that…
3x + 2y = 18
and at the same time…
3x – 2y = 6
we write instead…
3x + 2y = 18
3x – 2y = 6
© 2007 Herbert I. Gross
next
In the language of sets, the system of
simultaneous linear equations defined by
3x + 2y = 18
means…
3x – 2y = 6
{(x,y): 3x + 2y = 18 and 3x – 2y = 6}.
If we pick values for x and y at random the
chances are that the ordered pair (x,y) will
satisfy neither the equation 3x + 2y = 18
nor the equation 3x – 2y = 6.
© 2007 Herbert I. Gross
next
Important Point
There will be some pairs that will
satisfy one equation but not the other.
However, there will be only one ordered pair
that satisfies both of the equations
3x + 2y = 18 and 3x – 2y = 6.
As we shall soon prove, the ordered pair
must be (4,3).
© 2007 Herbert I. Gross
next
The use of graphs
gives us an easy way
to visualize how the combination of both
constraints determines a unique pair of
numbers, x and y , that satisfy the system
of linear equations…
3x + 2y = 18
3x – 2y = 6
Geometric Note
Namely, suppose we let L1 denote the line
represented by the equation 3x + 2y = 18
and we let L2 denote the line represented by
equation 3x – 2y = 6.
© 2007 Herbert I. Gross
next
L1
(0,9)
Then any
pair of
numbers,
(x,y), that
satisfies
3x + 2y = 18
can be
represented
by a point
on L1.
(2,6)
(4,3)
(6,0)
(8,-3)
next
© 2007 Herbert I. Gross
L2
(8,9)
Similarly, any
pair of
numbers,
(x,y)
that satisfies
3x – 2y = 6
can be
represented
by a point on
L2.
(6,6)
(4,3)
(2,0)
(0,-3)
next
© 2007 Herbert I. Gross
Therefore, for a pair of numbers, (x,y), to
satisfy both equations, the pair of numbers
must correspond to a point that is
on both lines.
But since the two lines have different slopes
(i.e., they are not parallel), they can meet at
one and only one point; and it is this point
that represents the unique pair of numbers
that satisfies each of the two constraints.
If we were to plot the lines L1 and L2 we
would find that the one point at which they
meet is (4,3).
next
© 2007 Herbert I. Gross
L1
L2
(0,9)
If we now
plot the
lines L1 and
L2 we would
find that the
one point at
which they
meet is (4,3).
(8,9)
(2,6) (6,6)
(4,3) (4,3)
(2,0)
(0,-3)
(6,0)
(8,-3)
next
© 2007 Herbert I. Gross
Notice that if we let x = 4 and y = 3, we see
that this number pair is a solution of the
simultaneous equations defined by…
3x + 2y = 18
3x – 2y = 6
Namely…
3(4) + 2(3) = 18
3(4) – 2(3) = 6
© 2007 Herbert I. Gross
next
While the graphing method is nice, it is very
limited. Among other things, suppose we
had a linear equation in, say, five unknowns.
How would we graph this 5-dimensional
equation?
Therefore what we’d like to do is find an
algebraic way that does the same thing as
the graphing method. The reason for this is
that we can usually find ways of extending
an algebraic technique to higher
dimensional equations which cannot be
next
solved graphically.
© 2007 Herbert I. Gross
Other Graphing Problems
Even in the 2 dimensional case, there
can be a problem finding the exact
point of intersection of the two lines.
© 2007 Herbert I. Gross
next
Example 1
If the two
lines are
almost
parallel…
Where would
the exact
point of
intersection
be?
next
© 2007 Herbert I. Gross
Example 2
If the two lines
do not
intersect on the
grid lines…
Where would
the exact
point of
intersection
be?
next
© 2007 Herbert I. Gross
Still Other Graphing Problems
Even in a case where the lines seem to
meet at the point of intersection of two
grid lines; because of the thickness of
the lines, how would we distinguish
between, say, (4,3) and (4.02,2.93), etc.?
© 2007 Herbert I. Gross
next
One strategy for solving systems of
simultaneous equations makes use of the
fact that the sum of a number and its
opposite is 0. For example, 2y + -2y = 0; or
in more informal language, 2y – 2y = 0.
