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Modeling with systems of inequalities
Sal models a real-world context into an algebraic system of linear inequalities and graphs it. Created by Sal Khan.
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- When Sal is drawing the coordinate system around3:00, he takes a guess at which resolution to use on the axes. This is not a good way to do it. Better is to wait until the step he does around7:00, that is, figure out the maximum values that g and s can take while still satisfying the second inequality. Had he done that, he would not have had an unused half of the coordinate space.
Actually, better is to figure out the values for both inequalities first, and then figure out how to best draw them.(15 votes)
- His stated goal is to solve by graphing, so it would be cheating for him to use algebra to find the answer and then solve by graphing!
Now if he wanted to publish his results somewhere, he would probably readjust his scales and use a graphing utility to give straight lines and perfect intercepts.(33 votes)
- Why are we able to make G equal to 0? I am having some trouble understand why and ho we make G or maybe y equal to zero? What is the process, why is it allowed?(8 votes)
- The condition specifies that you buy 15 ITEMS. Notice how it does not specify that you must buy one game or that you must buy 1 song. You just need to buy 15 items in total (which can mean having no songs or games), with the total bill being less than $25. That is why g can be 0.
In this video, Sal sets g to 0, and then S to 0, is because he wants to find the s and g intercepts (i.e what the s value is when g is 0, and vice versa).(12 votes)
- Did he forget the 18 on the "s" axis or am I imagining things?(0 votes)
- He is moving in jumps of four(1 vote)
- is this the same a linear programming ?(4 votes)
- This is a part of linear programming. The second part of linear programming is using the vertices of the unshaded area (On the graph in the video) to find the optimal value that can be obtained with use of an objective function.(4 votes)
- At3:20why did you put g as zero? could it have been different like instead of zero could you have put five?(2 votes)
- You set g equal to zero and s equal to zero so you can form a line. With the second inequality he does the same thing, sets g equal to zero and s equal to zero so he can form a line. By finding the x and y intercepts of the inequality (by setting g and s equal to zero) you can then draw a line. By doing that, he can find out where the inequalities overlap where or where they share the same solution set). He could have converted both inequalities into slope intercept form and graphed the line but it's unnecessary because you can just draw a line from the inequalities given.(8 votes)
- I tried switching the axes by putting the games on the y-axis and songs on the x-axis. However, I got a different slope which resulted in a different answer. Why is this so? Can someone help please?(2 votes)
- Think of it this way. The normal formula for slope = change in Y / change in X
If you reversed the 2 axis, you would have change in X / change in Y
Your slope should be the reciprocal of Sal's slope.
Hope this helps.(7 votes)
- Wait, you can't have a fraction or portion of a game, so why didn't Sal round the number of games down..?
Sal didn't round at this point in the calculation because he is trying to find the x and y intercepts in order to plot the line. He is not looking for an exact answer.(5 votes)
- At2:17, is it incorrect to switch the song axis and game axis? Does it matter which axis we have to put the variable on?(3 votes)
- You are correct. The choice of which axis should represent which variable was entirely arbitrary. Making the opposite choice would not have changed the problem.(2 votes)
- is it possible to use substitution and elimination to solve inequalities?(4 votes)
- In that case, you'll have to assume that the inequality is a gradient. That way when you plot the slope on the graph, you can use substitution of the value of "0" to check which region you should shade in.(0 votes)
- How do you represent only integer solutions on the graph?(2 votes)
- You need to solve the equations for integer values.
