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

Linear inequalities involve expressions in which two linear functions are compared using inequality signs (?, ?, >, <). They define regions on the coordinate plane rather than specific lines, and they are essential for describing constraints in optimization problems and systems analysis.

Graphical Representation of Inequalities

Graphically representing inequalities involves plotting the associated boundary lines of the inequalities on a coordinate plane and then shading the region that satisfies the inequality. This process helps in visually identifying the solution set, which is especially useful when working with systems of inequalities.

Systems of Inequalities

A system of inequalities consists of two or more inequalities that are considered simultaneously. The solution set of the system is the intersection of the regions defined by each individual inequality. This concept is fundamental in linear programming and optimization problems, as it enables the determination of common feasible regions.

Bounded vs. Unbounded Regions

The distinction between bounded and unbounded regions is crucial when analyzing the solution set of a system of inequalities. A bounded region is completely enclosed, implying that the solution set has finite limits, while an unbounded region extends infinitely in one or more directions. This concept is important in mathematical optimization and feasibility studies.

Transcript

00:01 Okay, so the purpose of this problem is to graph the solution set for this system of inequalities that is made of the equations y is less than or is less than x plus 6, 3x plus 2y is greater than or equal to 12, and x minus 2 y is less than or equal to 2.

00:18 We also want to find the coordinates of the vertices of the system of the solution of the system of inequalities, and we want to determine if it is a bounded or not bounded solution set.

00:30 So to start off, we're going to write these equations in point intercept form.

00:37 So point intercept.

00:42 And the reason to do this is because it is going to be easier to graph these equations in that form, as opposed to, in this case, with these two equations.

00:54 So for the first one, we're going to have that y is equal to, sorry, y is less than x plus 6.

01:05 The second one will require a little bit of algebra too.

01:08 Find the point intercept form.

01:10 So we'll have 3x plus 2y is greater than or equal to 12.

01:16 Then we're going to subtract 3x on both sides.

01:19 So we'll have 2y is greater than or equal to 12 minus 3x.

01:24 Finally we're going to divide both sides by 2 and have that this is 6 minus 3 halves x.

01:31 So this is the final equation.

01:37 The last equation it's going to be x minus 2 y is less than or equal to 2.

01:43 We're going to going to subtract again x on both sides minus 2 y is less than or equal to 2 minus x so we're going to divide by negative 2 and since we're dividing or multiplying by a negative number we need to flip the sign and in this case y is going to be greater than or equal to x divided by 2 minus 1 so this is a final answer for this equation as we can see we can graph this first equation knowing that it has a y intercept at 0 .6.

02:16 So in this grid right here, we're going to take the point 0 .6.

02:21 And we also want to find the point where y is equal to zero.

02:25 And in this case, it's easy to know that it's going to be the point x equals negative 6.

02:32 So we're going to join these two points with a dotted line.

02:41 Since we're not including all the values on the line y is equal to x plus 6, because we're using this sign, the less than sign.

02:51 Now we're going to test for this equation and i'm going to use the point 0 .0 to determine which region we should be shading.

03:00 If we substitute 0 .0 in the first equation, we're going to get that 0 is less than x.

03:08 Well, sorry, in this case it's going to be 0 again plus 6.

03:13 So we're going to have that 6 is greater than 0.

03:15 And this is true.

03:17 So we should be shading all of this region.

03:22 Next with the following equation, which is the green one, we know that it has a y intercept at y equals six, so it's going to be this point again.

03:34 And we're looking for the point where three minus three halves of x is equal to six.

03:40 And that's the point where y is equal to zero.

03:44 And if we solve for x by multiplying this six by two thirds or dividing by negative three halves, we're going to get that this is the point 4.

03:57 So this is the point x is equal to 4.

04:02 Oh, sorry, this should be negative 6.

04:05 So we're going to get the point x is equal to 4 and y is equal to 0.

04:10 Since we're using a greater than or equal sign, we're going to be joining these two lines by a solid line.

04:18 So this should be something like this.

04:23 We can now see that the point 0 .0, which is this one right here, is also on one side of the graph.

04:29 And i'm going to be testing with this same point since it's easy.

04:34 So i'm going to be substituting in the original one.

04:37 So it's going to be 3 times 0 plus 2 times 0 is greater than or equal to 12.

04:44 And it's going to give us that 0 is greater than or equal to 12, which is not a true statement.

04:51 So we're not going to be shading this region on the left hand side.

04:55 We're going to be shading this one on the right hand side.

04:57 So it's going to be the region that we should be shading for this graph.

05:03 Finally, with the blue one, we have the point at the y intercept at 0 comma negative 1.

05:10 It's going to be this one right here.

05:13 And we're looking for the point where y is equal to zero.

05:18 And that point is going to be where x divided by 2 by minus negative 1 is equal to 0.

05:25 It's just going to be x is equal to 1 by adding 1 in both.

05:30 Sides and by multiplying by two on both sides we're going to get that x is equal to 2.

05:36 So we'll have the point 2 comma 0.

05:41 Again we're going to be joining these two points with a solid line like this and again we can use the point 0 comma 0 to test whether or not this should the region on top of this graph should be the one that is going to be shaded or the one below it.

06:02 So we'll have the point 0 minus 2 times 0 is less than or equal to 2.

06:09 So this is just zero, it's less than or equal to two, and on top of the blue line.

06:18 Now we need to determine the region of the solution set.

06:21 So where does these three regions share a common area? so in this case, it's going to be this one, this one right here.

06:31 This region that i'm going to be shading black is a region where all of these three inequalities are true at the same time.

06:39 So for any given point in this region, we're going to have a solution for this system of inequalities.

06:46 Now to determine the vertices of this solution said we need to look at these two vertices.

06:53 These two points, the points where the blue line intersects the green one and the point where the red one intercepts the green one.

07:03 So as a reminder, this is the red one...

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