Lesson Explainer: Two-Variable Linear Inequalities | Nagwa Lesson Explainer: Two-Variable Linear Inequalities | Nagwa

Lesson Explainer: Two-Variable Linear Inequalities Mathematics • First Year of Secondary School

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In this explainer, we will learn how to graph two-variable linear inequalities.

We should already be familiar with graphing single-variable linear inequalities and identifying the regions that satisfy these, such as 𝑦>2 or π‘₯β‰€βˆ’5. Recall that we begin by drawing the boundary line for the region, the equation for which we obtain by temporarily replacing the inequality sign with an equals sign. For example, if we want to graph the inequality 𝑦>2, the equation of the boundary line is 𝑦=2. We recall also that the type of line we use to bound the region is dependent on the type of inequality we are graphing.

  • We use a solid line to represent a weak inequality; that is, an inequality involving either of the signs ≀ or β‰₯. This indicates that points on the boundary line itself satisfy the inequality.
  • We use a dotted or broken line to represent a strict inequality; that is, an inequality involving either of the signs < or >. This indicates that points on the boundary line do not satisfy the inequality.

We then shade the side of the line that contains all the points that satisfy the inequality. This is also known as the solution set of the inequality. In the figure below, we illustrate the graphical representation of the inequality 𝑦>2. The region is bounded by the line 𝑦=2, which has been drawn as a broken line due to the strict inequality, and we have shaded the region above this line. The 𝑦-coordinate of every point in this region is greater than 2.

In this explainer, we will extend our knowledge to include two-variable linear inequalities. The boundary lines will therefore be diagonal lines with equations given in the various forms of the equation of a straight line, such as π‘Žπ‘₯+𝑏𝑦=𝑐 or 𝑦=π‘šπ‘₯+𝑏. We should already be familiar with finding the equation of a straight line from its graph, which will be a key skill required here.

Let us first consider an example in which we identify the correct notation for an inequality that has been represented graphically.

Example 1: Identifying the Notation Used When Graphing Inequalities

Consider the inequalities shown in figure 1 and figure 2.

Which of the following is correct?

  1. Figure 1 shows 𝑦>π‘₯; figure 2 shows 𝑦β‰₯π‘₯.
  2. Figure 1 shows 𝑦β‰₯π‘₯; figure 2 shows 𝑦>π‘₯.

Answer

Examining the two graphs, we see that in both cases the equation of the boundary line is 𝑦=π‘₯, as for every point on the line the π‘₯- and 𝑦-coordinates are equal. We demonstrate this for just some of the points on the line in the figures below.

We note that in both graphs, the region above the boundary line is shaded. This means that for any given π‘₯-value, the 𝑦-coordinates of the points in the shaded region are greater than the value of 𝑦 on the boundary line. For example, if we consider when π‘₯ is equal to βˆ’4, we see that the coordinates of some of the points that lie in the shaded region are (βˆ’4,βˆ’2), (βˆ’4,0), (βˆ’4,2), and (βˆ’4,4). In every case, the value of the 𝑦-coordinate is greater than the value of the π‘₯-coordinate.

The two inequalities graphed are therefore 𝑦>π‘₯ and 𝑦β‰₯π‘₯, but we need to determine which inequality corresponds to which figure. The only difference between the two graphs is how the boundary line has been drawn; in figure 1 it is represented using a broken line, whereas in figure 2 it is represented using a solid line.

We recall that broken lines are used to represent strict inequalities; that is, inequalities involving the strictly less than or strictly greater than signs. Solid lines are used to represent weak inequalities; that is, those involving the less than or equal to signs, or those involving the greater than or equal to signs.

The boundary line in figure 1 is broken, indicating that the inequality graphed is a strict inequality. It must therefore represent the solution to 𝑦>π‘₯. In figure 2, however, the boundary line is solid, indicating that the inequality graphed is 𝑦β‰₯π‘₯.

Hence, the answer is A: figure 1 shows 𝑦>π‘₯; figure 2 shows 𝑦β‰₯π‘₯.

In our first example, the equation of the boundary line was given to us. Let us now consider an example where we identify an inequality from its graph, applying our knowledge of the equations of straight lines to first determine the equation of the boundary line.

Example 2: Identifying an Inequality from Its Graph

Write the inequality that has been graphed in the given figure.

Answer

We begin by finding the equation of the line that bounds the shaded region. Recall that the equation of a straight line in slope–intercept form is 𝑦=π‘šπ‘₯+𝑏, where π‘š represents the slope of the line and 𝑏 represents the 𝑦-intercept. From the figure, we identify that this line intercepts the 𝑦-axis at βˆ’3, and so 𝑏=βˆ’3. To determine the slope of the line, we recall the formula π‘š=π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯, where (π‘₯,𝑦) and (π‘₯,𝑦) are the coordinates of two points on the line.

