Graphing Systems of Inequalities: Videos & Practice Problems

Graphing two-dimensional inequalities involves a few straightforward steps that build on the concepts of one-dimensional inequalities. When dealing with inequalities like \( x \geq 1 \), the first step is to graph the corresponding line, which in this case is a vertical line at \( x = 1 \). This line divides the plane into two regions. To determine which side to shade, you need to identify the points that satisfy the inequality. For \( x \geq 1 \), you would shade the region to the right of the line, as all points in that area have x-coordinates greater than or equal to 1.

When graphing inequalities involving both x and y, such as \( y < x \), you start by graphing the line \( y = x \). Since the inequality is strict (less than), you would use a dashed line to indicate that points on the line are not included in the solution set. The next step is to determine which side of the line to shade. This can be done by testing a point not on the line, such as \( (0, 0) \). If you substitute these coordinates into the inequality and find a true statement, you shade the side of the line that includes that point. If the statement is false, you shade the opposite side.

For more complex inequalities like \( y > 2x - 4 \), the process is similar. First, graph the line \( y = 2x - 4 \) using a dashed line since the inequality is strict. Next, test a point, such as \( (0, 1) \). Substituting these values into the inequality shows that \( 1 > 2(0) - 4 \) simplifies to \( 1 > -4 \), which is true. Therefore, you would shade the region above the line, indicating that all points in that area satisfy the inequality.

In summary, when graphing linear inequalities, always start by graphing the corresponding line, determine whether to use a solid or dashed line based on the inequality symbol, and then test a point to decide which side of the line to shade. If the inequality can be rearranged into slope-intercept form \( y = mx + b \), you can quickly identify that for \( y > mx + b \), you shade above the line, and for \( y < mx + b \), you shade below the line. This method provides a clear visual representation of the solution set for the inequality.

In this exploration of systems of inequalities, we analyze three equations and their corresponding graphs. The first equation, \( y > -3x + 4 \), represents a line with a y-intercept of 4 and a slope of -3. To visualize this, we consider the line as if it were an equation of equality, which helps us identify the graph. Since the inequality is a 'greater than' symbol, the line is dashed, indicating that points on the line are not included in the solution set. The area above the line is shaded, representing all the solutions where y is greater than the line.

Next, we examine the second equation, \( x + y < 1 \). Rewriting this in slope-intercept form gives us \( y < -x + 1 \). This line has a y-intercept of 1 and a slope of -1. Similar to the first equation, the 'less than' symbol indicates a dashed line, and we shade below the line to represent the solutions where y is less than the line.

Finally, we look at the third equation, \( -3x - y \geq -4 \). By flipping the signs, we can rewrite it as \( 3x + y \leq 4 \). This transformation also flips the inequality symbol. In slope-intercept form, this becomes \( y \leq -3x + 4 \). Here, the y-intercept remains 4, and the slope is still -3. However, since we have a 'less than or equal to' symbol, the line is solid, indicating that points on the line are included in the solution set, and we shade below the line.

In summary, we matched each equation to its graph based on the characteristics of the lines and the direction of shading determined by the inequality symbols. Understanding how to manipulate equations into slope-intercept form and recognizing the implications of different inequality symbols are crucial skills in graphing systems of inequalities.

Graphing nonlinear inequalities, such as quadratic equations, follows a similar process to graphing linear inequalities. For example, consider the inequality \( y \geq x^2 - 1 \). This represents a quadratic function, where the graph is a parabola. The first step is to determine whether to use a solid or dashed line based on the inequality symbol. Since the symbol is "greater than or equal to," a solid line will be used to indicate that points on the line are included in the solution set.

The equation \( y = x^2 - 1 \) has a vertex at the point (0, -1) and opens upwards. To visualize this, sketch the parabola, ensuring the vertex is correctly placed. Next, to determine which side of the parabola to shade, select a test point. A convenient choice is (0, 2), which lies above the parabola. Substitute these values into the inequality:

\( 2 \geq 0^2 - 1 \)

This simplifies to:

\( 2 \geq -1 \)

This statement is true, indicating that the region containing the point (0, 2) satisfies the inequality. Therefore, shade the area above the parabola, which represents all points where \( y \) is greater than or equal to \( x^2 - 1 \).

