844 CHAPTER 8 Systems of Equations and Inequalities 8-52
(Figure 8.24). It
appears the value of
the objective function
for points (x, y) in this
corner are close to 4,
and it’s no accident
that at the corner
point (0, 4) the
maximum value is in
fact 4. Repeating this
procedure for the lower-right corner suggests the
maximum value near (2, 0) is also 4. Finally, press
to explore the region in the upper-right corner, where the
lines intersect. Move the cursor to this vicinity, locate it
very near the point of intersection [we stopped at
( and return to the home screen and eval-
uate (Figure 8.25). The value of the objective function is
0.957, 2.71624
GRAPH
near 5 in this corner of
the region, and at the
corner point (1, 3) the
maximum value is 5.
Exercise 1: The feasi-
ble region for the
system given to the
right has four corner
points. Use the ideas
here to explore the area
near each corner point of the feasi-
ble region to determine which point
is the likely candidate to produce
the minimum value of the objective
function Then
solve the linear programming
problem to verify your guess.
f 1x, y2 2x 4y.
When asked to solve a system of two equations in two
variables, we first select an appropriate method. In
Section 8.1, we learned three basic techniques: graphing,
substitution, and elimination. In this feature, we’ll explore
how these methods are
related using Example 2
from Section 8.1 where we
were asked to solve the
system
by graphing. The result-
ing graph, shown here in
Figure 8.26, clearly indi-
cates the solution is (3, 1).
As for the elimination
method, either x or y can be
easily eliminated. If the second equation is multiplied by
2, the x-coefficients will be additive inverses, and the sum
results in an equation with y as the only unknown.
The result is but remember, this is a system
of linear equations, and is still the equation of a
(horizontal) line. Since the system is equi-
valent to the original, it will have the same solution set. In
Figure 8.27, we note the point of intersection for the new
system is still (3, 1). If we eliminate the y-terms instead
e
4x 3y 9
y 1
y 1
y 1
R1 4x 3y 9
2R2 4x 2y 10
sum y 1
e
4x 3y 9
2x
y 5
(using , the
result is , which is
also the equation of a (ver-
tical) line. Creating another
equivalent system using
this line produces
and the graph shown in
Figure 8.28, where the
vertical and horizontal
lines intersect at (3, 1),
making the solution trivial.
Note: Here we see a
close connection to solving
general equations, in that
the goal is to write a series
of equivalent yet simpler
equations, continuing until
the solution is obvious.
As for the substitution
method, consider the
second equation written as
. This equation
represents every point (x, y) on its graph, meaning the rela-
tionship for the ordered pair solutions can also be written
. The same thing can be said for the line
, with its ordered pair solutions represented by
. At the point of intersection the y-coordinates
must be identical, giving . In other words,
we can substitute for y in the first equation, or
for y in the second equation, with both yielding the
4
3
x 3
2x 5
2x 5
4
3
x 3
1x,
4
3
x 32
4x 3y 9
1x, 2x 52
y 2x 5
e
x 3
y 1
x 3
R1 3R22
4
03
1.5
Figure 8.24
4
03
1.5
Figure 8.25
μ
2x 2y 15
x y 6
x 4y 9
x, y 0
Understanding Why Elimination and Substitution “Work”
y
543215 4 3 2 1
5
4
3
2
1
5
4
3
2
1
(3, 1)
y
543215 4 3 2 1
5
4
3
2
1
5
4
3
2
1
y
543215 4 3 2 1
5
4
3
2
1
5
4
3
2
1
Figure 8.26
Figure 8.27
Figure 8.28
College Algebra & Trignometry—
STRENGTHENING CORE SKILLS
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