In Exercises 27–62, graph the solution set of each system of ineq...

Question

In Exercises 27–62, graph the solution set of each system of inequalities or indicate that the system has no solution. 3x+y≤6, 2x−y≤−1, x>−2, y<4

Accepted Answer

Hello. Today we're gonna be solving the system of inequalities by using the method of graphing. So what we are given is a system of four inequalities. I'm gonna go ahead and label each of the inequalities and we're gonna analyze them individually. So if we take a look at the first inequality, we are given X is greater than or equal to one. Now we can go ahead and graph each of these by using the methods of graphing lines. See in this case here X greater than equal to one matches the form of X is equal to one. And that means that this is going to be a similar graph to a vertical line located at one on the X axis. Now, we have to be careful here because graphs of inequalities can be graphed either either using a dotted line or a solid line. In this case here, the first equation uses a greater than or equal to symbol. So that means that we want to include the values that are located on the line. Therefore, we're going to graph this vertical line using a solid line. Now, one more thing to note is that we want all the values of X that are greater or equal to one. And that means that we want all the values to the right of the region of the graph. So we're going to be shading in the right hand side of the graph. So this is going to be the individual graph for the first inequality. Now let's take a look at the second inequality. The second inequality is given to us as Y greater than equal to negative one. Now, just like before, this is similar to an equation of a line. This matches the form Y is equal to negative one. And this is the equation of a horizontal line located at negative one on the X axis. Now, just like before we have to determine if we're gonna graph this horizontal line using it in a dotted or a solid line. The inequality has a greater than or equal to symbol. So that means we're going to include the values that are located on the line. Therefore, we're going to be using a solid line. And this inequality is saying that we want all the values of Y that are greater than are equal to negative one. So that means we want the region above the horizontal line. So we're going to be shading the region above this line. Now let's take a look at the next graph. The third inequality is given to us as X plus six Y is less than 15. Now just like before, this matches the standard form of the equation of a line. This can be imagined as X plus six Y equal to 15. Now, normally we want graph this using the slope intercept form. So we want to go ahead and solve for the Y variable in or inequality, we can go ahead and subtract X to both sides of the inequality. And that is going to give us six, Y is less than negative X plus 15. Then if we divide both sides of the inequality by six, in order to solve for Y, we get Y less than negative 1/6 X plus 15/6. Now, one thing to note is that the value 15/6 is equal to 2.5. So this means that this is going to be the equation of a line starting at 2.5. Furthermore, this is going to have a negative slope of negative 1/6. This means that the second dot is going to be located at 1.5 comma six. Now the question is, do we want to connect these dots using the solid or dotted line? Well, what the inequality statement is saying is you want all the Y values that are less than this equation of X? So what this means is that we don't want to include the values that are on the line itself. So we're gonna graph this using a dotted line. Now, the question here is do we want to shape the region above or below the dotted line? The easiest way to figure this out is to pick a testing point to plug into the, into the inequality. And if the testing point makes, makes the inequality true, then we want to shade that region. The easiest point to pick in this case is going to be the origin. So if we pick 00 to plug into the inequality, we get zero is less than negative 1/6 times zero, which is zero plus 15/6. And this inequality is saying is zero, less than 15/6. This is a true statement. Therefore, we want the region that lies below the dotted line. Now let's take a look at the final inequality. The final inequality given to us is two X plus Y is less than or equal to five. Now, just like before we want to represent this using the equation of a line, we're gonna go ahead and solve for our Y variable in this case. And we're gonna subtract both sides of the inequality by two X that is going to give us Y less than or equal to negative two X plus five. This is the equation of a line in slope intercept four which starts at five on the Y axis and has a negative slope of negative two. Now again, do we want to graph this using a solid or dotted line? Well, in this case here, we have the less than or equal to symbol. And that means we're going to include the values that are located on the line. Therefore, we're going to graph this negative line using a solid line. And just like before, we're going to take a testing point to see if we want to shade in the lower or upper region of the line. Again, we're going to use the testing 0.0 comma zero. And if we plug this into the inequality, we get zero is less than or equal to negative two times zero, which is zero plus five. And simplifying this is going to give us zero is less than equal to positive five, which is a true inequality. So that means we're going to shade the region that is below this solid line. So now that we have a rough idea of how each of the inequalities are gonna be graphed. Let's go ahead and combine all of them into a single graph. So we're going to draw a big axis. Now, the first inequality was a vertical line located at one and it was a vertical line that was graphed using a solid line. The second inequality was a solid line located at negative one on the Y axis. The next graph was a negative slope line starting at 2.5 on the graph and it was shaped using a dotted line. So it has a slightly smooth negative slope and finally, the last equation was given to us starting at five and using a solid line with a pretty steep slope of negative two. And so if we combine the shaded regions, we're going to get this rough geometric shape. And the answer that roughly matches this problem is going to be B now, just to clarify, we have a horizontal line which starts at negative one on the y axis, we have a vertical line which starts at one on the X axis. We have the slope of the line which is defined as Y less than, or equal to negative two X plus five. And it's a bit hard to see here, but we have a dotted line on the top which is defined by the third inequality which is Y less than negative 1/ X plus 15/6. And with that being said, the answer to this problem is going to be B so I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

In Exercises 27–62, graph the solution set of each system of inequalities or indicate that the system has no solution. 3x+y≤6, 2x−y≤−1, x>−2, y<4

Key Concepts

Inequalities

Graphing Systems of Inequalities

Feasible Region

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