Graph the solution set of each system of inequalities. 3x + 5y ≤ ...

Question

Graph the solution set of each system of inequalities. 3x + 5y ≤ 15x^2 + y^2 < 9

Inequalities for graphing: 4y - 6x ≤ 15 and x² + y² < 16.

Accepted Answer

Hello everybody. I have night today today we're going to look at this question that states provide the graph of the solution set of the following system where our system includes two inequalities. Inequality number one is four, Y minus six, X is less than or equal to 15. And inequality number two is X squared plus Y squared is less than 16. Now, the first thing I'll note between these two inequalities is that inequality number one is non strict and inequality number two is strict. Now you determine this by looking at the sign of each inequality. So in inequality, number one, we have a less than or equal to sign. So that or equal to means that the inequality is not strict or inequality. Number two, where we only have a less than that inequality is strict. Now, when we have a non strict inequality, we will therefore have a solid graph or a solid line in our, within our graph. And when we have a strict inequality, we will therefore have a dashed line connecting the points in our graph. So that is something that we should know right before we even start this question. So now we can move on from that and actually begin working. So we'll start with inequality number one and work our way towards inquire number two. So the first thing I'll do is just replace the less than or equal to sign with a simple equal sign. So we'll have four Y minus six X is equal to 15. And now I'm going to want to try to isolate the Y. So I have an equation that's Y equals and we'll have four, Y is equal to six X plus 15 right there. I move the six X over from the left hand side to the right hand side. And now if I divide everything by four, we'll have that Y is equal to three divided by two, multiplied by X plus 15, divided by four. Now, why did I set it as a Y equals equation? Well, because we know of the line, the equation for a line. So we have the line equation is Y equals M X plus B. So we just put our inequality in our formula into that same form. This form the B is the Y intercept. And I'll write that as Y dash I N T. So this means that hour, Y intercept would be zero comma divided by four because the B in our equation is 15, divided by four. So just by rearranging our equation, we were able to get our first point in our graph, the Y intercept. So now, we want to focus on the X intercept and I'll write it as X dash I N T. Now, for the X intercept, there's no shortcut about looking at the formula and just obtaining it from that, we have to set our Y equal to zero. So we have four multiplied by zero instead of four Y. Now we have four multiply by zero minus six X, it's equal to 15, which means that negative six X is equal to 15, which means that X is equal to negative 15 divided by six, which is equal to negative five divided by two, which means that our X intercept will be negative five divided by two comma zero. So just like that, now we have two points for a graph and we can now move on to a test point and we need a test point because we have an inequality in an inequality, we will have multiple answers. So in art, in this case, we have four Y minus six X is less than, or equal to 14. So anything that's less than or equal to 15 will function as an answer. So we need to shade our region in our graph that will contain all the possible answers. So you can test any point that you want. I would, I like testing the origin zero, comma zero. So that's what I'll do. I'll have four multiplied by zero minus six, multiplied by zero. And all I did there was substitute the Y and the X with a zero since I'm testing zero, comma zero and we have that is less than, or equal to 15. So now we have that zero is less than, or equal to 15, which is true. Now, this means that the origin will fall within the region that we will shade for this inequality. And that's important to know once we get to graphic. So now we're done with an equality number one and we can move on to an equality number two, which is X squared plus Y squared equals 16. Now, before we move on, we want to note that the equation of a circle centered at the origin is X squared plus Y squared equals R squared R being the radius. If we look at our equation and try to get into this form, we have X squared plus Y squared equals what we have equal to 16. Is there anything that we can square to get 16? Yes, there is. So there is four. So we have X squared plus Y squared equals four squared because four squared is 16. So therefore our graph will be for this inequity will be a circle centered at the origin with a radius of four. And we obtain that just by looking at our formula X squared plus Y squared equals four squared. So now we do the same thing we did within a equality number one, we test a point and once again, I'll test the origin. So we have X squared plus Y squared is less than 16. So if we test the origin, we have zero squared plus zero squared is less than 16. So zero is less than 16, which is true, meaning that the origin will also be part of the region we shade in for inequality number two. So now we can go ahead and start graphing. So we draw a Y axis and an X axis and we can label each axis. So we'll have a dash on the Y axis. I'll, I'll draw a dash representing and seven. And I'll label the fifth one and I'll do the same thing on the negative side of the Y axis, negative one negative two, negative three, negative four, negative five, negative six and negative seven. And I'll label negative five and I'll do the same thing on the X axis. I'll label 67 and on the negative side, negative one negative two, negative three, negative four, negative five, negative six and negative seven. Now that I have my axis ready, I can go ahead and start plugging in the points that I know. So first, we'll focus on inquiry number one, which we know is the line we know that the Y intercept is zero, comma 15 divided by four and 15, divided by four falls within three and four. So we put a point in between three and four closer to four. And we know that the X intercept is negative five divided by two comma zero and negative five divided by two is negative 2.5. So we go and be in right in the middle between negative two and negative three and we put a point there and then we connect those two points. Now remember that for this equation, we had a non strict inequality, meaning that we can use a solid line to connect those two points. And now remember that we tested the origin and we saw that it does fall within the region that we will shade in. So we will first shade very lightly under the line and we'll shade lightly at first because we'll see what part of that region falls or it is shared with the second inequality to see which part they both share. So whatever region they share together, that's the part that we really care about. So we shade lightly at first and then we can focus on our circle or our second inequality. So we have a circle centered at the origin. So we have center at the origin, it has a radius of four. So we know that we'll have a point at zero, comma four or four, comma zero zero, comma four negative four, comma zero N zero, comma negative four. Now this equation or this inequality was a strict inequality. So we have to use a dashed line to connect the dots. Now, because we're doing this free handed, this is probably not going to look like a perfect circle. But let's try our best. So we use dashed lines to connect our dots and we go around each point that we know trying to make it as best of a circle as we can. And now we also noted that the test point, the origin is also part of our circle equation. So we shade in within that circle. And we noticed that the region that inter inter laps with the region that we shaded for our line, we can very clearly notice it since we lightly shaded the one for the line. So now that region that they both share, we can shade it in dark because that is the region that we care about, that's the region where both inequalities are true. So just like that, we have been able to graph both of our inequalities, one being a line and one being a circle and we were able to shade in the region with it, which they both have in common as our final answer. So I hope that was helpful. And until next time.

Graph the solution set of each system of inequalities. 3x + 5y ≤ 15x^2 + y^2 < 9

Key Concepts

Linear Inequalities

Quadratic Inequalities

Graphing Systems of Inequalities

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