# Video: Pack 2 β’ Paper 3 β’ Question 19

Pack 2 β’ Paper 3 β’ Question 19

03:59

### Video Transcript

Find the three inequalities that define the shaded region.

In order to find the inequalities that define this region, we must first identify the equations of each of the given lines. Probably, the easiest line to identify is the horizontal line. This line passes through the π¦-axis at negative three. In fact, every coordinate on this line has a π¦-value of negative three. This means it has the equation of π¦ is equal to negative three.

We can see that the shaded region sits above the line π¦ is equal to negative three. It is also a solid line. Therefore, this inequality must be π¦ is greater than or equal to negative three. Remember a dotted line is just greater than or less than, whereas the solid line is greater than or equal to or less than or equal to.

The other two equations are a little trickier. Remember the equation of any straight line is given as π¦ is equal to ππ₯ plus π, where π is the gradient and π is the π¦-intercept. If we begin by looking at the slightly steeper line, we can see that it has a π¦-intercept of positive three. The gradient is a little trickier. The gradient is sometimes called rise over run or change in π¦ over change in π₯.

If we choose two coordinates on this line as shown, we can construct a right-angled triangle. The change in π¦ is given by the height of the triangle which is positive four. The change in π₯ is the width of the triangle, positive one. Its gradient is, therefore, four divided by one which is four. The equation of this line is, therefore, π¦ is equal to four π₯ plus three.

The shaded region lies to the left of this line. In order to decide which way we want the inequality symbol to go, we choose a coordinate on this left side of the line. Letβs choose negative two and negative three. In this coordinate, π₯ has a value of negative two and π¦ has a value of negative three. So we can substitute these values into the equation for our straight line. This gives us negative three on the left-hand side of the equation and four multiplied by negative two plus three on the right. That gives us a value of negative five.

Negative three is greater than negative five and the line is a solid line. So our inequality becomes π¦ is greater than or equal to four π₯ plus three. Our final line clearly has a π¦-intercept of negative one. Letβs construct another right-angled triangle to help us find the gradient.

By marking two of the coordinates we can choose, we can see the change in π¦ is the height of this triangle β itβs positive two. The change in π₯ is the width of the triangle which is four. The gradient of this line is, therefore, two divided by four which is a half. The equation of this straight line is, therefore, π¦ is equal to a half π₯ minus one. This time the shaded region lies underneath or to the right of the line. So we choose a point on this side. We can choose the same one as last time β negative two and negative three.

Substituting these values in, we get negative three on the left-hand side of our equation and a half multiplied by negative two minus one on the right. That gives us negative two. Negative three is less than negative two and itβs a solid line. So our inequality becomes π¦ is less than or equal to a half π₯ minus one.

The three inequalities that define the shaded region are π¦ is greater than or equal to negative three, π¦ is greater than or equal to four π₯ plus three, and π¦ is less than or equal to a half π₯ minus one.