Graph the solution set of each system of inequalities.

Question

$e^{x}-y\le1$

$x-2y\ge4$

Accepted Answer

Welcome back everyone. In this problem, we are given the system of inequalities, E to the power of X plus Y is greater than or equal to four and X minus two Y is greater than, or equal to three. We want to graph its solution set here. We have our graph, a possible solution for air. A possible solution. Answer choice B A possible solution in answer choice C and the D A says our solution set is empty. Now below our answers, I've put a blank graph, OK, where we are going to graph our inequalities because by graphing each of the inequalities and identifying where they intersect that is going to be our solution set. So let's go ahead and try to do that. Let's start with our first inequality. OK? Where we know that E to the power of X plus Y is greater than or equal to four. Now how are we going to represent that? Well, first, let's let E to the X plus Y be equal to four. OK? Which means that Y is going to be equal to negative E to the X plus four. Now we can draw the graph of E to the X, we know it's going to be an Asymptote. Normally it's an Asymptote that has a Y intercept of 01. But no, since where it's negative E to the X, OK, then that means it's going to be a reflection in the X axis of Y equals E to the X. And since we're adding four, that means it's translated four units up. So now instead of X decreasing as it approaches the X axis, OK. In this case, X is going to decrease as it approaches the line X equals or sorry, Y equals four. OK. So as it approaches this dotted line, OK, then our Asymptote is going to decrease, but it never actually touches that line since it's reflected in the X axis, we know it's going to be flipped vertically. OK? And now if we keep all of those in mind to plot our graph, OK, then our line for Y equals negative E to the X plus four is probably going to look something like this. OK. I'm gonna try and fix this up a bit. I don't think I'm going to get it perfectly, but I think we can try to get it as close as possible to what it should look like. So let me bring this a bit over here. OK? And let's just bring this down a little bit. OK? And we know then that it should look, should look something along that line. As I said, it's not gonna be perfect. OK? It's not gonna be perfect, but we're gonna get to try and get it as close as possible. OK? But we know at least at this point it's going to be asymptotic. So it's not going to touch it, but it's going to come as close as it can. And this is, this should be straight actually. OK. Not that we have that, you know that we know that we have that then we can go ahead and start to plot our next graph. OK? Now remember our next graph, it's going to be X minus Rx inequality X minus Sui is greater than or equal to three. But guess what? I forgot to do something. First, we need to share the region, we need to figure out which area should be shaded. Now, in this case here, let's go back to E to the extra size greater than equal to four. Let's try to use a test point to figure that out. So let's test it with 00. OK? Now, when we test it with 00, the question is, is E to the zero plus zero greater than or equal to four. Well, E to the zero is 11 plus zero is one, so is one greater than our equal to four. No. So that means this 0.00 is not in our shaded region. It doesn't satisfy our inequality. So the inequality must be on the other side. Of our line and we know it's a sudden line because it's great. It's greater than, or equal to. OK. So that means our shaded region is above our line right here. OK. So let's represent that on our graph and our graph is gonna look something like this right now that we have our graph, we can go ahead and do our next inequality X minus two Y is greater than our equal to three. Now this is a linear inequality that means it's going to form a line. So if we can draw our line using two points, then we should be able to figure out which region or which area is going to be stated. So let's let X minus two Y be equal to three. OK? Now at X equals zero, let's choose a random point. Then we will have zero minus two, Y being equal to 32 Y uh sorry, negative two Y will be equal to three, which means that Y is going to be equal to negative three divided by two. Thus zero and negative three halves is going to be a point on our line. If we go to our graph here and go to zero, negative three halves is negative 1.5. So a point will be on our X axis on our Y axis. Sorry, my apologies. Next at Y equals zero. OK. Then we're going to have X minus two multiplied by zero, equal to 32, multiplied by zero is zero. So that means X equals three. Thus 30 is also going to be a point on the line if we put 30 in our graph and then we connect our points. Thus we have the line for X minus Y equals three. So the next question we need to answer is which area is going to be shaded? Well, again, let's take a test point. Let's use the same test point for a 00. If it satisfies the in the, in the inequality, then it's going to be in the shaded region. But if it is it is not, then the shaded region is on the opposite part of that line. Now when we test 00, OK, then we're really going to get zero minus two multiplied by zero and is zero minus two multiplied by zero, greater than or equal to three. Well, zero minus two, multiplied by zero is zero and zero is not greater than our equal to three. Therefore, this is not our shaded region above our line. Thus, the shaded region to satisfy our inequality must be below our line. So our shaded region must be below our line. OK? I know if I go ahead and put that on our graph here. OK. And that means our shared region is going to be right here. Now we have the shaded region for both our inequalities. So now the question is, where is the solution set. Well, the solution set is the region where both inequalities are satisfied. And I like to draw it like this because I like to think of it as the area where all of our colors overlap. Now, if you look closely at our graph here, you'll notice that our colors overlap in this dark shaded region. OK? Or rather this region that's a boned by our dark lines. I'm gonna just put that here. OK. So thus, this tells us then that this region OK? This region is or should be the area that satisfies our solution set. OK. So let's shade that region in black. So now all we need to do is to compare our region to our answer choices. Remember it might our own, our sketch might not be exact because of the limitations we have with our lines. But our answer should look like this shaded region here. Now, when you compare which one is the correct answer by comparison, you'll probably notice that A is the correct answer. Therefore, a is the graph that represents our solution set? Thanks a lot for watching everyone. I hope this video helped.

Graph the solution set of each system of inequalities.﻿ex−y≤1e^{x}-y\le1ex−y≤1﻿﻿x−2y≥4x-2y\ge4x−2y≥4﻿

Watch next

Graph the solution set of each system of inequalities.
$e^{x}-y\le1$
$x-2y\ge4$