We can choose any two points on the line to use in calculating the slope, but it is most straightforward to use points with integer coordinates. We have already identified that the line passes through the point (0,βˆ’3), and from the figure we see that the point (1,1) also lies on the line.

Substituting these values for (π‘₯,𝑦) and (π‘₯,𝑦) into the slope formula gives π‘š=1βˆ’(βˆ’3)1βˆ’0=41=4.

Substituting 𝑏=βˆ’3 and π‘š=4 into the slope–intercept form of the equation of a straight line, we find that the equation of the boundary line is 𝑦=4π‘₯βˆ’3.

However, we are not asked simply to find the equation of the line, but to find the inequality that describes the shaded region. We note that the boundary of the region has been drawn using a broken line and so the inequality will be strict; it will involve either the strictly greater than (>) or strictly less than (<) sign.

Next, we observe that the region shaded lies below the line. This means that for any given π‘₯-value, the 𝑦-coordinates of all the points that lie in the shaded region are smaller than the 𝑦-value on the line. The 𝑦-value on the line is equal to the value of the expression 4π‘₯βˆ’3 and hence the inequality that has been graphed is 𝑦<4π‘₯βˆ’3.

As we have just seen, we need to determine the inequality represented by a graph based on which side of the boundary line has been shaded. We saw in the previous example one way to do this: by considering whether, for a given value of π‘₯, a corresponding value of 𝑦 in the shaded region is smaller or larger than the 𝑦-value on the line. An alternative method is described below.

How To: Determining the Inequality Sign Needed

  • Identify whether the boundary line itself is solid, in which case a weak inequality sign is needed, or whether it is broken, in which case a strict inequality is needed.
  • Choose any point in the shaded region.
  • Substitute the π‘₯- and𝑦-coordinates of this point into each side of the equation of the boundary line and evaluate each expression.
  • Determine which side has the greater value. Use the correct inequality sign to represent this, bearing in mind whether a weak or strict inequality is needed.

Let us now consider an example in which we use the coordinates of a point to determine the inequality sign needed.

Example 3: Determining the sign of an Inequality

Fill in the blank: The point (4,βˆ’2) belongs to the solution set of the inequality 3π‘₯+2𝑦8.

  1. <
  2. >
  3. β‰ 
  4. ≀

Answer

This problem requires us to determine which inequality sign belongs in the gap. We can do this by evaluating the expression on the left-hand side at the point (4,βˆ’2) and then comparing it to the value on the right-hand side.

When π‘₯=4 and 𝑦=βˆ’2, the left-hand side is equal to 3π‘₯+2𝑦=(3Γ—4)+(2Γ—βˆ’2)=12βˆ’4=8.

This is equal to the value on the right-hand side. However, we are told that this is an inequality, not an equation. Whatever inequality sign we choose must allow the two sides to be equal if (4,βˆ’2) is to be included in its solution set. From the four options presented, only option D is an inequality sign that allows the two sides to be equal for some values of π‘₯ and 𝑦.

The answer is D: ≀.

In the two examples that follow, we will practice determining the inequality represented by a graph by first calculating the equation of the boundary line and then considering which side of the line has been shaded. Our next example will involve a boundary line that has a noninteger slope. This does not make the inequalities involved any more complicated, but care will be needed when determining the equation of the boundary line. If, for example, we are using points with noninteger coordinates to determine the slope, we must ensure we look carefully at the scale used on the coordinate axes.

Example 4: Identifying an Inequality from Its Graph

Which inequality has been graphed in the given figure?

  1. 𝑦β‰₯βˆ’15π‘₯βˆ’3
  2. π‘¦β‰€βˆ’15π‘₯βˆ’3
  3. 𝑦β‰₯βˆ’3π‘₯βˆ’15
  4. 𝑦<βˆ’15π‘₯βˆ’3
  5. 𝑦>βˆ’3π‘₯βˆ’15

Answer

We begin by determining the equation of the boundary line. Recall that the equation of a straight line in slope–intercept form is 𝑦=π‘šπ‘₯+𝑏, where π‘š represents the slope and 𝑏 represents the 𝑦-intercept. From the figure, we identify that the line crosses the 𝑦-axis at βˆ’3, and so 𝑏=βˆ’3.

To determine the slope of the line, we require the coordinates of two points that lie on the line. We have already identified the point (0,βˆ’3), but there do not appear to be any other points that have integer coordinates in the figure. We therefore need to consider the scale that has been used on the axes. On both the π‘₯- and 𝑦-axes, there are five small squares between each integer value, and so each subdivision on these axes represents 0.2. From the figure, we identify that when π‘₯=βˆ’3, the 𝑦-coordinate on the line is 2 small squares below βˆ’2, and so the 𝑦-coordinate here is βˆ’2.4. Substituting the points (0,βˆ’3) and (βˆ’3,βˆ’2.4) into the slope formula gives π‘š=βˆ’3βˆ’(βˆ’2.4)0βˆ’(βˆ’3)=βˆ’3+2.40+3=βˆ’0.63=βˆ’0.2.

