Graph the system of inequalities and indicate the region (if any) of solutions satisfying all equations. 3 x − 2 y > 6 3x-2y>6 3 x − 2 y 3x-2y 3 x − 2 y < − 4 -4 − 4

Welcome back, everyone. So in this practice problem, we're gonna graph this system of inequalities. We've got 3x minus 2y is greater than 6. 3x minus 2y is less than negative 4. And so we're gonna graph this and indicate the region. I'm gonna go ahead and do this in green highlighter of solutions that satisfy all of the inequalities. Remember, these types of inequality these types of problems are pretty straightforward. We're just gonna have to take each one of these inequalities and sort of go through the 3 steps that we've been using for any inequality. And then later on, we'll see the shading, the regions that will overlap each other, and that will be our solution. Alright? Let's get started. So let's take a look at this first inequality. These inequalities are unfortunately in standard form, so it's a little bit sort of difficult to see what they look like. So we're going to have to convert them to slope intercept form just to be able to graph them. So So I'm going to go ahead and do that over here. So this is going to be 3x minus 2y is greater than 6. Remember, the idea here is we're just going to have to isolate for y. So I'm gonna subtract 3x from both sides, 3x over here, and I'm gonna get negative 2y is greater than negative 3x+6. So The way I get negative sorry. The way I get y by itself is by dividing by negative 2 on both sides to cancel, and then I'm gonna have to divide everything on the right side by negative 2. Now the key thing to remember here is that remember the rules of inequalities. Whenever you divide by a negative number, you have to flip the sign over here. So now what happens is we're gonna have y is less than because this symbol over here will flip. Alright. So that's the most important thing here. So it flips because you're dividing by negative 2. Alright. So this becomes y is less than. Now the negatives will cancel, so you'll just get 3 halves x. And then 6 divided by negative 2 becomes minus 3. Alright? So that's what this inequality looks like. The sign flips. And so let's take a look here. So this is gonna be 3 halves x minus 3. So this is gonna be going through the y intercept of negative 3, and it's gonna have a slope of up 3 over 2. Rise over run. Right? So go up 3 over 2 and then up 3 over 2, and we'll connect these points. Now is this a solid or dashed line? Well, actually, if you take a look for both of them, we're dealing with symbols that don't have the bar underneath. So both of these are actually just gonna be dashed lines. Alright? So what this is gonna look like is we're gonna have something that goes through negative 3, and it's gonna be up 3 over 2. So we're just gonna connect all these points. Alright? Alright. So that's what the red equation looks like. Now let's do the same exact thing for the blue equation. So we've got 3x minus 2y is now less than negative 4. So same thing. So subtract the 3x, you'll get negative 2y is less than negative 3x minus 4. And now we divide by negative 2. Now remember, dividing by negative 2 is gonna flip the sign just like it did for the red equation. So now instead of a less than symbol, the sign is now gonna flip. So now you have a greater than symbol. This is really important here. So you remember your rules of, inequalities. And now what happens is we're gonna have the same 3 halves x. And now this negative 4 divided by negative 2 will become positive 2. Alright? So now what this looks like is it's gonna look like going through the points or the y intercept of positive 2, but it'll have the exact same slope. So you go up 3 over 2. You can also go down 3 into the left 2. And so if you connect these points, you're actually gonna see here that we've ended up with some parallel lines. So these things have the same exact slope. Right? So these look like this, but different y intercepts. Alright? So what about the shadings for each one of these? Well, well, what we would do here is we would test the point on either side of the line by plugging x, y into the inequalities. Right? So let me actually just go ahead and do that for the red equation. I'm gonna go ahead and use this point over here, which is 0 comma negative 4. Alright? So 0 comma negative 4. I'm gonna test this point out here. 0 comma negative 4. So what this says here is this is gonna be 3 times 0 minus 2 times negative 4. Is that greater than negative 6? Alright? So if we took a look here, we're we're gonna or sorry. Is that greater than positive 6? So we're gonna see here that this turns to 0. So 3 times 0 minus 2 times negative 4. What we have here is, actually, this is gonna become 8. So is 8 greater than 6? In fact, that is a true statement. So because of that, we're gonna shade this side of the line that includes that point. By the way, another way we could have done this is by using our shortcut for lines. We see here that we have a less than symbol, so we're gonna shade everything that's underneath or below that line, and we're would have gotten the same answer. Alright? So I'm gonna use that shortcut actually for the blue equation. So here we have a greater than symbol, which means that I'm gonna shade everything that is above that line. So for the blue equation, if I were to test out this point over here, for example, we would actually see that this would be a true statement as well. So that means that this and you can pause the video and do that yourself if you want to. So what you're going to see here is that this is going to be the area of solutions for the blue equation. So really, that's doing all of steps 1, 2, and 3 for each one of these equations kind of altogether. Now the last thing we have to do is to shade the overlap of all shaded regions. Now, unfortunately, if you take a look here, what happens is that usually, in sort of our problems that we've seen before, we've ended up with situations where at least some part of the graph was overlapping or some part of the regions were overlapping. But here, we actually see that all the solutions for the red equation are underneath. All the solutions for the blue equation are above the blue equation. And because these are parallel lines, they'll actually never actually intersect. And so unfortunately, what happens here is that there is no solution. So there's no solution. And usually, you're going to get these kinds of situations whenever you have parallel lines. Alright? So a little bit of a trick problem, but not really because you go through all the steps, and you'll just actually find that in some situations, you may not have a solution at all. Alright? Hopefully, this made sense here. Let me know if you have any questions, and I'll see you in the next video.

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Graph the system of inequalities and indicate the region (if any) of solutions satisfying all equations.3x−2y>63x-2y>63x−2y3x-2y3x−2y < −4-4−4

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