To apply this strategy to the system of
simultaneous linear equations defined by…
3x + 2y = 18
3x – 2y = 6
notice that by the “equals added to equals”
principle, we can add the two left hand sides
of the two equations and equate this to the
next
sum of the two right hand sides.
© 2007 Herbert I. Gross
That is…
3x + 2y = 18
+ 3x – 2y = 6
6x (+0) = 24
6
6
x=4
This illustrates the power of logical thinking.
Simply by adding both equations and then
dividing by 6, we were able to prove that if
the system of equations defined by the
above is to be satisfied, then x must equal 4.
© 2007 Herbert I. Gross
next
From a different point of view, we have
proven that if x ≠ 4, then it is impossible to
find values for x and y for which the
equations 3x + 2y = 18 and 3x – 2y = 6
are both satisfied.
Once we know that x = 4, we may replace x
by 4 in either equation to find the value of y.
For example, if we replace x by 4 in the
equation 3x + 2y = 18, we obtain the
equation 3(4) + 2y = 18; from which it
follows that y = 3.
© 2007 Herbert I. Gross
next
As a check, we may replace x by 4 and
y by 3 in the equation 3x – 2y = 6 to obtain
the true statement…
3x – 2y = 6
3(4) – 2(3) = 6
12 – 6 = 6
6=6
© 2007 Herbert I. Gross
next
That is, we obtained y = 3 using only the
equation 3x + 2y = 18. If we had made a
mistake, our answer would not have worked
in the equation 3x – 2y = 6.
As usual, notice that there is more than one
way to solve a system of equations.
For example, we could have used the
equation 3x + 2y = 18 to express y in terms
of x. Namely, if we subtract 3x from both
sides of the equation, we get the equivalent
equation 2y = 18 – 3x from which it follows
y = 18 – 3x
that…
next
2
© 2007 Herbert I. Gross
We may then replace y by y = 18 – 3x
2
in equation 3x – 2y = 6 to get the equation
3x – 2 ( 18 – 3x ) = 6
2
which is a linear equation in x.
© 2007 Herbert I. Gross
next
One way we can solve this equation is by
the following sequence of steps…
3x – 2 ( 18 – 3x ) = 6
2
3x – (18 – 3x) = 6
3x + -1(18 + -3x) = 6
3x + -18 + 3x = 6
6x – 18 = 6
6x = 24
© 2007 Herbert I. Gross
x=4
next
The method described in the above note is
often referred to as the method of
substitution. While this method works very
nicely for a two dimensional linear system
of equations, it becomes quite
cumbersome when we are involved with
linear systems that have a greater
dimension. On the other hand, as we shall
soon see, our first method generalizes very
nicely to the case of the general system of
linear equations in more than two
unknowns.
© 2007 Herbert I. Gross
next
Adding in the case in which
the multipliers of either x or y
(these multipliers are usually called
coefficients) are opposites eliminates one of
the variables.
–
In a similar way, subtracting
in the case in which the coefficients
of either x or y are equal also eliminates
one of the variables.
© 2007 Herbert I. Gross
next
With respect to this example, if we subtract
3x – 2y = 6 from 3x + 2y = 18, the x terms
will cancel. That is, recalling that to
subtract, we add the opposite, we obtain…
–
3x + 2y = 18
– 3x – 2y = – 6
0 + 4y = 12
4
4
y=3
© 2007 Herbert I. Gross
next
Caution
Be careful when you subtract!
For example, when we subtract the equation
3x – 2y = 6 from the equation 3x + 2y = 18,
we are subtracting -2y from 2y.
That is, by the “add the opposite” rule
2y – -2y = 2y + 2y.
Don’t make the mistake of calling the
answer 0.
© 2007 Herbert I. Gross
next
A Helpful Strategy…
Because it is less confusing to add than to
subtract, it may be helpful to you if you
multiply both sides of either one of the
equations by -1.
Since equals multiplied by equals are
equal, this will not change the solution
set of the system, but it manages to
convert a subtraction problem into an
addition problem.