As a fraction, the slope is equal to βˆ’15. The equation of the boundary line is therefore 𝑦=βˆ’15π‘₯βˆ’3.

Next, we note that the boundary line has been drawn using a solid line, indicating a weak inequality. We now choose any point in the shaded region; the simplest point is the origin. Substituting 𝑦=0 into the left-hand side of the equation of the boundary line gives 0. Substituting π‘₯=0 into the right-hand side of the equation of the boundary line gives βˆ’15π‘₯βˆ’3=ο€Όβˆ’15Γ—0οˆβˆ’3=βˆ’3.

As 0>βˆ’3, we find that for points in the shaded region, the value of 𝑦 is greater than the value of the expression βˆ’15π‘₯βˆ’3.

Combining this with the requirement for a weak inequality to include points on the boundary line itself, we find that the inequality that has been graphed is 𝑦β‰₯βˆ’15π‘₯βˆ’3.

We could also have determined the correct inequality sign by observing that the region shaded lies above the boundary line.

In each example we have considered so far, our answers have been given in a form in which one variable, 𝑦, is isolated on one side of the inequality. However, this does not have to be the case and indeed we may sometimes be asked to give our answer in a different form. We can rearrange an inequality to any equivalent form using the balancing method. We must be careful when doing this to follow the usual rules for rearranging or solving inequalities; specifically, if we multiply or divide both sides of an inequality by a negative value, we must reverse the direction of the inequality sign. Let us now consider an example in which we give our answer in the form π‘Žπ‘¦+𝑏π‘₯>𝑐.

Example 5: Determining the Inequality Represented by a Given Graph

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

Answer

The solution set of an inequality is the set of all points that satisfy that inequality. In the figure, this corresponds to all points in the colored region.

We begin by determining the equation of the boundary line. It looks from the figure as if the line intersects the 𝑦-axis at βˆ’1.5, but we cannot be sure due to the scale used. We recall that the general equation of a straight line in point–slope form is π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯), where π‘š represents the slope of the line and (π‘₯,𝑦) are the coordinates of any point on the line.

From the figure, we identify the points (βˆ’3,βˆ’6) and (1,0) that both lie on the line and that we will first use to calculate the slope. We recall the formula for the slope of a line, π‘š=π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯.

Substituting the coordinates of the two points gives π‘š=βˆ’6βˆ’0βˆ’3βˆ’1=βˆ’6βˆ’4=32.

Substituting this value of π‘š and the point (1,0) into the point–slope form of the equation of a straight line gives π‘¦βˆ’0=32(π‘₯βˆ’1).

To simplify, we first multiply both sides of the equation by 2 and then distribute the parentheses on the right-hand side as follows: 2𝑦=3(π‘₯βˆ’1)2𝑦=3π‘₯βˆ’3.

We have therefore found the equation of the boundary line. We note also from the figure that the boundary line has been drawn using a broken line and so the inequality is strict, containing either the strictly less than (<) or strictly greater than (>) sign. To test which sign we should use, we can choose any point in the colored region. The simplest point is the origin (0,0).

Substituting 𝑦=0 into the left-hand side of the equation of the boundary line gives 2𝑦=2Γ—0=0.

On the right-hand side, we have the expression 3π‘₯βˆ’3, which is equal to (3Γ—0)βˆ’3=βˆ’3.

As 0>βˆ’3, the correct inequality is 2𝑦>3π‘₯βˆ’3. We do not have to give our answer in this particular form; any rearrangement of this inequality is also correct. By subtracting 3π‘₯ from each side, we obtain 2π‘¦βˆ’3π‘₯>βˆ’3.

In the previous example, we could have chosen to collect all three terms on the same side of the inequality and give our answer as 2π‘¦βˆ’3π‘₯+3>0. Alternatively, we could have collected all three terms on the other side of the inequality and given our answer as 3π‘₯βˆ’2π‘¦βˆ’3<0. Any of these alternative forms are acceptable, provided the correct steps have been followed when performing the rearrangement.

Let us finish by recapping some key points from this explainer.

Key Points

  • The solutions to two-variable linear inequalities can be represented using graphs.
  • The solution set of an inequality is the set of all points that satisfy that inequality and is represented on a graph by a shaded region.
  • To determine the inequality that has been graphed, we first determine the equation of the boundary line, using either the point–slope or the slope–intercept form of the equation of a straight line.
  • The boundary lines for strict inequalities are represented using broken lines, and those for weak inequalities are represented using solid lines.
  • To determine the inequality sign needed, we can consider the coordinates of any point that lies in the shaded region. By substituting the coordinates into each side of the equation of the boundary line, we can determine which side has the greater value and hence the inequality sign needed.

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