### Video Transcript

The region π
represents the solutions to the inequalities π₯ plus π¦ is less than or
equal to three, π¦ is greater than one-third π₯ plus one, π₯ is greater than
negative two. Draw the region π
on the grid.

So weβve been given three inequalities and weβre asked to draw the region π
, which
represents the solution to all three inequalities simultaneously. We first need to draw our lines which represent the boundary of this region π
. And to do this, we change each inequality sign to an equal sign.

So we have π₯ plus π¦ equals three, π¦ equals a third π₯ plus one, and π₯ equals
negative two. And these are the equations of the three lines which we need to draw. Letβs consider the line π₯ equals negative two first of all. Lines with the equation π₯ equals some constant are vertical lines. And as our line is π₯ equals negative two, it will pass through negative two on the
π₯-axis.

Before we can draw this line then, we need to recall what type of line weβre
drawing. And to do this, we look at the type of inequality sign that we have. If we have a weak inequality sign, thatβs less than or equal to or greater than or
equal to, then we represent this with a solid line because points that are on the
line itself do satisfy the inequality.

If however we have a strict inequality, so thatβs strictly less than or strictly
greater than, then we represent this with a dotted line because points that lie on
the line itself do not satisfy the inequality.

In this case, the inequality sign was a strict inequality. So weβre going to draw π₯ equals negative two as a dotted line. So here is that first line.

Now, letβs consider the second inequality for which we join the line π¦ equals
one-third π₯ plus one. Now this is going to be a diagonal line. And we can compare it with the general form of the equation for a straight line: π¦
equals ππ₯ plus π, where π represents the gradient and π represents the
π¦-intercept.

We can identify that the π¦-intercept of this line is one, which means it is going to
cross the π¦-axis at this value. So we can plot our first point. The gradient of the line is one-third, which means for every one unit we move to the
right the line moves one-third of a unit up.

However, this isnβt particularly easy to draw. So what we can do instead is say that for every three units we move to the right, the
line moves one unit up. If we move three units right and one unit up of our π¦-intercept, we get the point
three, two. So we can plot this point.

We can also go the other way from our π¦-intercept, three units left and one unit
down, to give the point negative three, zero. So now, we have three points that lie on this line. As the original inequality was a strict inequality, weβre again going to represent
this with a dotted line. So we have our second line.

For the first inequality, the line weβre drawing is equation π₯ plus π¦ equals
three. And we can rearrange this into the form π¦ equals ππ₯ plus π by subtracting π₯ from
each side, which gives π¦ equals negative π₯ plus three.

The line has a π¦-intercept of three, so we can plot this point, and a gradient of
negative one, meaning that for every one unit we move to the right, the line moves
one unit down, so moving one unit right and one unit down of our π¦-intercept. And then again we can plot some more points that lie on our line.

The original inequality was a weak inequality. So this time weβre going to be drawing the line as a solid line. Thereβs actually another method that we could use to plot the two diagonal lines if
you werenβt that comfortable with the gradient intercept method.

We can choose different π₯-values, substitute them into the equation of the line, and
work out their corresponding π¦-values. For example, when π₯ is equal to zero, π¦ is equal to negative one multiplied by zero
plus three which is equal to three.

This tells us the point zero, three lies on the line π¦ equals negative π₯ plus
three. We knew this using our first method because three was the π¦-intercept of the
line. We can also find the π¦-values for other π₯-values. For example, when π₯ is equal to one, π¦ will be equal to negative one multiplied by
one plus three which is equal to two. So the point with coordinates one, two also lies on this line.

Weβve actually got all the information we need to plot the line. But it is always sensible to have at least three points so that we can check our
working. Once you have your three points, you can plot them and then join them together to
draw this line.

So now that we have all three lines on our grid, we can see that the three lines
enclose a triangular region. So itβs probably the case that this triangular region is the region π
. But we need to check this by considering which side of each line we should be
shading.

We do this one line at a time. For example, for the line π₯ equals negative two, this was representing the
inequality π₯ is greater than negative two, which means we need to shade the side of
the line where the π₯-values are bigger than negative two, which means weβre shading
the right of the line.

The line π¦ equals a third π₯ plus one was representing the inequality π¦ is greater
than a third π₯ plus one, which means for any given π₯-value, the corresponding
π¦-value needs to be bigger than the value on the line. So weβre going to be shading above this line.

We rearrange the equation π₯ plus π¦ equals three to give the equation π¦ equals
negative π₯ plus three. So if weβd rearrange the inequality, weβd have π¦ is less than or equal to negative
π₯ plus three.

This means that for any π₯-value, the π¦-value needs to be less than or equal to the
value on the line, which means weβre going to be shading below the line. As weβve drawn the line as a solid line, points on the line itself are also
included.

We can see that by shading to the right of π₯ equals negative two above π¦ equals a
third π₯ plus one and below π¦ equals negative π₯ plus three, we do indeed shade the
triangular region thatβs bounded by all three lines, which we can label as π
.

Now, there is actually another method that we could use to confirm that it is indeed
the triangular region we want to shade. And what we do is pick any point that lies inside this region. So for example, Iβm going to choose the point with coordinates negative one,
three.

We then need to check whether or not this point satisfies each of the three
inequalities. In the case of the third inequality, the π₯-coordinate of this point is negative
one. And as negative one is greater than negative two, this point certainly satisfies the
third inequality.

For the second inequality, we need to check that the π¦-coordinate which is three is
bigger than one-third of the π₯-coordinate. Thatβs one-third multiplied by negative one plus one. One-third multiplied by negative one is negative one-third. And then weβre adding one to this, which we can also think of as one minus
one-third.

One minus one-third is two-thirds. And as three is greater than two-thirds, this means that for this point the π¦-value
is greater than one-third of the π₯-value plus one. So it does indeed satisfy the second inequality.

For the first inequality, we need to check whether the sum of the π₯- and
π¦-coordinates is less than or equal to three. So we have negative one plus three. Negative one plus three is two. You can also think of this as three minus one. And as two is less than three, itβs certainly less than or equal to three. And therefore, this point also satisfies the first inequality.

This confirms that the point negative one, three is indeed in the region π
because
it satisfies all three inequalities. And therefore, the triangular region that weβve shaded is the solution to all three
inequalities.