Determine graphically the solution set for each system of inequalities and indicate whether the

step 1 So we have the following system of inequalities. And the purpose of the problem is to find out t

Determine graphically the solution set for each system of...

Key Concepts

Inconsistent Systems

In some systems of inequalities, the conditions imposed can be contradictory, resulting in an empty intersection where no single point satisfies all the inequalities. This situation is described as an inconsistent system, meaning there is no feasible solution set.

Bounded and Unbounded Regions

A region is bounded if it is contained within a finite area, meaning every point in the region lies within some fixed distance from a central point. In contrast, an unbounded region extends infinitely in at least one direction. Determining the nature of the feasible region helps in many applications, such as optimization.

Linear Inequalities

Linear inequalities are expressions that involve a linear function set in an inequality relationship. They define regions on the coordinate plane (half-planes) where the inequality holds true, with the boundary line determined by the corresponding equality.

Graphical Representation

Graphically solving linear inequalities involves plotting their corresponding boundary lines and then shading the half-planes that satisfy the inequalities. This visual method helps in understanding and identifying the regions where the inequalities overlap.

Feasible Region

The feasible region is the common area where all the shaded regions of the individual inequalities intersect. It represents the set of all points that simultaneously satisfy every inequality in the system.

Transcript

00:01 So we have the following system of inequalities.

00:03 And the purpose of the problem is to find out the region that they are sharing.

00:08 That means the resolution region of the system of inequalities, the vertices of the solution region, and to determine whether or not the region is bounded or unbounded.

00:18 Moreover, we're going to graph it.

00:20 And, first of all, to start, we're going to determine whether or not the graph of y is equal to x square, intercepts the line x equals zero and the line y equals four and this is used to determine whether or not we'll have vertices at these two points so at x equals zero y is equal to x square so y is equal to x square so y is equal to zero since this is true we'll have the point zero comma zero at y equals four we're going to have x square equals four so we're going to find the solution to this.

01:13 So taking the square root, we're going to have the points plus or minus two.

01:18 However, since we are restricted only for values of x greater than zero, we're going to say that the x coordinate is going to be two.

01:28 So our other point is going to be two, four, or our vertices.

01:36 So we have the vertices at zero 0 .0 and 2 .4.

01:45 Now, we know how to craft these two equations.

01:48 And i have already written down these two vertices over here and also the equations.

01:53 So since all of these inequalities have a greater than or less than an equal to or less than or equal to sign, we're going to use solid lines to represent each one of these.

02:07 So the red line is going to be x squared and zero.

02:11 That is just going to be a line across x is equal zero, which is this one right here.

02:17 Should be a straight line.

02:20 Green line is just going to be y is less than four.

02:25 So it's going to be a line, horizontal line that goes all the way across y equals 4, which is about here.

02:32 And next we're going to have the parabola represented by the blue line.

02:36 So we know that a parabola squares the value.

02:39 So in this case, we're going to test with two values, with x is equal to 2.

02:47 And x is equal to 3.

02:50 At x equals to 2, just to give us an idea on how the parabola looks like, we're going to have 2 square, we're going to have 4.

03:00 So we're going to have the point 2 .4.

03:03 At x equals 3, we're going to have 3 square.