Graph the solution set of each system of inequalities.
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Question

Graph the solution set of each system of inequalities.

4x - 3y ≤ 12

y ≤ x²

Accepted Answer

Hello everybody. I have you doing night today. Going to again this question that states provide the graph of the solution set of the following system where our system includes two inequalities. With the first inequality being four, Y minus two, X is less than or equal to 15. And inequality number two is Y is greater than or equal to negative X squared plus two. Now, the first thing I'll note of both of our inequalities is that they're both non strict inequalities. Now this occurs when we have a less than or equal to or a greater than or equal to. If we had a less than or a greater than without the or equal to, then there would be considered strict inequalities. So because we have non strict inequalities, we will therefore have solid line within our graph. Now this will become more clear. Once we actually get to the graphing part, I'll be able to show it more clearly. So going back to the inequalities we have at hand, the first thing I'll do is focusing on inequality. Number one, I'm going to change the less or equal to sign with just an equal sign just for the purpose of solving. So we'll have four Y minus two X is equal to 15. And I'm going to want to get this into a line equation. So I, I see that I have a, a Y and an X and I can rearrange this to resemble a line. The equation of a line where I'll have the Y on one side and the X on the other, I'll, I'll want to solve this as a Y equals equation. So I'll have four wine, it's equal to. And then the first thing I'll do is move the minus two X over to the right hand side of the equation. So I'll have that four, Y is equal to two X plus 15. And now to isolate the Y I divide everything by four. I don't have that Y is equal to one half, X plus 15 divided by four. No, the line equation, if I recall, I know that it will be Y equals M X plus B where B is equal to the Y intercept. And I'll write Y I N T is an abbreviation. So this means that we will have a Y intercept at zero comma 15 divided by four. That'll be the one to set up our equation. And we obtain that simply by just rearranging our equation. So now we can go ahead and continue looking for X intercept and I write X dash I N T as an abbreviation. Now for this there's no shortcut, shortcut by just looking at the equation. So what we have to do is set our Y equal to zero. So if we do that, we have that four multiplied by zero minus two, X is equal to 15 or that negative two X is equal to 15 or that X is equal to negative 15 divided by two, which means that we have a, an X intercept of negative 15 divided by two comma zero. So that'll be the X intercept. And now we can continue. And because this is an inequality, we know that we're going to have to shade in a region of our graph. And this is because inequalities don't have one single answer. So in this case, we have four, Y minus two, X is less than or equal to 15. So anything that is less than or equal to 15 would be an answer to inequality number one. So we have to shade in that region that would apply here. So we can test a point in our graph, any point that you want. I like testing the origin sort testing zero comma zero. And to do that, all we do is plug in zero for our Y and zero for our X. So we have four multiplied by zero minus two, multiplied by zero is less than or equal to 15, which follows that zero is less than or equal to 15, which is true. So this means that the origin will be part of the shaded region of inequality number one. So now we can move on to quality number two. And we'll start out by doing the same thing changing the greater than or equal to sign with an equal sign. And we have Y is equal to negative X squared plus two. Now, this is the same thing as saying Y is equal to negative open parentheses, X minus zero, close parentheses squared plus two. And I want to write it like this because this will resemble something I can recall that is called the vertex form of a quadratic function. Now that vertex form can be written as Y is equal to a multiplying an open parentheses, X minus H closed parentheses squared plus K where H comma K is the vertex. Now if we compare that information with our equation, we'll see that the age. So we'll have our vertex, we see that the H is zero and the K is two. So we have a vertex at zero comma two and we know that we have a quadratic function because we have an X squared. So this will look what this will have the general shape of a parabola. So after I find out the vertex, I can go ahead and drop some points. And to do this, basically, I'll just test different XS X values and see what Y would give me. So let's say I test X equals negative one. We'll have Y is equal to negative open parentheses, negative one, close parentheses squared plus two. And that will be equal to one. So that will give us the point negative one comma one. And that'll be a point that we can use. Now, we can also set X equal to one positive one, which means that Y will be equal to negative open parentheses, one closed parentheses squared plus two. And that will once again equal one, which means we have a point at one comma one. So now we have a vertex and two different points. Now once again, because we have an inequality, we have to test a point to see what region we will shade. And once again, I'll test the origin zero comma zero. So by doing that, we'll have zero, it's greater than, or equal to negative open parenthesis, zero, close parentheses squared plus two. And this follows that zero is greater than or equals two, which is false. So for this inequality, the origin will not be part of the, of the shaded region. So we can now go ahead and graph. So we draw a Y axis and then X axis and we can start plugging in the points that we know. So for our inequality number one, we know that it is a line which had a one intercept of zero comma 15 divided by four and 15 divided by four is between three and four. So I'll label my wax draw dashes at 123456 and seven. And I'll label the fifth one and I'll do the same thing on my X axis 67. And on level eight as well as well, on the negative side will have a negative or five 678. Now, on the negative side of the Y axis, I will also label 1234 and negative five. So now I know where different points points will lie. So we're looking at inequality number one, our Y that was zero comma 15 divided by four. So zero comma somewhere between three and four. So I go to the area between three and four and I, I'll place a dot there and the X intercept was negative 15 divided by two comma zero and negative 15 divided by two is around negative 7.5. So I go to that area and I place a dot there and then I can connect the two points. Now, remember we said that we have non strict inequalities, meaning we'll use a solid line so we can draw a solid line connecting these two points and we will extend the past both of them. I remember we said we will shade the region under it because this includes the origin. So we shade under the line. Now for inequality number two, we noted that we have a vertex at zero comma two. So we place a point there and we had the point at negative one, comma one and one comma one. And once again, we have a non strict inequality. So we use a solid line to connect those points and extend past them having an upside down Parabola. And now for this inequality, we said that we do not include the region. So we draw or we shade in the region outside of the para and that region that intersects or the region that both inequalities share in common is the region that we care about. So we shaded in very darkly to emphasize that this will function as the answer. So the region that both once we shade in the region that both inequalities share in common, that will function as the answer to this question to the solution set. And just like that, we were able to graph both of our inequalities and shade in their prospective regions showing the one that they have in common. So I hope that was helpful. And until next time.

Graph the solution set of each system of inequalities.
4x - 3y ≤ 12
y ≤ x²

Key Concepts

Graphing Linear Inequalities

Graphing Quadratic Inequalities

Solution Set of a System of Inequalities

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