Lesson Objectives

- Learn how to graph a system of linear inequalities
- Learn how to graph a system of nonlinear inequalities

## Graphing Systems of Linear and Nonlinear Inequalities

We previously learned how to graph a linear inequality in two variables. Let's review this procedure and then move into solving systems of linear inequalities.

Example 1: Graph each $$2x + y ≥ -2$$ $$x - 2y ≥ -6$$ Let's solve each inequality for y: $$y ≥ -2x - 2$$ $$y ≤ \frac{1}{2}x + 3$$ Let's graph our top inequality first: $$y ≥ -2x - 2$$ Now let's graph our bottom inequality: $$y ≤ \frac{1}{2}x + 3$$ The solution for the system is the overlap of the two graphs: Example 2: Graph each $$2x - y ≤ 5$$ $$4x + y > 7$$ Let's solve each inequality for y: $$y ≥ 2x - 5$$ $$y > -4x + 7$$ Let's graph our top inequality first: $$y ≥ 2x - 5$$ Now let's graph our bottom inequality: $$y > -4x + 7$$ The solution for the system is the overlap of the two graphs:

Example 3: Graph each $$6x + y < -5$$ $$-18x - 3y < 9$$ Let's solve each inequality for y: $$y < -6x - 5$$ $$y > -6x - 3$$ Before we graph anything, we can see that our boundary lines are going to be parallel. The slopes are the same and they each have a different y-intercept. Although we may still have a solution when parallel boundary lines occur, we should be prepared for a scenario in which there is "no solution".

Let's graph our top inequality first: $$y < -6x - 5$$ Now let's graph our bottom inequality: $$y > -6x - 3$$ When we graph the two inequalities on the same coordinate plane, we see there is no overlap. This means there is no section of the coordinate plane that satisfies both inequalities and, therefore, there is no solution for the system. No Solution ∅

Example 1: Graph each inequality $$(x - 1)^2 + (y + 3)^2 < 25$$ We begin by graphing the boundary. We will replace the less than symbol "<" with an equality symbol "=": $$(x - 1)^2 + (y + 3)^2=25$$ This is the graph of a circle, with a center at (1,-3) and a radius of 5. Since we have a strict inequality, we will graph our boundary as a dashed circle. Now, we will choose a test point. If (0,0) is not on the boundary, it is usually the easiest to use. Let's plug a 0 in for x and a 0 in for y in the original inequality: $$(0 - 1)^2 + (0 + 3)^2 < 25$$ $$(-1)^2 + (3)^2 < 25$$ $$1 + 9 < 25$$ $$10 < 25 \require{color}\hspace{.25em}\color{green}{✔}$$ Since the test point works as a solution to the original inequality it lies in the solution region. Since (0,0) is inside of the boundary, we will shade everything inside of the boundary or circle.

### Graphing a Linear Inequality in Two Variables

- Solve the inequality for y
- Graph the boundary line
- Replace the inequality symbol with an equality symbol
- Graph the resulting equation
- The boundary line is solid for a non-strict inequality
- The boundary line is dashed for a strict inequality

- Shade the solution region
- When the inequality is solved for y:
- We shade above the boundary line for a greater than
- We shade below the boundary line for a less than

### Graphing a System of Linear Inequalities

- Graph each inequality of the system
- Shade the solution region for the system
- The solution region for the system is the area of the coordinate plane that satisfies both inequalities
- We can think of the solution region as the overlap of the graphs

Example 1: Graph each $$2x + y ≥ -2$$ $$x - 2y ≥ -6$$ Let's solve each inequality for y: $$y ≥ -2x - 2$$ $$y ≤ \frac{1}{2}x + 3$$ Let's graph our top inequality first: $$y ≥ -2x - 2$$ Now let's graph our bottom inequality: $$y ≤ \frac{1}{2}x + 3$$ The solution for the system is the overlap of the two graphs: Example 2: Graph each $$2x - y ≤ 5$$ $$4x + y > 7$$ Let's solve each inequality for y: $$y ≥ 2x - 5$$ $$y > -4x + 7$$ Let's graph our top inequality first: $$y ≥ 2x - 5$$ Now let's graph our bottom inequality: $$y > -4x + 7$$ The solution for the system is the overlap of the two graphs:

### Systems of Linear Inequalities with No Solution

In some cases, we will see a system of linear inequalities with no solution. This will occur when the boundary lines are parallel and there is no overlap between the two graphs. Let's take a look at an example.Example 3: Graph each $$6x + y < -5$$ $$-18x - 3y < 9$$ Let's solve each inequality for y: $$y < -6x - 5$$ $$y > -6x - 3$$ Before we graph anything, we can see that our boundary lines are going to be parallel. The slopes are the same and they each have a different y-intercept. Although we may still have a solution when parallel boundary lines occur, we should be prepared for a scenario in which there is "no solution".