It is important to remember that the parabola acts as a boundary. While shading above the curve, ensure that you do not include points that lie outside the bounds of the parabola itself. The solution set includes all points above the curve, effectively capturing the area inside the "bowl" formed by the parabola.

In summary, graphing nonlinear inequalities involves determining the type of line to use, sketching the corresponding curve, testing points to identify the correct shading region, and ensuring that the shaded area accurately reflects the solution set defined by the inequality.

To graph the inequality \( x^2 + (y - 1)^2 \leq 9 \), we start by determining the type of line to use. Since the inequality includes a "less than or equal to" symbol, we will use a solid line. This indicates that points on the boundary are included in the solution set.

Next, we rewrite the inequality as an equation to find the boundary: \( x^2 + (y - 1)^2 = 9 \). This equation resembles the standard form of a circle, \( (x - h)^2 + (y - k)^2 = r^2 \), where \( (h, k) \) is the center and \( r \) is the radius. Here, we identify \( h = 0 \) and \( k = 1 \), giving us a center at the point \( (0, 1) \). The right side, \( 9 \), indicates that \( r^2 = 9 \), so the radius \( r \) is \( \sqrt{9} = 3 \).

To graph the circle, we plot the center at \( (0, 1) \) and then extend 3 units in all directions: up to \( (0, 4) \), down to \( (0, -2) \), right to \( (3, 1) \), and left to \( (-3, 1) \). Connecting these points forms a circle with a radius of 3.

After graphing the circle, we need to determine where to shade. To do this, we can test a point to see if it satisfies the inequality. A convenient point to test is \( (2, 0) \). Substituting this into the inequality gives:

\( 2^2 + (0 - 1)^2 \leq 9 \)

\( 4 + 1 \leq 9 \)

\( 5 \leq 9 \), which is true.

This means the area that includes the point \( (2, 0) \) should be shaded. However, we must ensure that we are shading the correct region. Testing another point, such as \( (2, 2) \), which is also below the circle but outside it, can help clarify:

\( 2^2 + (2 - 1)^2 \leq 9 \)

\( 4 + 1 \leq 9 \)

\( 5 \leq 9 \), which is also true.

However, testing a point like \( (4, 1) \) gives:

\( 4^2 + (1 - 1)^2 \leq 9 \)

\( 16 + 0 \leq 9 \)

\( 16 \leq 9 \), which is false.

This indicates that the area outside the circle does not satisfy the inequality. Therefore, we conclude that the correct shading is the area inside the circle, as only points within this region will satisfy the inequality \( x^2 + (y - 1)^2 \leq 9 \).

Graphing a system of inequalities involves plotting multiple inequalities on the same coordinate plane and identifying the region where their shaded areas overlap. This overlapping region represents the solution set that satisfies all inequalities simultaneously.

To begin, each inequality is graphed individually. For instance, consider the inequality \(y \leq -x + 4\). This is represented by a solid line because the inequality includes equality (the "less than or equal to" symbol). The line intersects the y-axis at the point (0, 4) and has a slope of -1. After graphing the line, the area below it is shaded, indicating all points that satisfy the inequality.

Next, for the inequality \(y > 2x + 1\), a dashed line is drawn since the inequality does not include equality (the "greater than" symbol). This line has a y-intercept of 1 and a slope of 2. The area above this line is shaded, representing the points that satisfy this inequality.

To find the solution to the system, the overlapping shaded regions from both inequalities are identified. This intersection is where all conditions are met, and it represents the solution set for the system of inequalities.

In cases where one or more inequalities are nonlinear, such as parabolas, the same principles apply. For example, the inequality \(y \geq x^2 - 4\) represents a parabola opening upwards with a vertex at (0, -4). The area above this parabola is shaded, indicating the solutions that satisfy this inequality. When combined with another linear inequality, the overlapping region again provides the solution set.

It is also important to note that some systems of inequalities may have no solutions, which occurs when the shaded regions do not intersect at all. Understanding how to graph and analyze these inequalities is crucial for solving systems effectively.