Lesson Video: Solving Systems of Linear Inequalities Mathematics • 9th Grade

In this video, we will learn how to solve systems of linear inequalities by graphing them and identify the regions representing the solution.

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Video Transcript

In this video, we will learn how to solve systems of linear inequalities by graphing them and identify the regions representing the solution. We should already be comfortable with using and interpreting inequality notation, plotting straight-line graphs given their equations, and graphing single linear inequalities.

A system of linear inequalities is a set of two or more linear inequalities in several variables. For example, 𝑦 is greater than or equal to one, π‘₯ is greater than zero, and π‘₯ plus 𝑦 is less than or equal to four is a system of linear inequalities in the two variables π‘₯ and 𝑦. Such systems are used in practical contexts when one problem requires a range of solutions, and there is more than one constraint on those solutions.

Let’s begin by recalling some of the basics of how we represent linear inequalities graphically. We’ll consider the inequalities 𝑦 is less than five and π‘₯ is greater than or equal to three. We recall that when representing inequalities on a graph, we first need to determine the equation of the boundary line of the region. And to do this, we swap the inequality sign for an equals sign. So the boundary line for the inequality 𝑦 is less than five is 𝑦 is equal to five. We then draw this line, recalling that lines of the form 𝑦 equals some constant are horizontal lines. We also need to recall that we use a solid line if we have a weak inequality and we use a dashed line if we have a strict inequality. Here, the inequality is strict. It’s 𝑦 is strictly less than five, so we draw the line 𝑦 equals five as a dashed line.

We now have the boundary line for the graphical solution to this inequality. And we need to decide which side of the line to shade. If we want 𝑦 to be less than five, then we need to include all points in the coordinate plane for which the 𝑦-coordinate is less than five. That’s all points below the boundary line, so we shade this side of the line.

We’ve now represented the solution to the first inequality graphically. So let’s now consider the second inequality: π‘₯ is greater than or equal to three. Once again, we replace the inequality with an equals sign to find the equation of the boundary line. The equation of the boundary is π‘₯ equals three, which we recall is a vertical line through the value three on the π‘₯-axis.

As we have a weak inequality, π‘₯ is greater than or equal to three, we represent this boundary with a solid line. We then need to decide which side of the line to shade. And as we want π‘₯ to be greater than or equal to three, we need to shade the region to the right of the boundary. Remember that the solid line indicates that points on the boundary line itself are included in the solution to this inequality. So, we’ve represented the solution to each inequality graphically.

If we want to treat this as a system of linear inequalities, then this means we need to find the region that satisfies both inequalities simultaneously, which is the intersection of the two regions we’ve shaded. The region below the orange line and to the right of the pink line therefore represents the solution to this system of inequalities. It contains all the points in the coordinate plane such that 𝑦 is less than five and π‘₯ is greater than or equal to three. Let’s now consider some examples.

State the system of inequalities whose solution is represented by the following graph.

From the diagram, we can see that the shaded region is the first quadrant of the coordinate plane, which contains all points such that both π‘₯ and 𝑦 are positive. The boundary lines for the region, which are the positive π‘₯- and 𝑦-axes, are drawn as solid lines. So points along the boundaries are also included. This means that zero values of both π‘₯ and 𝑦 are included in the region. Using a bit of logic, we can therefore express this region using the inequalities π‘₯ is greater than or equal to zero and 𝑦 is greater than or equal to zero.

Let’s confirm this by representing each of these inequalities graphically ourselves. The inequality π‘₯ is greater than or equal to zero is represented by the 𝑦-axis, where π‘₯ is equal to zero, and all the points to the right of the 𝑦-axis. The inequality 𝑦 is greater than or equal to zero is represented by the π‘₯-axis, where 𝑦 equals zero, and all points above the π‘₯-axis. The intersection of these two regions is the part that is shaded in both orange and pink. That’s the whole of the first quadrant, including the positive π‘₯- and 𝑦-axes, which matches up with the graph we were given. So, we can conclude that the system of inequalities represented by the given graph is π‘₯ is greater than or equal to zero and 𝑦 is greater than or equal to zero.

Let’s now consider a slightly more complex example.

State the system of inequalities whose solution is represented by the following graph.

We can see that this system of inequalities contains horizontal, vertical, and diagonal boundary lines, all of which we will need to find the equations of. First though, we identify that the shaded region is in the first quadrant, which contains all nonnegative values of π‘₯ and 𝑦. This can be represented by the inequalities π‘₯ is greater than or equal to zero and 𝑦 is greater than or equal to zero.

Next, let’s consider the vertical lines. The equations of these lines are π‘₯ equals three and π‘₯ equals six. The region between these lines has been shaded, so this corresponds to π‘₯ being between three and six. But we need to be clear about which inequality signs to use at each end of the interval. As the line drawn at π‘₯ equals three is a solid line, this corresponds to a weak inequality. So we have π‘₯ is greater than or equal to three. As the line drawn at π‘₯ equals six is a dashed line, this corresponds to a strict inequality. So we have π‘₯ is less than six. This can be combined into the double-sided inequality π‘₯ is greater than or equal to three and less than six.

Next, we consider the horizontal lines. All horizontal lines have equations of the form 𝑦 equals some constant. So the equations of these lines are 𝑦 equals two and 𝑦 equals six. The blue region is between the two lines, so 𝑦 is between two and six. As these lines are solid, both inequalities are weak inequalities. So this region can be represented as 𝑦 is greater than or equal to two and less than or equal to six.

