### Video Transcript

In this video, we will learn how to
graph two-variable linear inequalities. You should be familiar with the
four different inequalities. We have greater than, greater than
or equal to, less than, or less than or equal to.

Letβs start with a recap of how we
graph a one-variable, or single-variable, inequality. We can take the example of π¦ is
greater than two. If we demonstrated this on a number
line, we would have a hollow circle above two and an arrow pointing to the right to
indicate all the values which are greater than two. The fact that this circle on a
number line is hollow or empty is really important. If it was colored in or filled in,
then that would indicate the inequality greater than or equal to in this
occasion. When the inequality is a strict
inequality, like greater than or less than, then we would fill in the circle.

We have to consider this when we
are graphing as well. When it comes to representing an
inequality on a graph, we begin by temporarily replacing the inequality with an
equals sign. So here we start by considering π¦
equals two. This can be shown on the graph, the
point where π¦ is equal to two. But before we draw the line
representing π¦ equals two, we need to consider the inequality. Because this inequality π¦ is
greater than two is a strict inequality, then the line that we draw will be a dotted
or dashed line.

And then, finally, we will need to
represent the region where π¦ is greater than two. If we considered a point or
coordinate below the line, letβs say this coordinate zero, one, at this point the
π¦-value is one. But we know that one is less than
two. So this wouldnβt be in the region
where π¦ is greater than two. Taking another coordinate, this
time the coordinate one, four which is above the line, the π¦-value at this point is
four. Four is greater than two. And so this is part of the region
where π¦ is greater than two, and so we can represent π¦ is greater than two by
shading in the region that we wish. But a word of warning, sometimes in
examinations, weβre asked to shade the region that we want and sometimes weβre asked
to shade the region that we donβt want. So we must always be careful to
read the question.

Graphing two-variable linear
inequalities can be done by using exactly the same process. The dotted line represents the
strict inequalities, either greater than or less than. But the complete full line
represents the weak inequalities, greater than or equal to or less than or equal
to. The difference is that instead of
having just one variable of π₯ or π¦, we will have equations given in the different
forms of the equation of a straight line, for example, ππ₯ plus ππ¦ equals π or
π¦ equals ππ₯ plus π. Because we are working with
graphing equations such like this, then we should be confident with finding the
equation of a straight line from its graph. We can now look at the first
example. In this example, we will identify
the correct notation for an inequality which has been represented on a graph.

Consider the inequalities shown in
Figure 1 and Figure 2. Which of the following is
correct? Option (A) Figure 1 shows π¦ is
greater than π₯; Figure 2 shows π¦ is greater than or equal to π₯. Or option (B) Figure 1 shows π¦ is
greater than or equal to π₯; Figure 2 shows π¦ is greater than π₯.

The first thing we need to do when
we are identifying a graphed inequality is to work out the equation of the line,
whether that line is dotted or a complete line. Both the lines given in Figure 1
and Figure 2 would represent the same equation. On both graphs, we can identify the
coordinates one, one; two, two; three, three, and so on. Because every π¦-value is equal to
the π₯-value, then the equation of either of these lines can be given as π¦ equals
π₯. We notice that in both graphs, that
is, in both figures, itβs the region above the line which has been shaded. The difference is that in Figure 1
we have a dotted line and in Figure 2 we have a complete line.

Dotted lines indicate a strict
inequality, so they will either be greater than or less than, whereas complete lines
represent weak inequalities. It will either be greater than or
equal to or less than or equal to. So Figure 1 will have the graph of
π¦ is greater than π₯ or π¦ is less than π₯. Figure 2 will represent π¦ is
greater than or equal to π₯ or π¦ is less than or equal to π₯. Letβs take Figure 1 and pick a
coordinate which is in the shaded region. So letβs pick the coordinate zero,
four. The π¦-value of this is four, and
the π₯-value is zero. The inequality that we would use
between four and zero is that four is greater than zero. Because this coordinate is in the
shaded region, then the region must be representative of π¦ is greater than π₯.

In the same way, we could pick the
same coordinate in Figure 2, the coordinate zero, four. We know that four is greater than
zero, but this time we know that it has to be a weak inequality: four is greater
than or equal to zero. This graph represents the
inequality π¦ is greater than or equal to π₯. We can therefore highlight that
itβs the statement in option (A) which is correct. Figure 1 shows π¦ is greater than
π₯; Figure 2 shows π¦ is greater than or equal to π₯.

Letβs look at another example.

Which inequality has been graphed
in the given figure?

When we have an inequality
represented on a graph, we will always have a line, whether thatβs a complete line
or a dotted line, along with a shaded region. The first thing we will need to do
is identify the equation of the straight line. We can remember that the equation
of a straight line can be given as π¦ equals ππ₯ plus π, where π represents the
slope, or gradient, and π is the π¦-intercept. The π¦-intercept is usually very
easily identified from the graph. Itβs the point where the line
crosses the π¦-axis. On this graph, this happens at the
coordinates zero, negative three. And so the π¦-intercept is equal to
negative three. The slope of a line can be found by
the rise over the run. And we can work this out using two
coordinates π₯ one, π¦ one and π₯ two, π¦ two by the slope is equal to π¦ two minus
π¦ one over π₯ two minus π₯ one.

To make this process easier for
ourselves, we should select coordinates which are easily identifiable. Usually, these will have integer
values. We can see that the coordinates
zero, negative three lie on the line but so do the coordinates one, one. It doesnβt matter which coordinates
we designate with the π₯ one, π¦ one or π₯ two, π¦ two values. So letβs take π₯ one, π¦ one to be
the coordinates zero, negative three. The slope will therefore be equal
to one minus negative three over one minus zero. This simplifies to four over one,
which of course is equal to four. Since we know that the slope π is
four and the π¦-intercept is negative three, we have the equation of the line as π¦
equals four π₯ minus three.

