# Video: Pack 1 • Paper 3 • Question 23

Pack 1 • Paper 3 • Question 23

03:34

### Video Transcript

Solve the system of inequalities by shading the correct region on the grid.

We’ve been given three inequalities and asked to represent the solution to this system of inequalities on the coordinate grid. To do so, we first need to draw three lines on the grid, the lines with equations 𝑦 equals negative two, 𝑥 equals negative two, and 𝑦 is equal to negative two 𝑥 plus three.

Let’s consider what each of these lines look like. An equation of the form 𝑦 equals constant is a horizontal line. Remember, when we’re representing inequalities on coordinate grids, the type of line we draw depends on the type of inequality. For strict inequalities, we use a dotted line. And for weak inequalities, we use a solid line.

As our first inequality is a weak inequality, we draw the line 𝑦 equals negative two as a solid line. As we’re looking for where 𝑦 is greater than or equal to negative two, this means that we’re going to be shading above the line. As we’ve drawn the line as a solid line, points on the line will also be included in our solution. Lines with the equation 𝑥 is equal to a constant are vertical lines.

As the second inequality is a weak inequality, we draw the line 𝑥 equals negative two as a dotted line. The inequality is 𝑥 is greater than negative two, which means we’re going to be shading points to the right of the line.

Let’s consider what the final line looks like. If we compare this to 𝑦 equals 𝑚𝑥 plus 𝑐, the general form of the equation of a straight line, then we can see that the 𝑦-intercept of this line is positive three. The gradient of this line is negative two, which means, for every one unit we move to the right, the line moves two units down.

As this inequality is a weak inequality, we again draw a dotted line for the line 𝑦 is equal to negative two 𝑥 plus three. Now we need to consider which side of this line we want to shade. To do so, let’s consider any point that doesn’t lie on the line. So I’m going to consider the origin, the point with coordinates zero, zero.

Let’s substitute zero for 𝑥 and zero for 𝑦 into the equation of the line. This gives zero on the left-hand side and negative two multiplied by zero plus three on the right-hand side. The right-hand side just simplifies to three. What’s the relationship between zero and three? Well, it’s that zero is less than three, which means that this point does satisfy the inequality 𝑦 is less than negative two 𝑥 plus three. And therefore, this point is in the solution region that we’re looking for. This means that we want to shade points that are to the left of the diagonal line we’ve drawn.

Putting all of this information together for the three lines, so that’s shading points above 𝑦 equals negative two to the right of 𝑥 equals negative two and below the diagonal line, we see that the region that solves this system of inequalities is the triangular region here.

We’ve already tested that the third inequality does hold true for the origin, which is a point in our region. But if you wanted to be extra confident in your answer, you could pick another random point inside the region, for example, the point with coordinates negative one, two, and confirm that the three inequalities are indeed all true for this point.