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

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

Key Concepts

Linear Inequalities in Two Variables

This concept deals with inequalities that involve a linear expression in two variables, typically x and y. Instead of finding a precise solution, the goal is to identify a set of points on the coordinate plane that satisfy the condition defined by the inequality.

Graphical Representation of Linear Inequalities

This refers to the method of plotting the boundary line of the inequality on a graph and then shading the appropriate side of the line. The line itself may be solid or dashed depending on whether the inequality is inclusive (? or ?) or strict (> or <), respectively.

Boundary Lines and Half-Planes

In graphing linear inequalities, the boundary line is where the expression equals a constant, and it divides the plane into two regions (half-planes). One half-plane represents all points where the inequality holds true, while the other does not.

Feasible Region

This concept involves finding the common area where all inequalities in a system overlap. The feasible region represents all possible solutions that satisfy the system of inequalities. It is determined by the intersection of the individual half-planes corresponding to each inequality.

Bounded vs Unbounded Regions

A bounded region is completely enclosed by boundaries, meaning it is contained within a finite area. An unbounded region extends infinitely in at least one direction. Determining whether the solution set is bounded or not is crucial in optimization problems and feasibility studies.

Transcript

00:01 So in this problem, we're going to solve the following system of inequalities.

00:05 We also want to find the solution set for the system as a graph, and we want to find all the vertices and determine whether the area that is going to be shared by all of these three inequalities is bounded or not.

00:18 To start off, i'm going to take the first inequality, which i'm going to label with rep, and i'm going to write it as a plain equation.

00:27 So that will be y is equal to x -q.

00:30 So this is just a cube graph.

00:33 So we know that it has a y intercept of 0 and an x intercept of 0.

00:39 And we know that 2 cube is just 8.

00:42 And we can take these two points and just join them.

00:48 The same is going to happen on the other hand, but with a negative sign.

00:52 So negative 2 cube is just negative 8.

00:55 And this is the shape of a cube graph.

00:59 Now i'm going to take a point in this graph and i'm going to use negative 1 .0.

01:03 And i'm going to test test with negative 1 .0 and i want to substitute this point into this first inequality and check whether or not it satisfies it or not.

01:20 So it's just going to be 0.

01:23 Oh, sorry.

01:25 0 is equal to negative 1.

01:30 I was thinking of the point negative 1 and i said this is not the point that i was looking for, but this one right here.

01:40 So we have 0.

01:42 Is greater than or equal to negative 1.

01:47 So in this case, this point does satisfy the inequality.

01:52 So we're going to be looking to shade this region over here.

02:02 As a reminder, we're joining the points on the graph with a solid line because we're including the values above this graph.

02:10 So on the line y equals x q.

02:15 Next, i'm going to be on has a y intercept of 4, 0 .4, which is, up here and it has a x intercept when y is equal to zero so for y is equal to zero zero is equal to two x plus four negative four is equal to two x i'm just solving for x in this case so dividing both sides by four by two sorry we're going to get that x is equal to negative two so in this case we have the point negative two comma four which is this one over here just going to be joining the graph again we're going to use the same test point which is on one side of the graph test with minus 1 comma 0 in this case we're going to get that 0 is less than or equal to 2 times negative 1 plus 4 0 is less than or equal to negative 2 plus 4 so 0 is less than or equal to 2 and this is true so we're going to be shading this region over here and i'm going to use this type of lines.

03:49 Finally, the line x plus y is greater than zero.

03:55 This is just the line y is equal to negative x.

03:59 This goes across the zero and it's just a line that it's with negative slope of one, just going to be something like this.

04:14 And again, i want to test with the point zero negative 1 .0, sorry.

04:21 In this case, we have that negative 1 is greater than or equal to 0.

04:26 So this statement is not true.

04:29 Negative 1 is not greater than 0.

04:32 So this is wrong.

04:34 So we're looking to graph this region above.

04:41 I left this region in blank on purpose.

04:43 Since you can see, this is the region in which all of these three shared a common solution.

04:50 And this is our solution set.