all right today in unit three we’re going to be looking at solving systems of inequalities so we want you to be able to solve where an inequality is involved remember that’s less than greater than those types of symbols all right so the first one on your note sheet a little word problem here says during one heartbeat blood pressure reaches a maximum pressure and a minimum pressure which are measured in millimeters of mercury it is expressed as a maximum over a minimum for example your blood pressure might be 20 over 80 no more blood pressure for people under 40 years old ranges from 100 to 140 millimeters of mercury and maximum 60 to 90 okay we want to express this as a set of inequalities so first we got to figure out what are we going to call these things so let’s call a let’s use X for our maximum blood pressure and let’s use Y for minimum blood pressure so if we read here normal blood pressure ranges from 6th or 100 to 140 that means that our max here should be our maximum blood pressure should be greater than or equal to 100 but we also want it less than or equal to 1 40 doing the same thing here for minimum it’s got to be between 60 and 90 so minimum we’re going to use why why should be greater than or equal to 60 and we want it less than or equal to 90 now we have a system here of inequalities that describes the blood pressure what you could do now let’s graph something like this out and you get a chart like this okay the x-axis here that’s our maximum like we said our y-axis that’s our minimum if we were to graph these lines out we would get it remember when you graph a x equals equation these are vertical lines and when you graph y equals ones those are horizontal lines okay so if you go at here’s the 100 and the 140 those are both vertical lines if it’s less or greater than 100 that’s kind of graphing everything over here if it’s less than 140 that would be kind of graphing everything less than that line that would be in here notice this the overlap and shading happens between these two lines okay if we do the same thing with the Y lines greater than 60 that’s kind of everything above this horizontal line less than 90 that’s everything below this line so here again the overlapping occurs in the middle well since the overlapping occurs in the middle for both of these the part that all four of these lines would have in common is this box right here as long as your blood pressure falls in that box you have normal or healthy blood pressure so this is the overlapping of all four lines and that’s what we’re going to be doing today when we graph these out where does all the overlaps shading occur we want to find that location all right so going on here this next one says classy catering needs at least 15 food servers and five bussers to cater a large part in order to make a profit they can have no more than 34 food servers and seven bussers working at an event right a system of inequalities for this all right so let’s make it easy on ourselves let’s use F for the food servers and let’s use be before the bussers okay so let’s see if we can write some inequalities here we know classic a Turing needs at least that’s a keyword we’ve had before 15 food servers at least means equal to or more so one inequality we can get from that is that F has to be greater than or equal to 15 okay and they need at least five bussers so again we want to use greater than or equal to bussers we need to have a least five in order to cater apart all right but to make profit they can’t have they couldn’t have unlimited servers they got to pay all these people so they said no more than we’ve had that keyword to that means less than or equal to ok so for food servers we have to be less than or equal to 34 for bussers we have to be less than or equal to 7 otherwise we’re paying out too many people now we have a system of inequalities that

represents the situation and for this problem that’s all they wanted us to do is to get to that point so that would be the answer later we’re going to have to probably graph a situation like that out as well similar to how we did in the last problem okay for number two now they give us some inequalities they want us to graph it out and figure out where is that region where all the overlap happens okay so first one we want to do we want to graph one of these lines let’s graph this one right here let’s graph that first one lucky for us this is in slope intercept form so we know on this one we can start at four okay that’s our y-intercept our slope here is negative two if it’s a whole number put it over one hey that’s saying down to write one so we want to keep doing that down to write one down to write one down to write one okay keep going till you run out of space now we have to connect these now remember when you’re graphing inequalities there’s two types lines that you want to use if it’s greater than or equal to or less than or equal to you want to use solid line if it’s just greater than or less than then you want to use a dashed or dotted line so here we should be connecting these with a solid line okay but again we’re not done here if it was equal if this was an equal sign we’d be done but it’s an inequality which means there’s many many many many many solutions that make that true we want to say decide which side of this line is going to make this true we know the points on the line make it true which side of the gret this is going to make it true so to do that pick a test point and the easiest test point it we always set a zero zero unless your line actually goes through the zero zero point here doesn’t so we can use 00 so let’s put 0 in for Y and let’s put a 0 in for our X and see what we get well if we put 0 and for both we get that 0 is greater than or equal to 4 that is a false statement 0 is not bigger than 4 which means our test point here at 0-0 does not come out to be true any point we pick on this side of the line is not going to come out true so that means the other side of the line must make them all true so in this one we’d want to shade everything over here okay so that’s for the first line now we got to do the same thing for the second light we want to graph it on the same graph again it’s in slope intercept form so start at negative 2 the slope is there’s nothing in front of this X so we know it’s a one and just like we did before we can put it over one it’s a whole number so 1 over 1 up 1 over 1 up 1 over 1 up 1 over 1 okay or we could go down on left foot no one left work this is going to be a solid line because it’s or equals 2 so the boundary line itself does include answers and then we want to check which side of the line makes this one true they get plugins your own zero and we can check we plug in zero and zero here I get that 0 is less than or equal to negative 2 that is again false zero is not less than the negative numbers it’s bigger than the negative numbers so the origin again it comes out to be false any point we pick over on this side is going to come out false so we want to shade on the opposite side we want to shade everything down here ok so now looking at this graph our answers our solutions are in the region that is overlapped and shading okay in here it’s right here any point in this region of the graph makes it true so what you want to do is heavily shaded in that section that is you any point in that region is your solution we’re going to refer to this as the feasible region that means any point in that feasible region feasible means something that’s possible is a possible answer try another one on this one number three we want to graph these to the top one we’re going to come back to that in a second let’s do one that you’re a little bit more familiar with why don’t you pause the video and try graphing this one Kerwick make sure you shade for it and see what you get alright let’s test out and see how you did so for this y is greater than or equal to X minus 1 it’s in slope intercept form start at negative 1 slope is there’s a one in front of their 1 over 1 is the slope up 1 over 1 or you can go down one left one it’s or equals