© 2007 Herbert I. Gross
next
For example, if we multiply both sides of
the equation 3x – 2y = 6 by -1 (and if we
recall that multiplying by -1 changes the
sign of each term), we see that the
system…
3x + 2y = 18
3x – 2y = 6
is equivalent to the system…
3x + 2y = 18
-3x + 2y = -6
© 2007 Herbert I. Gross
next
Now if we add these
two equations…
we obtain 4y = 12
or y = 3.
3x + 2y = 18
-3x + 2y = -6
4y = 12
If you had wanted to you could have
multiplied both sides of 3x + 2y = 18 by -1
and obtained -3x + -2y = -18.
Now when you add,
the sum becomes
-4y = -12 and the
solution is still y = 3.
© 2007 Herbert I. Gross
-3x
+ -2y = -18
3x + -2y = 6
-4y
= -12
next
However, since most people
prefer to work with positive
rather than negative coefficients,
it is often helpful to make our
choice based on which one
allows us to work with positive
coefficients.
© 2007 Herbert I. Gross
next
To help ensure that you understand our
discussion, try to do the following example.
If you think you know how to solve the
problem do it before you read our solution.
Otherwise you may read our solution first.
For what values of x and y is it true that…
2x + 3y = 16
4x – 3y = 14
© 2007 Herbert I. Gross
next
By adding equals to equals (i.e., the sum
S of the two left sides must equal the sum
o
of the two right sides), we see that…
l
2x + 3y = 16
u
+ 4x – 3y = 14
t
i
6x (+0) = 30
o
6
6
n
x=5
Once we know that x = 5, we may replace
x by 5 in either of the two equations and
then solve for y.
next
© 2007 Herbert I. Gross
For example, if we replace x by 5 in the
equation 2x + 3y = 16, we obtain the
1-dimensional equation…
2x + 3y = 16
2(5) + 3y = 16
10 + 3y = 16
3y = 6
y=2
© 2007 Herbert I. Gross
next
As a check we can now replace x by 5 and
y by 2 in the equation 2x + 3y = 16 and see
whether we get a true statement.
2x + 3y = 16
2(5) + 3(2) = 16
10 + 6 = 16
16 = 16
© 2007 Herbert I. Gross
next
So we replace x by 5 and y by 2 in the
equation 4x – 3y = 14 to obtain…
4x – 3y = 14
4(5) – 3(2) = 14
20 – 6 = 14
14 = 14
…which is a true statement.
© 2007 Herbert I. Gross
next
Notice that the check is
Notes
part of the solution.
Namely, once we knew the value of x, we
found the value of y from the equation
2x + 3y = 16.
To make sure we didn’t make a numerical
error we check the result in the equation
that we didn’t use to find the value of y.
For example, once we knew that x = 5, we
could have solved for y using the equation
4x – 3y = 14. In this case we would have
checked the result by using the equation
next
2x + 3y = 16.
© 2007 Herbert I. Gross
The big question that
Notes
remains, however, is
how likely is it that the coefficients of y in
each of the two equations will be either the
same or opposites?
The answer is that while it is very unlikely
that in a pair of simultaneous equations
chosen at random that the coefficients of
either x or y will be opposites, we can
always replace the given system by an
equivalent system in which the coefficients
will be opposites.
© 2007 Herbert I. Gross
next
Definition
By two equations being equivalent we
mean that they have the same solution set.
For example, if we multiply both sides of
the equation 2x = 6 by 5, we obtain the
equation 10x = 30. The solution for each
of these equations is x = 3.
Thus, the equations 2x = 6 and 10x = 30
are equivalent.
next
© 2007 Herbert I. Gross
Let’s practice this technique by solving the
system of equations…
3x + 4y = 25
2x + 3y = 18
We have a choice of solving the system for
either x or y. If we elect to solve for x it
means that we want to eliminate y.
To do this we will replace the two equations
above by equivalent equations in which the
coefficients of y are opposites.
© 2007 Herbert I. Gross
next
We might start by observing that the
coefficients of y in equation below are
4 and 3 and that 12 is the least common
multiple of 3 and 4.