Finally, we consider the orange region below the diagonal line. This line passes through eight on the π‘₯-axis and eight on the 𝑦-axis. We won’t go into the detail of how to find its equation here, but its equation can be written as 𝑦 equals eight minus π‘₯. As the shaded region is below the line and the line itself is solid, this corresponds to the inequality 𝑦 is less than or equal to eight minus π‘₯. We could leave this inequality in this form, or we can rearrange it by adding π‘₯ to both sides to give π‘₯ plus 𝑦 is less than or equal to eight.

So we’ve found the system of inequalities whose solution is represented by the graph. Their solution set is the intersection of the individual shaded regions, which is the green triangular region. The system of inequalities is π‘₯ is greater than or equal to zero. 𝑦 is greater than or equal to zero. π‘₯ is greater than or equal to three and less than six. 𝑦 is greater than or equal to two and less than or equal to six. And π‘₯ plus 𝑦 is less than or equal to eight.

Let’s now consider another example.

Which of the points below belongs to the solution set of the system of linear inequalities represented by the figure shown? (a) One, three; (b) six, eight; (c) four, three; or (d) three, six.

It’s important to realize that we aren’t being asked to find the system of linear inequalities represented by the figure shown but rather to determine which of four given points lies in their solution set. From the figure, we can see that there are essentially three types of linear inequalities that have been graphed. There are two vertical lines, and the region between these lines is shaded in red. There are two horizontal lines, and the region between these lines is shaded in blue. Finally, there is a diagonal line, and the region below this line is shaded in orange. The region that satisfies each of these inequalities simultaneously is the intersection of the three shaded regions, which is the triangle shaded in green.

To determine which of the four given points lies in the solution set of the system of inequalities, we just need to plot each of these points on the graph and identify any points that are inside or on the solid boundaries of this region. When we do this, we find that the only point that lies in the shaded region and hence belongs in the solution set of the system of linear inequalities is the point four, three.

Here’s another example.

Find the solution set of the linear inequalities shown below.

So there are two linear inequalities graphed on the same coordinate plane. And we’re asked to find the solution set of this pair of linear inequalities. This means we need to find the region that satisfies both inequalities simultaneously, which is the intersection of the two individual regions. But it’s clear from looking at the graph that these two regions don’t overlap. The two diagonal lines are parallel, so they never meet. The red region is above the higher diagonal line, and the blue region is below the lower line. As a result, there is no intersection between these two regions. And so the solution set of this pair of linear inequalities is the empty set, πœ™.

Let’s consider one final example.

Which region on the graph contains solutions to the set of inequalities 𝑦 is greater than two, 𝑦 is greater than or equal to negative π‘₯, and π‘₯ is less than one?

These inequalities have already been graphed for us, and our job is to identify which of the regions marked with letters represents the solution to all three inequalities simultaneously. We’ll do this by first identifying the solution set to each inequality separately. The boundary line for the inequality 𝑦 is greater than two is 𝑦 is equal to two, which is a horizontal line through the value two on the 𝑦-axis. That’s the dashed line drawn in blue. It’s a dashed line because the inequality is 𝑦 is strictly greater than two. As the inequality is 𝑦 is greater than two, the solution set to this inequality will be the region above the dashed line. So this immediately narrows the options down to G, D, and F.

The boundary line for the inequality 𝑦 is greater than or equal to negative π‘₯ is the line 𝑦 equals negative π‘₯, which is a diagonal line through the origin with a slope of negative one. That’s the solid line drawn in green, and a solid line has been used here because it is a weak inequality. As the inequality is 𝑦 is greater than or equal to negative π‘₯, the solution set to this inequality is the region on and above the line. We can therefore rule out region G.

For the final inequality, the boundary line is π‘₯ equals one, which is a vertical line through one on the π‘₯-axis. This is of course the final line drawn on the coordinate grid, the vertical dashed line drawn in red. As the inequality is π‘₯ is less than one, the region satisfying this inequality is to the left of the boundary line. This rules out option F. So, we can conclude that the region that contains the solutions to the system of all three inequalities, which is the intersection of the three individual regions, is region D.

Let’s now summarize the key points from this video. For a vertical line parallel to the 𝑦-axis of the form π‘₯ equals some constant π‘˜, the region to the right of the line contains all the points for which π‘₯ is greater than π‘˜ and the region to the left of the line contains all the points such that π‘₯ is less than π‘˜. For a horizontal line of the form 𝑦 equals π‘˜, the region above the line corresponds to 𝑦 is greater than π‘˜ and the region below the line corresponds to 𝑦 is less than π‘˜. For a diagonal line of the form 𝑦 equals π‘šπ‘₯ plus 𝑏, the region above the line corresponds to where 𝑦 is greater than π‘šπ‘₯ plus 𝑏 and the region below the line corresponds to where 𝑦 is less than π‘šπ‘₯ plus 𝑏.

We use dashed lines to represent strict inequalities, which means the boundary line itself is not included in the solution set, and solid lines to represent weak inequalities, which means the boundary line is included. The solution set for a system of linear inequalities is the region that satisfies every inequality simultaneously, which is the intersection of the individual regions. We also saw that if the regions do not overlap, then there are no solutions to the system of linear inequalities, in which case we write the solution set as the empty set πœ™.

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