Video: AQA GCSE Mathematics Higher Tier Pack 1 β€’ Paper 2 β€’ Question 26

AQA GCSE Mathematics Higher Tier Pack 1 β€’ Paper 2 β€’ Question 26

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Video Transcript

The region 𝑅 represents the solutions to the inequalities π‘₯ plus 𝑦 is less than or equal to three, 𝑦 is greater than one-third π‘₯ plus one, π‘₯ is greater than negative two. Draw the region 𝑅 on the grid.

So we’ve been given three inequalities and we’re asked to draw the region 𝑅, which represents the solution to all three inequalities simultaneously. We first need to draw our lines which represent the boundary of this region 𝑅. And to do this, we change each inequality sign to an equal sign.

So we have π‘₯ plus 𝑦 equals three, 𝑦 equals a third π‘₯ plus one, and π‘₯ equals negative two. And these are the equations of the three lines which we need to draw. Let’s consider the line π‘₯ equals negative two first of all. Lines with the equation π‘₯ equals some constant are vertical lines. And as our line is π‘₯ equals negative two, it will pass through negative two on the π‘₯-axis.

Before we can draw this line then, we need to recall what type of line we’re drawing. And to do this, we look at the type of inequality sign that we have. If we have a weak inequality sign, that’s less than or equal to or greater than or equal to, then we represent this with a solid line because points that are on the line itself do satisfy the inequality.

If however we have a strict inequality, so that’s strictly less than or strictly greater than, then we represent this with a dotted line because points that lie on the line itself do not satisfy the inequality.

In this case, the inequality sign was a strict inequality. So we’re going to draw π‘₯ equals negative two as a dotted line. So here is that first line.

Now, let’s consider the second inequality for which we join the line 𝑦 equals one-third π‘₯ plus one. Now this is going to be a diagonal line. And we can compare it with the general form of the equation for a straight line: 𝑦 equals π‘šπ‘₯ plus 𝑐, where π‘š represents the gradient and 𝑐 represents the 𝑦-intercept.

We can identify that the 𝑦-intercept of this line is one, which means it is going to cross the 𝑦-axis at this value. So we can plot our first point. The gradient of the line is one-third, which means for every one unit we move to the right the line moves one-third of a unit up.

However, this isn’t particularly easy to draw. So what we can do instead is say that for every three units we move to the right, the line moves one unit up. If we move three units right and one unit up of our 𝑦-intercept, we get the point three, two. So we can plot this point.

We can also go the other way from our 𝑦-intercept, three units left and one unit down, to give the point negative three, zero. So now, we have three points that lie on this line. As the original inequality was a strict inequality, we’re again going to represent this with a dotted line. So we have our second line.

For the first inequality, the line we’re drawing is equation π‘₯ plus 𝑦 equals three. And we can rearrange this into the form 𝑦 equals π‘šπ‘₯ plus 𝑐 by subtracting π‘₯ from each side, which gives 𝑦 equals negative π‘₯ plus three.

The line has a 𝑦-intercept of three, so we can plot this point, and a gradient of negative one, meaning that for every one unit we move to the right, the line moves one unit down, so moving one unit right and one unit down of our 𝑦-intercept. And then again we can plot some more points that lie on our line.

The original inequality was a weak inequality. So this time we’re going to be drawing the line as a solid line. There’s actually another method that we could use to plot the two diagonal lines if you weren’t that comfortable with the gradient intercept method.

We can choose different π‘₯-values, substitute them into the equation of the line, and work out their corresponding 𝑦-values. For example, when π‘₯ is equal to zero, 𝑦 is equal to negative one multiplied by zero plus three which is equal to three.

This tells us the point zero, three lies on the line 𝑦 equals negative π‘₯ plus three. We knew this using our first method because three was the 𝑦-intercept of the line. We can also find the 𝑦-values for other π‘₯-values. For example, when π‘₯ is equal to one, 𝑦 will be equal to negative one multiplied by one plus three which is equal to two. So the point with coordinates one, two also lies on this line.

We’ve actually got all the information we need to plot the line. But it is always sensible to have at least three points so that we can check our working. Once you have your three points, you can plot them and then join them together to draw this line.

So now that we have all three lines on our grid, we can see that the three lines enclose a triangular region. So it’s probably the case that this triangular region is the region 𝑅. But we need to check this by considering which side of each line we should be shading.

We do this one line at a time. For example, for the line π‘₯ equals negative two, this was representing the inequality π‘₯ is greater than negative two, which means we need to shade the side of the line where the π‘₯-values are bigger than negative two, which means we’re shading the right of the line.

The line 𝑦 equals a third π‘₯ plus one was representing the inequality 𝑦 is greater than a third π‘₯ plus one, which means for any given π‘₯-value, the corresponding 𝑦-value needs to be bigger than the value on the line. So we’re going to be shading above this line.

We rearrange the equation π‘₯ plus 𝑦 equals three to give the equation 𝑦 equals negative π‘₯ plus three. So if we’d rearrange the inequality, we’d have 𝑦 is less than or equal to negative π‘₯ plus three.

This means that for any π‘₯-value, the 𝑦-value needs to be less than or equal to the value on the line, which means we’re going to be shading below the line. As we’ve drawn the line as a solid line, points on the line itself are also included.

We can see that by shading to the right of π‘₯ equals negative two above 𝑦 equals a third π‘₯ plus one and below 𝑦 equals negative π‘₯ plus three, we do indeed shade the triangular region that’s bounded by all three lines, which we can label as 𝑅.

Now, there is actually another method that we could use to confirm that it is indeed the triangular region we want to shade. And what we do is pick any point that lies inside this region. So for example, I’m going to choose the point with coordinates negative one, three.

We then need to check whether or not this point satisfies each of the three inequalities. In the case of the third inequality, the π‘₯-coordinate of this point is negative one. And as negative one is greater than negative two, this point certainly satisfies the third inequality.

For the second inequality, we need to check that the 𝑦-coordinate which is three is bigger than one-third of the π‘₯-coordinate. That’s one-third multiplied by negative one plus one. One-third multiplied by negative one is negative one-third. And then we’re adding one to this, which we can also think of as one minus one-third.

One minus one-third is two-thirds. And as three is greater than two-thirds, this means that for this point the 𝑦-value is greater than one-third of the π‘₯-value plus one. So it does indeed satisfy the second inequality.

For the first inequality, we need to check whether the sum of the π‘₯- and 𝑦-coordinates is less than or equal to three. So we have negative one plus three. Negative one plus three is two. You can also think of this as three minus one. And as two is less than three, it’s certainly less than or equal to three. And therefore, this point also satisfies the first inequality.

This confirms that the point negative one, three is indeed in the region 𝑅 because it satisfies all three inequalities. And therefore, the triangular region that we’ve shaded is the solution to all three inequalities.

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