3x + 4y = 25
2x + 3y = 18
So we can multiply both sides of 3x + 4y = 25
by 3 to obtain the equivalent equation…
3(3x + 4y) = 3(25)
9x + 12y = 75
© 2007 Herbert I. Gross
next
And we can multiply both sides of
2x + 3y = 18 by -4 to obtain the equivalent
equation…
-4(2x + 3y) = -4( 18)
-8x
+ -12y = -72
In other words, the system of equations
defined by…
3x + 4y = 25
2x + 3y = 18
is equivalent to the system of equations
defined by…
9x + 12y = 75
-8x
© 2007 Herbert I. Gross
+ -12y = -72
next
We sometimes use the
Notation
symbol “~” to indicate
that two systems are equivalent.
In terms of this notation…
3x + 4y = 25
9x + 12y = 75
-8x + -12y = -72
2x + 3y = 18
x
= 3
The beauty of the latter system is that the
coefficients of y are now opposites.
Hence, when we add the two equations
in the latter system the terms involving y
cancel, and we are left with x = 3.
next
~
© 2007 Herbert I. Gross
Once we know that x = 3,
we may go to either of
the two equations in either of the
systems…
3x + 4y = 25
9x + 12y = 75
-8x + -12y = -72
2x + 3y = 18
Notes
~
(It doesn’t matter which because the
systems are equivalent) to determine the
value of y.
© 2007 Herbert I. Gross
next
However, we want to
practice the method of
solving for y by first eliminating the terms
involving x.
Notes
3x + 4y = 25
2x + 3y = 18
This time, however, we want to eliminate
the terms that involve x.
© 2007 Herbert I. Gross
next
Noticing that the coefficients
Notes
of x are 2 and 3, we see that
6 is the least common multiple.
Hence, we can multiply both sides of the
first equation by 2… and both sides of the
second equation by -3… to get the
equivalent system of equations…
2(3x + 4y) = 2(25)
-3(2x
-6x
6x + 8y = 50
+ 3y) = -3( 18)
+ -9y = -54
6x + 8y = 50
© 2007 Herbert I. Gross
-6x
+ -9y = -54
next
Notes
If we now add these
two equations…
6x + 8y = 50
-6x
+ -9y = -54
-y = -4
we obtain -y = -4 or y = 4.
As a check, if x = 3 and y = 4,
3x + 4y = 3(3) + 4(4) = 25.
and
2x + 3y = 2(3) + 3(4) = 18
© 2007 Herbert I. Gross
next
Notes
In solving the system
of equations…
3x + 4y = 25
2x + 3y = 18
we chose to multiply the top equation by 2
and the bottom equation by -3. It would
have been just as logical to have multiplied
the top equation by -2 and the bottom
equation by 3 to obtain the equivalent
system…
-6x + -8y = -50
6x + 9y = 54
next
© 2007 Herbert I. Gross
Notes
If we now add these
two equations…
we obtain immediately
that y = 4 .
-6x
+ -8y = -50
6x + 9y = 54
y=4
Either way we obtain the correct answer but
most of us would rather work with positive
coefficients than negative ones. Thus, a
good strategy would be to choose our signs
in such a way that we wind up dealing with
positive coefficients.
next
© 2007 Herbert I. Gross
Notes
One way to do this is to
ignore the signs until
the last step.
For example, suppose we want to
eliminate y from the system of
equations…
3x + 5y = 30
4x + 2y = 40
© 2007 Herbert I. Gross
next
Notes
We begin by multiplying both sides of the
top equation by 2 and both sides of the
bottom equation by 5 to obtain the
equivalent system of equations…
2( 3x ) + 2( 5y ) = 2( 30 )
5( 4x ) + 5( 2y ) = 5( 40 )
6x + 10y = 60
8x + 10y = 200
© 2007 Herbert I. Gross
next
Notes
Since 8 is greater than 6, we can make sure
when we add that the coefficient of x is
positive if we multiply the top equation by -1
to obtain…
-6x
+ -10y = -60
8x + 10y = 200
In this way, when we add we obtain the
equation 2x = 140 or x = 70.
© 2007 Herbert I. Gross
next
This concludes our lesson.
In essence, Lesson 21,
part 2 may be viewed as
a continuation of
Lesson 21 part 1 in the
sense that it introduces
the cases in which the
dimension of the linear
system is greater than 2.
© 2007 Herbert I. Gross
next