Graph the solution set of each system of inequalities. 2x + y ...

Question

Graph the solution set of each system of inequalities. 2x + y > 2 x - 3y < 6

Inequalities 4x + y > 8 and 6x - y < 2 displayed in a mathematical format.

Accepted Answer

Hello everybody everything night today that we are going to be looking at this question that states provide the graph of the following solution set of the following system where the system is or X plus Y is greater than eight. And we'll call that equation number one. And we also have equation number two, which is six X minus Y is less than two. Now, the first thing I'll notice here is that both of these inequalities are strict and this is because they have a less than, or a greater than sign if it was less than, or equal to or greater than, or equal to. Those are non strict. Now because they're strict, we will therefore have a dashed line in our graph, our graph will be dashed and this will make a bit more sense once we actually start graphing. So let's shift our focus back into our inequalities and let's break this down one by one. So first, let's deal with inequality number one within equality. Number one. The first thing I'm gonna do is replace the greater than sign with an equal sign just for the sake of solving. So I'll have four X plus Y, it's equal to eight. And now this reminds me of the equation for a line R Y equals M XX plus B. So I want to get this form, our equation number one into that same form of Y equals M X plus B. So to do that, I'll leave the Y on the left hand side and I'll move the four X from the left hand side to the right hand side to have Y equals negative four, X plus eight. Now, in the Y equals M X plus B formula, we know that B equals the Y intercept. And all right, that is Y dash I N T. So we have that our Y intercept in our equation is the beat, which is eight, which means we have a 1% at zero comma eight. Now, for the X intercept, there's no shortcut. But what we can do is set our Y equal to zero. So we'll have zero is equal to negative four X plus eight. So now we solve for X, we move the eight over from the right hand side to the left hand side to have negative eight is equal to negative four XX. And if we divide both sides by negative four to isolate the X, we'll have that X is equal to two or the 0.2 comma zero. So now we have a Y intercept and an X intercept. And now, because we have an inequality, we know that we're going to have to shade a region of our graph because there's not one single answer to an inequality because it's anything greater than eight in this case. So we're going to test, we can test any point we want, but I'll test the origin zero comma zero, which will leave us with. So all we do now is plug in zero for our X and zero for our Y and our inequality. So we'll have four, multiplied by zero plus zero is greater than 84, multiplied by zero is zero plus zero, is zero. So we're left with zero is greater than eight, which is false. So that means that the origin will not be part of the shaded region in our graph. So now we can shift focus to inequality number two and I'll do the same steps. So first, I'm going to change our, less than sign with an equal sign. So we'll have six X minus Y is equal to two. And then I'm going to try to get it in the Y equals M X plus B form. So we'll have negative Y, I'm gonna move the six X over from the left hand side to the right hand side and we'll have negative Y, it's equal to negative six plus two. And if we multiply everything by negative one, we'll have that Y is equal to six X minus two, which means that my Y intercept for this equation, we'll be at zero, comma negative two. And then we do the same thing for that we did for the X intercept, we set our Y equal to zero. So we'll have that zero is equal to six X minus two. And if we move the two over from the right hand side to the left hand side, we'll move, we'll see we'll have that two equals six X. So now, now we divide both sides by six and we'll have two divided by six, which is one third equals X, which means our X intercept is at one divided by three comma zero. So now we have two points for our inequalities and we'll do the same thing. So we'll test the point because again, we have an inequality. So we need to see what region we're gonna shade. And once again, I'll test zero comma zero, which means my inequality will turn while plug in zero for the X and zero for the Y, which means I'll have six, multiplied by zero minus zero is less than two, which means we'll have that zero is less than two, which is true. Meaning that the origin will be part of the shaded region for equation number two. So now what we're left with to do is graph. So we draw a Y axis and will draw an X axis and we'll draw points that we know. So on the X axis, I'll draw a dash in what I'll label X of one X two N X three and for the Y axis, I'll label in two. So I'll do two four, six, eight and 10. And on the negative side of the Y axis are labeled the same thing. So I'll have negative two, negative four and negative six. And now I'll plug in points that I know. So first let's do line number one. So we know that our Y intercept will be at zero comma eight. So we put a point there and our X inter set will be at two comma zero. So we put a point there. Now, before combining those points, we have to keep in mind what we said at the beginning of this equation or, or, or this question where we have strict inequalities and we will have dashed lines. So to connect those points, I will draw dashed lines connecting them and I will extend them past each side. And that is equation number one. Now, we know that we said the origin was not part of this equation. So we have to shade the region above the line. So we don't include the origin. Now we move on to equation number two where we have a Y intercept at zero, comma negative two and an X intercept at one third comma zero. So it's between zero and one will go to half and go a little under half and place a dot there. And the same thing we said that this was a dashed inequality. So we'll use a dash line to connect them and will extend past it upwards and downwards. And now we said that the origin was part of this equation. So we would shade in the region above this line. Now, what we care about here is the region where both lines intersect. So we see that there is this kind of V shape where both regions are shaded and that's the region that we care about here. So we shade that we shade that in that region in much darker. And that is the region that is, that would work or the set of inequalities that we have. And just like that, we were able to break down our inequalities one by one graph, each one of them. And then by shading in the regions, we were able to find the common region that they had and shade that in. So I hope that was helpful. And until next time.

Graph the solution set of each system of inequalities. 2x + y > 2 x - 3y < 6

Key Concepts

Graphing Linear Inequalities

Finding Intersection of Solution Sets

Using Test Points to Determine Shading

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