two so we want to choose a solid line connect these with now to check the shading if you plug in 0 0 as a test point you’re going to get that 0 is greater than negative 1 that’s true so we want to shade on the side our origins at which would be here so you should shade everything on this side ok now we got to graph this other one the other one’s a little bit harder we haven’t maybe graft one like this before what does this line look like well we just got done a few chapters ago looking at absolute value okay absolute value means the distance from zero so we want all the points that are less than three away from zero so that can mean that why here is less than three but it could also mean that it’s greater than the opposite of that which is negative three so we want to graph out all the points in this region okay to do that you could kind of split it in half look at each part separately here one thing that would say is that why needs to be less then three so on our graph let’s graph that one out what does y less than three look like well why less than three why lines that’s our intercept okay there’s no slope it’s kind of like plus 0x so there’s no slope with this one but we still want to start at and intercept the three here okay and if it’s a slope of zero that means it’s the straight-across horizontal line okay so we want to make a horizontal line going across through that three-point notice our symbol here it doesn’t include the line so this would be a dotted line that we want to use it so that is for that part of it okay and it’s less than so we want to shade everything less than this line so anything below this line should get shaded it because all the numbers less than three would be below okay let’s graph the other part of this line right here says that negative 3 is less than Y if we were to switch that around so that the wise on the Left make the threes on the right notice our symbols pointing at the negative 3 so let’s have a continued pointing at the negative 3 this is saying why should be greater than negative 3 okay that’s another one in slope intercept form so if we start at negative 3 again a slope of zero on this one and we should connect it with a dotted line okay this one set this is a greater than symbol here we all vote numbers greater than negative 3 are going to be a pulse so we want to graph everything above the line here okay so which region is the region where every all three of these overlap it would be right here okay this is where all the overlapped shading or cursor so that’s the section you want to have Li shaded when you’re doing this so that would be the is any point in that section includes our answer alright so when you’re doing this user probably for the first when you’re shading in each line shaded lightly and then when you’re done kind of shade the darker and for the final one all right let’s try another problem here it says solve these ones by graphing so we want to graph each line out here these ones shouldn’t be too hard they’re both in slope-intercept form so let’s start with that first one here plus four one two three four right here slope is 1 over 4 that means up 1 over 4 or we could go down one left floor ok it’s going to be a boundary line that’s dotted here and then we need to figure out which side to shade on on this one if you picked put it or put 0 and 0 in you’re going to get 0 is less than four that’s true we want to shade on the side of the origin here so let’s shade on that side so do some shading here I’m still show you everything here ok for that one for our next line start at negative 2 slope is again up 1 over 4 so we know that these two lines are parallel these two lines are not going to intersect and that’s one we want to use a solid likes it as it were equals two so we want to see where this one shades and where the overlap happens so if you pick a test

