### Video Transcript

On the grid, shade the region that satisfies all three inequalities. π¦ is greater than negative three π₯ plus four, π₯ plus π¦ is less than six, and π¦
is greater than or equal to three π₯.

Before looking at the specific inequalities in this question, it is worth noting that
if our inequality sign is greater than or equal to or less than or equal to, we need
to draw a bold line. However, if the inequality sign is just greater than or just less than, we need to
draw a dotted line. In our question, the first two lines will be dotted lines, and the third line will be
a bold line.

There are lots of ways of approaching this question. One way is to rewrite the equations in the form π¦ equals ππ₯ plus π, as this will
enable us to work out the gradient π and the π¦-intercept π of each of the
inequalities. An alternative method would be to construct a table for each of the inequalities in
order to find some coordinates that we could plot on our grid.

If we consider the first inequality, π¦ is greater than negative three π₯ plus four,
and the three π₯ values negative two, zero, and two, we can substitute these into
the inequality to calculate the corresponding values of π¦. Substituting in π₯ equals negative two gives us negative three multiplied by negative
two plus four. Negative three multiplied by negative two is equal to six. Adding four to this gives us 10. Therefore, the corresponding value of π¦ is 10.

Substituting zero into the inequality gives us negative three multiplied by zero plus
four. This is equal to four. Therefore, when π₯ is equal to zero, π¦ is equal to four. Finally, when we substitute π₯ equals two into the inequality, we get negative three
multiplied by two plus four. This is equal to negative two. Therefore, when π₯ is equal to two, π¦ is equal to negative two.

Plotting these three points and joining them with a dotted line gives the line shown
on the grid. This line has an intercept of plus four and a gradient of negative three. As our inequality stated that π¦ is greater than negative three π₯ plus four, weβre
going to have to shade above this line, as shown by the pink arrows. We can create a similar table for our second inequality: π₯ plus π¦ is less than
six.

In order to calculate our corresponding values of π¦, it might be easier to rearrange
this inequality so it is in the form π¦ is less than six minus π₯ or π¦ is less than
negative π₯ plus six. Substituting in π₯ equals negative two gives us six minus negative two. This is equal to eight. In the same way, substituting π₯ equals zero gives us six minus zero, and
substituting π₯ equals two gives us six minus two.

This gives us two further values. When π₯ equals zero, π¦ is equal to six. And when π₯ equals two, π¦ is equal to four. We can once again plot these three coordinates and join them up with a dotted
line. As π₯ plus π¦ needed to be less than six, we need to shade underneath or below this
dotted line. Our final inequality is π¦ is greater than or equal to three π₯.

Substituting π₯ equals negative two into this inequality gives us a corresponding
value of π¦ of negative six. In the same way, π₯ equals zero gives us a value of π¦ equal to zero, and π₯ equals
two gives us a value of π¦ equal to six. We now need to plot these three points and join them with a bold line. As our inequality sign in this case was greater than or equal to, we need to shade
above the line.

We have now identified the region that needs to be shaded in orange. The borders of the region are denoted in pink, one bold line, and two dotted
lines. This is the region that satisfies the three inequalities: π¦ is greater than negative
three π₯ plus four, π₯ plus π¦ is less than six, and π¦ is greater than or equal to
three π₯.