### Video Transcript

In this video, we will learn how to
solve systems of linear inequalities by graphing them and identify the regions
representing the solution. We should already be comfortable
with using and interpreting inequality notation, plotting straight-line graphs given
their equations, and graphing single linear inequalities.

A system of linear inequalities is
a set of two or more linear inequalities in several variables. For example, π¦ is greater than or
equal to one, π₯ is greater than zero, and π₯ plus π¦ is less than or equal to four
is a system of linear inequalities in the two variables π₯ and π¦. Such systems are used in practical
contexts when one problem requires a range of solutions, and there is more than one
constraint on those solutions.

Letβs begin by recalling some of
the basics of how we represent linear inequalities graphically. Weβll consider the inequalities π¦
is less than five and π₯ is greater than or equal to three. We recall that when representing
inequalities on a graph, we first need to determine the equation of the boundary
line of the region. And to do this, we swap the
inequality sign for an equals sign. So the boundary line for the
inequality π¦ is less than five is π¦ is equal to five. We then draw this line, recalling
that lines of the form π¦ equals some constant are horizontal lines. We also need to recall that we use
a solid line if we have a weak inequality and we use a dashed line if we have a
strict inequality. Here, the inequality is strict. Itβs π¦ is strictly less than five,
so we draw the line π¦ equals five as a dashed line.

We now have the boundary line for
the graphical solution to this inequality. And we need to decide which side of
the line to shade. If we want π¦ to be less than five,
then we need to include all points in the coordinate plane for which the
π¦-coordinate is less than five. Thatβs all points below the
boundary line, so we shade this side of the line.

Weβve now represented the solution
to the first inequality graphically. So letβs now consider the second
inequality: π₯ is greater than or equal to three. Once again, we replace the
inequality with an equals sign to find the equation of the boundary line. The equation of the boundary is π₯
equals three, which we recall is a vertical line through the value three on the
π₯-axis.

As we have a weak inequality, π₯ is
greater than or equal to three, we represent this boundary with a solid line. We then need to decide which side
of the line to shade. And as we want π₯ to be greater
than or equal to three, we need to shade the region to the right of the
boundary. Remember that the solid line
indicates that points on the boundary line itself are included in the solution to
this inequality. So, weβve represented the solution
to each inequality graphically.

If we want to treat this as a
system of linear inequalities, then this means we need to find the region that
satisfies both inequalities simultaneously, which is the intersection of the two
regions weβve shaded. The region below the orange line
and to the right of the pink line therefore represents the solution to this system
of inequalities. It contains all the points in the
coordinate plane such that π¦ is less than five and π₯ is greater than or equal to
three. Letβs now consider some
examples.

State the system of
inequalities whose solution is represented by the following graph.

From the diagram, we can see
that the shaded region is the first quadrant of the coordinate plane, which
contains all points such that both π₯ and π¦ are positive. The boundary lines for the
region, which are the positive π₯- and π¦-axes, are drawn as solid lines. So points along the boundaries
are also included. This means that zero values of
both π₯ and π¦ are included in the region. Using a bit of logic, we can
therefore express this region using the inequalities π₯ is greater than or equal
to zero and π¦ is greater than or equal to zero.

Letβs confirm this by
representing each of these inequalities graphically ourselves. The inequality π₯ is greater
than or equal to zero is represented by the π¦-axis, where π₯ is equal to zero,
and all the points to the right of the π¦-axis. The inequality π¦ is greater
than or equal to zero is represented by the π₯-axis, where π¦ equals zero, and
all points above the π₯-axis. The intersection of these two
regions is the part that is shaded in both orange and pink. Thatβs the whole of the first
quadrant, including the positive π₯- and π¦-axes, which matches up with the
graph we were given. So, we can conclude that the
system of inequalities represented by the given graph is π₯ is greater than or
equal to zero and π¦ is greater than or equal to zero.

Letβs now consider a slightly more
complex example.

State the system of
inequalities whose solution is represented by the following graph.

We can see that this system of
inequalities contains horizontal, vertical, and diagonal boundary lines, all of
which we will need to find the equations of. First though, we identify that
the shaded region is in the first quadrant, which contains all nonnegative
values of π₯ and π¦. This can be represented by the
inequalities π₯ is greater than or equal to zero and π¦ is greater than or equal
to zero.

Next, letβs consider the
vertical lines. The equations of these lines
are π₯ equals three and π₯ equals six. The region between these lines
has been shaded, so this corresponds to π₯ being between three and six. But we need to be clear about
which inequality signs to use at each end of the interval. As the line drawn at π₯ equals
three is a solid line, this corresponds to a weak inequality. So we have π₯ is greater than
or equal to three. As the line drawn at π₯ equals
six is a dashed line, this corresponds to a strict inequality. So we have π₯ is less than
six. This can be combined into the
double-sided inequality π₯ is greater than or equal to three and less than
six.

Next, we consider the
horizontal lines. All horizontal lines have
equations of the form π¦ equals some constant. So the equations of these lines
are π¦ equals two and π¦ equals six. The blue region is between the
two lines, so π¦ is between two and six. As these lines are solid, both
inequalities are weak inequalities. So this region can be
represented as π¦ is greater than or equal to two and less than or equal to
six.

