Graph the solution set of each system of inequalities. x + y ≥...

Question

Graph the solution set of each system of inequalities. x + y ≥ 0 2x - y ≥ 3

Graph showing the solution set for the inequalities x - y ≥ 0 and 4x + 5y ≥ 9.

Accepted Answer

Hello everybody. Everything. All right. Today, today we're going to be looking at this question that states provide the graph of the solution set of the following system where we have two equations in our system. And equation number one will be X minus Y is greater than, or equal to zero. And equation number two will be four, X plus five, Y is greater than, or equal to nine. Now let's first, let's go equation by equation and let's first focus on equation number one. We have X minus Y is greater than, or equal to zero. Now let's just change the greater than, or equal to sign to just an equal sign for the sake of solving the, the solution. And we'll have X minus Y equals zero. So we have X minus Y equals zero. And if we move everything over we'll have X is equal to why? Now, what does this mean? We have a, a solution for our first equation? We have that X is equal to Y. So if we plug in different points, let's say we plug an X for a zero or a zero for our X, we'll have a Y of zero as well. If we plug in a one for our X, we'll have a Y of one. If we plug in a two for our X, we'll have a Y of two and so on. So any value that we plug in for the X will be the same value for the Y. So we'll have a straight line going zero ze- zero comma 01, comma 12, comma two. And also for the negative values negative one, comma negative one, negative two, comma negative two and so on. So a drop a general shape of that. So we draw a Y axis and an X axis and I'll draw a dash line at what could be and A Y of one A Y of two N A Y F three and I'll do the same for the negative values. So Y of negative one A Y or negative two A Y of negative three and the X values as well have 123 and negative one, negative two, negative three. And now we can plot our line. So we know we're gonna have a point at zero, comma 01, comma 12, comma two and so on. And we'll have one up uh other points at negative one comma negative one, negative two, comma negative two in negative three, comma negative three and so on. So I can connect all those points and make that line and extend it as it goes infinitely, it's aligned So it'll go infinitely to the right and up and to the left and down. Now we have an inequality, a non strict inequality. So it's something we should note of our inequality here because it's a greater than, or equal to sign, that means we have a non strict inequality and we use a solid line when we connect our points. If it was strict inequality, it would be different. So now that we have that with a solid line, we use a test point. So I will test 0.0 comma two. So you test a point that does not lie on the line. So if I test zero comma two, I'll have X minus Y is greater than, or equal to zero or zero minus two is greater than or equal to zero. And that will leave us with negative two is greater than or equal to zero, which is false. Therefore, the 0.0 comma two does not fit within our inequality. N zero comma two lies above our line. So for our inequality, we would shade the region under our line. And that is it for inequality number one. And now we can move on to inequality. Number two. So we had four X plus five Y is greater than, or equal to nine. And once again, I'll change it with an equal sign, I'll change the greater than or equal to sign with an equal sign. So I have four X plus five Y equals nine. And once again, it's good to know that this inequality is also non strict, meaning we will use a solid line in our graph. So I will plug in the same point. So let's say if X equals zero, we have four times zero plus five, Y equals nine, meaning that five, Y equals nine and Y equals nine divided by five or the 50.0 comma nine divided by five. Now, if we have a Y of zero, we do the same thing and we'll have four, X plus five, multiplied by zero. Now equals nine, which means that four, X is equal to nine and X is equal to nine divided by four or the 0.9 divided by four comma and zero. So we've obtained two other points for our inequality are inequality number two. So we can also graph this inequality. We'll set a, we draw a Y axis and an X axis and we draw dashes at an X of one to and three. And why is that one? Two and three? Now, for our X intercept it is at nine divided by four comma 09, divided by four is a bit above two. So we'll put a point there and our Y intercept is zero comma nine divided by five, which is a bit less than two on our Y axis. So we'll put a point there and then we can connect the two dots and make a line and once again, we use a solid line because this is a non strict inequality. Now, once again, we can test a point and in this case, I'm going to test the origin. So zero comma 200 comma zero, which will leave us with four multiplied by zero plus five, multiplied by zero is greater than or equal to nine. Meaning that four multiplied by zero is zero plus five, multiplied by zero is zero. So we have zero is less than greater than, or equal to nine, which is also false. So the origin won't be part of our inequality and we shade the region above our line. So now we have two graphs for each inequality. And what we do now is combine both graphs. So once again, I'll draw a Y axis and an X axis and I'll draw dashes and X of one to three and I'll put four and on the Y axis once again, I'll draw points at one to three and four. And then we can draw the, our first inequality. We know it's gonna be aligned with points at one, comma 12, comma 23, comma 34, comma four and so on. And our other inequality we know will start a little under two and will intersect the X axis a bit above two on the X axis. Now, where do they intersect? Well, if we plug in a one or, or X, we know that an inequality one will have one comma one, the 10.1 comma one. And if we plug in one for our two, our equation number two will have four, multiplied by one is equal to five Y or four multiplied by one plus five, Y is equal to nine. And that will leave us with four plus five, Y is equal to nine. And if we move the four over, we'll have five, Y is equal to nine minus four, which is five and we'll have that Y equal one or the 0.1 comma one. So both inequalities share the 10.1 comma one between them. So that is where the inequalities will intersect. Now we know that for our line equation, our equation number two, number one line, we shade under the region under the line. And for our line for equation number two, we shade above the line. So we can see that our lines will intersect there is a region that both lines share in common and that is the line or the region that we will shade. Now in that region will look like a starting at the 0.1 comma one and will extend to the right with almost kind of like a sideways V pointing, right. So that will be the graph for the system, we draw each function individually and shade their individual regions and then we graph them together and shade the regions that they both share in common. So I hope that was helpful and until next time.

Graph the solution set of each system of inequalities. x + y ≥ 0 2x - y ≥ 3

Key Concepts

Graphing Linear Inequalities

System of Inequalities

Testing Points to Determine Solution Regions

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