point let’s pick 0 again we get 0 is greater than negative 2 that’s also true we would get one state on this shade on the side of the origin so that means we should be shading above this one and it looks like the overlap in shading happens everywhere between these two lots so this is the side you’d want to shade in dark that’s where all the overlap happened so that would be all the solutions now when you get parallel lines like this the it would be the one time where you could maybe not get a solution if you were to graph two parallel lines and maybe one ends up getting shaded above and one gets shaded below okay and no overlap happens that would be the one case where you may not you might not end up with any solutions with these systems of inequalities that’s where you would say since no point makes both of them no points make both of them through the beat no solution all right this next one now we get two lines and these ones we want to graph to but there in that in slope intercept line or slope intercept form so if they’re not in slope intercept form there’s kind of two ways you could go about doing this one would be that cover-up method we talked about so you could cover up one of the variables and see what the other one is so for example on this first line if we were to cover up say this one okay then it looks like X ends up equaling negative one so on the graph which we could do let’s come over here and find where negative one in and put a point okay then you could do the same thing with the Y so if you cover up the X what’s the login equal so if we divide this by negative 1 looks like why ends up being positive 1 so then what you’d want to do on this one is put a positive fine were positive 1 Y is then put it out there and then connect those with whatever type of line looks like this one’s supposed to be a solid pipe and it looks like the pattern I here is up on over one so if you just continue that the line would look like this and we connect them with a solid line so one way to grab those is that cover-up method that we’ve previously done for this one that again we need to choose a test point so if we put 0 and for both x and y here 0-0 is you zero is zero greater than or equal to negative one yes 0 is bigger than all the negatives so we want to shade on the side of the origin here so let’s say there everything in this side championship ok for the second line we could do cover up again let’s try another method though for this on the other way you could do this if it’s not in slope-intercept form is get it in slope intercept form so if we’re going to get this one in slope intercept form we want to get y by itself ok so one thing we could do to get that minus 3x to the other side that cancels out so now i get negative y is less than or equal to negative 3x plus 4 ok but we’re not we don’t y by itself yet we still got this negative means that there’s a negative one here let’s divide everything by negative 1 ok if we divide everything by negative 1 that cancels out we’ve got plain old y negative 3 divided by negative 1 is positive 3x positive 4 divided by negative 1 is negative 4 also this is an inequality recall that when you’re dealing with inequalities and you multiply or divide by a negative we also need to switch the sign so the sign should go the other way on this one so now we want to graph out this line right here on our graph too ok so we want to start at the y intercept of negative 4 put a point and then we want to do slope on this one it’s three any whole number you can put over 1 so we want to go up three write one up three write one up three right what this one also has the or equals to so choose a solid line for a boundary line ok pick a test point let’s go back to zero zero again is 0 greater than negative 4 yes you’re bigger than all the negatives so again we want to shade on the side of

the origin so let’s do that here and if we shade in we shade in here everything on this side it looks like we get this area and green here that’s where overlap is that should be the part you shade in that’s where our solutions are so this is our feasible region right okay so if it’s not slope intercept lines then try to get them insulator so plans there’s slope intercept form all right this last one we want to again find out graph this one out there’s also an extra thing they want us to do here they also want us to find the coordinates of the vertices and we’ll talk about that in a second let’s graph this out first the first ones in slope intercept form be careful here they kind of throw a little curveball at you this one the X isn’t first so if you want it first if you used to that maybe rewrite it here our y-intercept is at three our slope here is negative 1 over 1 so go down one left one keep repeating that can repeat on the other side and this one we want to connect with a dotted or dashed line here because we don’t we don’t want to include the points on the line itself right now we got a shape ok so pick a side and shade on and on this one if we put in 0 0 it looks like 0 is less than 3 that’s true we want to shade on the side of the origin so everything here should get a change all right next line is y is greater than negative 1 ok that’s a slope intercept line to it’s just not they don’t really have the slope list it’ll be like plus 0x so we want to start at nay one and half a slope of 0 which means horizontal this one is a solid line put the cookie or equals 2 and it’s greater than all the numbers greater than negative 1 are going to be above the line so we want to shade everything here that would be above that line so if we do that be everything you understate all this alright the final thing now that we need to do is graph the last one this is not a slope intercept form like because there’s no y involved is just x greater than negative 1 this is a vertical line find where negative 1 is on the x-axis right here okay so it’s a vertical line through there we don’t want to include the line doesn’t have the or equals to so it’s a vertical dash or dotted line and it’s greater than negative 1 all the points greater than negative 1 are to the right so we want to graph everything so graph shading this in everything is right we get this okay and you can now see where our feasible region is it would be this triangle right in here so this anything in this triangle is our answer okay so that’s the graphing it the last thing they wants to do is figure out what’s the coordinates of the vertices that’s where the lines over where the lines bordering this feasible region intersect each other okay so there if it’s a triangle there’s going to be three points two of the points are pretty easy to see one’s right there so we got one point at negative 1 negative 1 that’s one ordered pair that border or that is a vertice of this feasible region okay another one poker is over here ok so then again find it now we know that for sure that’s going to be at negative 1 here now you get to see where these intersect it looks like it’s 1234 over there so for negative 1 would be another one the last one is where it intersects here so find the ordered pair there we know it’s going to be at negative 1 because it’s from this vertical line here and then it looks

like it appears to be up for so negative 14 so those are the vertices of feasible region all right try out your worksheet doing so