Let's graph our top inequality first: $$y < -6x - 5$$ Now let's graph our bottom inequality: $$y > -6x - 3$$ When we graph the two inequalities on the same coordinate plane, we see there is no overlap. This means there is no section of the coordinate plane that satisfies both inequalities and, therefore, there is no solution for the system. No Solution ∅

### Nonlinear Inequalities

A second-degree inequality is an inequality with at least one variable raised to the second power and no variable with an exponent larger than 2. When we graph a second-degree inequality, we use the following steps:- Graph the boundary, the boundary separates the solution region from the non-solution region
- To graph the boundary, replace the inequality symbol with an equality symbol. Graph the resulting equation.
- The boundary is solid for a non-strict inequality "≥" or "≤"
- The boundary is dashed or broken for a strict inequality ">" or "<"

- Use a test point to find and shade the solution region
- Recall that the test point can be any point that is not on the boundary
- If the test point works in the original inequality, the test point lies in the solution region
- If the test point fails in the original inequality, the test point lies in the non-solution region

Example 1: Graph each inequality $$(x - 1)^2 + (y + 3)^2 < 25$$ We begin by graphing the boundary. We will replace the less than symbol "<" with an equality symbol "=": $$(x - 1)^2 + (y + 3)^2=25$$ This is the graph of a circle, with a center at (1,-3) and a radius of 5. Since we have a strict inequality, we will graph our boundary as a dashed circle. Now, we will choose a test point. If (0,0) is not on the boundary, it is usually the easiest to use. Let's plug a 0 in for x and a 0 in for y in the original inequality: $$(0 - 1)^2 + (0 + 3)^2 < 25$$ $$(-1)^2 + (3)^2 < 25$$ $$1 + 9 < 25$$ $$10 < 25 \require{color}\hspace{.25em}\color{green}{✔}$$ Since the test point works as a solution to the original inequality it lies in the solution region. Since (0,0) is inside of the boundary, we will shade everything inside of the boundary or circle.

## How to Graph a System of Non-Linear Inequalities

When we encounter a system of non-linear inequalities, we will begin by graphing each inequality separately. Once this is done, we will shade the overlap between the graphs. This overlap is the section of the graph that satisfies all inequalities of the system. Let's look at an example.

Example 2: Graph each non-linear system of equations $$-2x + y ≤ -3$$ $$y ≥ x^2 - 3x - 4$$ To solve this system, we will graph each inequality separately. Let's begin by graphing our first inequality. $$-2x + y ≤ -3$$ Solve for y: $$y ≤ 2x - 3$$ We will replace the inequality symbol with an equality symbol. This will be our boundary line. $$y=2x - 3$$ Since we solved the inequality for y, we don't need to use a test point. Since y is less than or equal to 2x - 3, we know we need to shade below the line. Alternatively, if we used a test point of (0,0), we would see this point would lie in the non-solution region. $$-2(0) + (0) ≤ -3$$ $$0 ≤ -3 \hspace{.25em}\color{red}{✖}$$ Let's shade below the line: Now, let's think about our second inequality. $$y ≥ x^2 - 3x - 4$$ We will replace our inequality symbol with an equality symbol. This will give us our boundary. $$y=x^2 - 3x - 4$$ Vertex: (3/2, -25/4)

x-intercepts: (4,0),(-1,0)

y-intercept: (0,-4)

Additional Points: (3,-4), (5,6), (-2,6) Let's graph our boundary: If we use the test point of (0,0), we will see that it lies in the solution region. $$0 ≥ (0)^2 - 3(0) - 4$$ $$0 ≥ -4 \hspace{.25em}\color{green}{✔}$$ Now, let's shade the solution region. This will be inside of the parabola. Now, to find the solution for the system, let's overlay the two graphs on top of each other. The solution will be the overlap of the two solution regions. This area of the coordinate plane will satisfy both inequalities of the system. The area in orange is the overlap of the two graphs. We can complete our graph of the system by shading this area only:

Example 2: Graph each non-linear system of equations $$-2x + y ≤ -3$$ $$y ≥ x^2 - 3x - 4$$ To solve this system, we will graph each inequality separately. Let's begin by graphing our first inequality. $$-2x + y ≤ -3$$ Solve for y: $$y ≤ 2x - 3$$ We will replace the inequality symbol with an equality symbol. This will be our boundary line. $$y=2x - 3$$ Since we solved the inequality for y, we don't need to use a test point. Since y is less than or equal to 2x - 3, we know we need to shade below the line. Alternatively, if we used a test point of (0,0), we would see this point would lie in the non-solution region. $$-2(0) + (0) ≤ -3$$ $$0 ≤ -3 \hspace{.25em}\color{red}{✖}$$ Let's shade below the line: Now, let's think about our second inequality. $$y ≥ x^2 - 3x - 4$$ We will replace our inequality symbol with an equality symbol. This will give us our boundary. $$y=x^2 - 3x - 4$$ Vertex: (3/2, -25/4)

x-intercepts: (4,0),(-1,0)

y-intercept: (0,-4)

Additional Points: (3,-4), (5,6), (-2,6) Let's graph our boundary: If we use the test point of (0,0), we will see that it lies in the solution region. $$0 ≥ (0)^2 - 3(0) - 4$$ $$0 ≥ -4 \hspace{.25em}\color{green}{✔}$$ Now, let's shade the solution region. This will be inside of the parabola. Now, to find the solution for the system, let's overlay the two graphs on top of each other. The solution will be the overlap of the two solution regions. This area of the coordinate plane will satisfy both inequalities of the system. The area in orange is the overlap of the two graphs. We can complete our graph of the system by shading this area only:

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