However, this isnβt the final
answer since we were asked to write the inequality represented by the shaded
region. Instead of the equals sign, we will
need one of the inequalities, greater than, greater than or equal to, less than, or
less than or equal to. So how do we know which one of
these it will be? Well, we can notice that the line
that we were given is a dotted line. This means it is a strict
inequality. It will either be greater than or
less than. The two other inequalities that
involve equals to will not be a possibility.

We can now test which inequality it
will be by selecting a coordinate in the shaded region, which does not lie on the
line. Letβs pick the coordinate three,
two. At these coordinates, the π₯-value
is three and the π¦-value is two. We will therefore need to compare
the value of π¦, which is two, with the value of four π₯ minus three, which is four
times three minus three. When we simplify the right-hand
side, we get nine. We, of course, know that two is
less than nine. And therefore, the missing
inequality must be less than. The answer is therefore that the
inequality which has been graphed is π¦ is less than four π₯ minus three.

Itβs worth pointing out that
everything which is not in the shaded region and not on the line represents the
inequality π¦ is greater than four π₯ minus three. If we checked this using a
coordinate, for example, the coordinate zero, zero, plugging this into the
inequality we would have π¦ is equal to zero and π₯ is equal to zero. On the left-hand side of the
inequality, we would have zero. And on the right-hand side, we
would have negative three. Since this region is π¦ is greater
than four π₯ minus three, then the region which is shaded in is π¦ is less than four
π₯ minus three.

We can now look at one final
example.

Determine the inequality whose
solution set is represented by the colored region.

The solution set of an inequality
is the set of all points that satisfy that inequality. In this figure, this corresponds to
all the points in the colored or shaded region. The first thing we will do is
identify the equation of this straight line, which is the boundary line between the
colored and the uncolored region. We sometimes use the equation of a
straight line in the form π¦ equals ππ₯ plus π to help us do this. However, sometimes there are graphs
which are not that easy to read, and this is one of them. The π¦-intercept, the point where
the line crosses the π¦-axis, is not particularly clear. It could be negative 1.5, but we
canβt say for certain.

In cases like these, it can be
useful to recall the pointβslope form of the line given by π¦ minus π¦ one equals π
times π₯ minus π₯ one, where the coordinates π₯ one, π¦ one is a point on the
line. The value of π indicates the slope
of the line. Letβs calculate the slope
first. To calculate the slope between two
coordinates π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two. We calculate π¦ sub two minus π¦
sub one over π₯ sub two minus π₯ sub one. When we are selecting two
coordinates to use for this formula, itβs helpful to find coordinates with integer
values. Do be careful with this graph
because the major grid lines represent two on the π₯- and π¦-axis. But the minor grid lines, thatβs
the smaller grids, represents one-half.

One coordinate that we could select
is the coordinate negative three, negative six. Another coordinate is that of one,
zero. We can select the coordinates one,
zero to have the π₯ sub one, π¦ sub one values and the coordinates negative three,
negative six with the values of π₯ sub two and π¦ sub two. The slope of this line can
therefore be calculated as negative six minus zero over negative three minus
one. When we simplify this, we get
negative six over negative four, which in turn can be simplified to three over
two. The slope of the line π is equal
to three over two.

We can substitute this value into
the pointβslope form of the equation of a line, along with any coordinates on the
line. We can use the same coordinates of
one, zero in this equation. This will give us π¦ minus zero is
equal to three over two times π₯ minus one. If we wanted to simplify this
equation even further, we can multiply both sides of the equation by two. This gives us two π¦ equals three
times π₯ minus one. Then distributing the parentheses
on the right-hand side would give us two π¦ equals three π₯ minus three.

Any of these three equations would
be valid equations of the straight line. However, we also need to identify
the inequality represented by the shaded region. Letβs clear some space and work
with this equation, two π¦ equals three π₯ minus three. Because we have an inequality, that
means that instead of an equals sign, weβll have one of the following: greater than,
less than, greater than or equal to, or less than or equal to. We therefore recall that we need to
check if this boundary line, the line of the equation, is dotted or a complete
line. Since the line here is a dotted
line, that means weβll have a strict inequality. It will either be greater than or
less than but none of the inequalities which involve the equal to portion.

In order to identify which one of
these two inequalities it will be, we can check a coordinate which lies in the
region which is represented. We could pick any coordinate in the
region, but letβs pick a nice easy coordinate. The coordinate zero, zero lies in
the shaded region. Remember that this means that the
π₯-value is zero and the π¦-value is zero. So letβs substitute these values
in. That means weβll be comparing two
times zero with three times zero minus three. When we simplify this, weβre
comparing zero and negative three. Which inequality would we select
here? Well, zero is greater than negative
three. That means that the region must be
two π¦ is greater than three π₯ minus three.

We can note that there are a number
of different ways in which we can represent this inequality. For example, we could have
subtracted three π₯ from both sides of this inequality. Or indeed, we could have collected
all three terms on one side of the inequality. Any alternative forms are
acceptable, provided we perform the correct steps when rearranging. So here we can give the answer that
the inequality represented by the colored region is two π¦ minus three π₯ is greater
than negative three.

We can now summarize the key points
of this video. Firstly, we saw that the solutions
to two-variable linear inequalities can be represented using graphs. Then, to determine the inequality
that has been graphed, we first determine the equation of the boundary line using
either the pointβslope or slopeβintercept forms of the equation of a straight
line. The boundary lines of strict
inequalities are broken or dashed lines. Weak inequalities are represented
by solid lines. Finally, to determine the
inequality sign, we consider the coordinates of a point in the shaded region. By substituting these coordinates
into each side of the equation of the line, we can determine which side has the
greater value and hence the inequality sign which we need.