### Video Transcript

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

As the line in our figure is broken or dotted, we know that this corresponds to strictly greater than or strictly less than. The reason we use a broken line to represent strict inequalities is because the points on the line itself are not included in the solution set. Our first step here will be to find the equation of the line which we know will be in the form π¦ equals ππ₯ plus π, where π is the π¦-intercept and π is the slope or gradient.

If we select any two points on the line, we can calculate the slope by finding the difference between the π¦-coordinates and dividing this by the difference in the π₯-coordinates. This is sometimes known as the rise over the run. The slope or gradient π is equal to π¦ two minus π¦ one divided by π₯ two minus π₯ one. Whilst it is not essential, it is useful to select two points with integer coordinates. In this case, we will choose the two coordinates five, six and negative five, negative nine. We will let the first of these have coordinates π₯ one, π¦ one and the second π₯ two, π¦ two.

Substituting in our values, π is equal to negative nine minus six divided by negative five minus five. This simplifies to negative 15 over negative 10. And dividing the numerator and denominator by negative five, we see that the slope or gradient of the line is three over two or three-halves. This means that our line has equation π¦ is equal to three over two π₯ plus π. It appears from the graph that our line intersects the π¦-axis at negative 1.5 or negative three over two. We can check this by substituting one of our known points into our equation. Substituting in the coordinates five, six gives us six is equal to three over two multiplied by five plus π.

The right-hand side simplifies to 15 over two or 7.5 plus π. Subtracting 15 over two from both sides of our equation, we see that π does indeed equal negative three over two or negative 1.5. The π¦-intercept is equal to negative three over two. We now have our equation written in slopeβintercept form. We see from the figure that all the points in the region that is shaded lie above this line, for example, negative two, eight.

To determine which of our inequality signs is correct, we can substitute these coordinates into our equation. We will then be able to work out whether the left- or right-hand side is greater. As π₯ is equal to negative two and π¦ is equal to eight, the left-hand side becomes eight and the right-hand side is equal to three over two multiplied by negative two minus three over two.

The right-hand side simplifies to negative three minus three over two. This is equal to negative nine over two or negative 4.5. Eight is clearly larger than this. Therefore, the left-hand side is greater than the right-hand side. We could repeat this for any other point in our shaded region, telling us that π¦ is greater than three over two π₯ minus three over two.

Whilst the slopeβintercept form is perfectly acceptable, in this question we will rearrange so that our inequality is in general form. We begin by multiplying both sides of the inequality by two. Two π¦ is therefore greater than three π₯ minus three. Next, we can subtract three π₯ from both sides. This gives us the inequality two π¦ minus three π₯ is greater than negative three.

The inequality whose solution set is represented by the colored region is two π¦ minus three π₯ is greater than negative three.