Finally, we consider the orange
region below the diagonal line. This line passes through eight
on the π₯-axis and eight on the π¦-axis. We wonβt go into the detail of
how to find its equation here, but its equation can be written as π¦ equals
eight minus π₯. As the shaded region is below
the line and the line itself is solid, this corresponds to the inequality π¦ is
less than or equal to eight minus π₯. We could leave this inequality
in this form, or we can rearrange it by adding π₯ to both sides to give π₯ plus
π¦ is less than or equal to eight.

So weβve found the system of
inequalities whose solution is represented by the graph. Their solution set is the
intersection of the individual shaded regions, which is the green triangular
region. The system of inequalities is
π₯ is greater than or equal to zero. π¦ is greater than or equal to
zero. π₯ is greater than or equal to
three and less than six. π¦ is greater than or equal to
two and less than or equal to six. And π₯ plus π¦ is less than or
equal to eight.

Letβs now consider another
example.

Which of the points below
belongs to the solution set of the system of linear inequalities represented by
the figure shown? (a) One, three; (b) six, eight;
(c) four, three; or (d) three, six.

Itβs important to realize that
we arenβt being asked to find the system of linear inequalities represented by
the figure shown but rather to determine which of four given points lies in
their solution set. From the figure, we can see
that there are essentially three types of linear inequalities that have been
graphed. There are two vertical lines,
and the region between these lines is shaded in red. There are two horizontal lines,
and the region between these lines is shaded in blue. Finally, there is a diagonal
line, and the region below this line is shaded in orange. The region that satisfies each
of these inequalities simultaneously is the intersection of the three shaded
regions, which is the triangle shaded in green.

To determine which of the four
given points lies in the solution set of the system of inequalities, we just
need to plot each of these points on the graph and identify any points that are
inside or on the solid boundaries of this region. When we do this, we find that
the only point that lies in the shaded region and hence belongs in the solution
set of the system of linear inequalities is the point four, three.

Hereβs another example.

Find the solution set of the
linear inequalities shown below.

So there are two linear
inequalities graphed on the same coordinate plane. And weβre asked to find the
solution set of this pair of linear inequalities. This means we need to find the
region that satisfies both inequalities simultaneously, which is the
intersection of the two individual regions. But itβs clear from looking at
the graph that these two regions donβt overlap. The two diagonal lines are
parallel, so they never meet. The red region is above the
higher diagonal line, and the blue region is below the lower line. As a result, there is no
intersection between these two regions. And so the solution set of this
pair of linear inequalities is the empty set, π.

Letβs consider one final
example.

Which region on the graph
contains solutions to the set of inequalities π¦ is greater than two, π¦ is
greater than or equal to negative π₯, and π₯ is less than one?

These inequalities have already
been graphed for us, and our job is to identify which of the regions marked with
letters represents the solution to all three inequalities simultaneously. Weβll do this by first
identifying the solution set to each inequality separately. The boundary line for the
inequality π¦ is greater than two is π¦ is equal to two, which is a horizontal
line through the value two on the π¦-axis. Thatβs the dashed line drawn in
blue. Itβs a dashed line because the
inequality is π¦ is strictly greater than two. As the inequality is π¦ is
greater than two, the solution set to this inequality will be the region above
the dashed line. So this immediately narrows the
options down to G, D, and F.

The boundary line for the
inequality π¦ is greater than or equal to negative π₯ is the line π¦ equals
negative π₯, which is a diagonal line through the origin with a slope of
negative one. Thatβs the solid line drawn in
green, and a solid line has been used here because it is a weak inequality. As the inequality is π¦ is
greater than or equal to negative π₯, the solution set to this inequality is the
region on and above the line. We can therefore rule out
region G.

For the final inequality, the
boundary line is π₯ equals one, which is a vertical line through one on the
π₯-axis. This is of course the final
line drawn on the coordinate grid, the vertical dashed line drawn in red. As the inequality is π₯ is less
than one, the region satisfying this inequality is to the left of the boundary
line. This rules out option F. So, we can conclude that the
region that contains the solutions to the system of all three inequalities,
which is the intersection of the three individual regions, is region D.

Letβs now summarize the key points
from this video. For a vertical line parallel to the
π¦-axis of the form π₯ equals some constant π, the region to the right of the line
contains all the points for which π₯ is greater than π and the region to the left
of the line contains all the points such that π₯ is less than π. For a horizontal line of the form
π¦ equals π, the region above the line corresponds to π¦ is greater than π and the
region below the line corresponds to π¦ is less than π. For a diagonal line of the form π¦
equals ππ₯ plus π, the region above the line corresponds to where π¦ is greater
than ππ₯ plus π and the region below the line corresponds to where π¦ is less than
ππ₯ plus π.

We use dashed lines to represent
strict inequalities, which means the boundary line itself is not included in the
solution set, and solid lines to represent weak inequalities, which means the
boundary line is included. The solution set for a system of
linear inequalities is the region that satisfies every inequality simultaneously,
which is the intersection of the individual regions. We also saw that if the regions do
not overlap, then there are no solutions to the system of linear inequalities, in
which case we write the solution set as the empty set π.