### Video Transcript

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.

To begin answering this question, we will identify the three lines on our graph that represent the equations π¦ equals two, π¦ equals negative π₯, and π₯ equals one. Any equation in the form π¦ equals π, where π is some constant, will correspond to a horizontal line. This means that the graph of the equation π¦ equals two passes through the π¦-axis at two. Any equation in the form π₯ equals π, where π is a constant, will correspond to a vertical line. This means that the graph of the equation π₯ equals one is a vertical line passing through one on the π₯-axis.

We note that both of these lines drawn on the graph are dashed. This is because the inequalities are strictly greater than and strictly less than. When the inequality sign is greater than or equal to or less than or equal to, we have a solid line. The equation π¦ equals negative π₯ tells us that every π¦-value will be equal to the negative of the π₯-value. This means that the coordinates one, negative one and two, negative two lie on this line. In the same way, the points negative one, one and negative two, two lie on the line π¦ equals negative π₯. The line also passes through the origin and has a negative slope or gradient as shown on the graph.

Now that weβve identified the three lines, we can consider the inequality signs. We are told that π¦ is greater than two. Therefore, our region must lie above the line π¦ equals two. We know that π₯ must be less than one. Therefore, our region is to the left of the line π₯ equals one. Finally, as π¦ is greater than or equal to negative π₯, our region must be above the line π¦ equals negative π₯. The only part of our graph that satisfies all three of these inequalities is region π·.

We can check this answer by selecting a point within region π·. For example, letβs consider the point negative two, three. This has an π₯-coordinate equal to negative two and a π¦-coordinate equal to three. The first inequality stated that π¦ was greater than two. And as three is greater than two, this inequality is satisfied. Substituting in the point to our second inequality, we have three is greater than or equal to negative negative two. This is the same as three is greater than or equal to two. As this statement is correct, the point also satisfies the second inequality.

Finally, we have π₯ is less than one. And as our value of π₯ is negative two, we have negative two is less than one. This is also correct. So the point negative two, three satisfies all three inequalities. And as this point lies in region π·, this suggests our answer is correct.