Use the second equation and calculate the solutions where S=0 and G=15, S=1 and G=14, S=2 and G=13, etc. Then, calculate additional points to fill in the gaps where the is money left over. Then, draw the graph along integer values on the horizontal and vertical lines. No short cuts here from point one to point two. You either go East and North or North and East. When you are done, inflation will have changed the values and you will have to do it over.(4 votes)
Luis receives a gift card worth $25 to an online retailer that sells digital music and games. Each song costs $0.89 and each game costs $1.99. He wants to buy at least 15 items with this card. Set up a system of inequalities that represents this scenario and identify the range of possible purchases using a graph. And that's why we have some graph paper over here. So let's define some variables. Let's let s equal the number of songs he buys. And then let's let g equal the number of games that he buys. Now if we look at this constraint right here, he wants to buy at least 15 items with this card. So the total number of items are going to be the number of songs plus the number of games. And that has to be at least 15. So it has to be greater than or equal to 15. So that's what that constraint tells us right there. And then the other constraint is the gift card is worth $25. So the amount that he spends on songs plus the amount that he spends on games has to be less than or equal to 25. So the amount that he spends on songs are going to be the number of songs he buys times the cost per song. Times $0.89 times-- so I will say 0.89-- times s. That's how much he spends on songs plus the cost per game, which is $1.99 times the number of games. This is going to be the total amount that he spends. And that has to be less than or equal to 25. Now if we want to graph these, we first have to define the axes, so let me do that right here. And we only care about the first quadrant because we only care about positive values for the number of songs and the number of games. We don't talk about scenarios where he buys a negative number of songs or games. So just the positive quadrant right here. Let me draw the axes. So let's make the vertical axis that I'm drawing right here, let's make that the vertical axis and let's call that the song axis. So that's the number of songs he buys. Let me make sure you can see that. That is the song axis. And then let's make this, this horizontal, that's going to be the number of games he buys. Let's bold it in. And just to make sure that we can fit on this page-- because I have a feeling we're going to get to reasonably large numbers-- let's make each of these boxes equal to 2. So this would be 4, 8, 12, 16, 20, so on and so forth. And this would be 4-- this obviously would be 0-- 4, 8, 12, 16, 20, and so on. So let's see if we can graph these two constraints. Well, this first constraint, s plus g is going to be greater than or equal to 15. The easiest way to think about this-- or the easiest way to graph this is to really think about the intercepts. If g is 0, what is s? Well, s plus 0 has to be greater than or equal to 15. So if g is 0, s is going to be greater than or equal to 15. Let me put it this way. So if I'm going to graph this one right here. If g is 0, s is greater than or equal to 15. So g is 0, s, 15, let's see, this is 12, 14, 15 is right over there. And s is going to be all of the values equivalent to that or greater than for g equal to 0. If s is equal to 0, g is greater than or equal to 15. So if s is equal to 0, g is greater than or equal to 15. So g is greater than or equal to 15. So the boundary line, s plus g is equal to 15, we would just have to connect these two dots. Let me try my best to connect these dots. So it would look something like this. This is always the hardest part. Let me see how well I can connect these two dots. Nope. Let me see. I should get a line tool for this. So that's pretty good. So that's the line s plus g is equal to 15. And we talk about the values greater than 15, we're going to go above the line. And you saw that when g is equal to 0, s is greater than or equal to 15. It's all of these values up here. And when s was 0, g was greater than or equal to 15. So this constraint right here is all of this. All of this area satisfies this. All of this area-- if you pick any coordinate here, it represents-- and really you should think about the integer coordinates, because we're not going to buy parts of games. But if you think about all of the integer coordinates here, they represent combinations of s and g, where you're buying at least 15 games. For example here, you're buying 8 games and 16 songs. That's 24. So you're definitely meeting the first constraint. Now the second constraint. 0.89s plus 1.99g is less than or equal to 25. This is a starting point. Let's just draw the line 0.89s plus 1.99 is equal to 25. And then we could think about what region the less than would represent. Oh, 1.99g. And the easiest way to do this, once again, we could do slope y-intercept all that type of thing. But the easiest way is to just find the s- and the g-intercepts. So if s is equal to 0 then we have 1.99g is equal to 25 or g is equal to-- let's get a calculator out for this. So if we take 25 divided by 1.99, it is 12.56. g is equal to 12.56. So when s is 0, let me plot this. When s is 0, g is 12.56. This is 12, this is 14. 12.56 is going to be right there, a little bit more than 12. That's that value there. And then let's do the same thing if g is 0. So if g is equal to 0, then we have-- so this term goes away-- we have 0.89s. If we use just the equality here, the equation-- is equal to 25 or s is equal to-- get the calculator out again. So if we take 25 divided by 0.89, we get-- it's equal to 28.08. Just a little over 28. So 28.08. So that is, g is 0, s is 28. So that is 2, 4, 24, 6, 8. A little over 28. So it's right over there. So this line, 0.89s plus 1.99g is equal to 25 is going to go from this coordinate, which is 0, 28. So that point right there. All the way down to the point 12.56,0. So let me see if I can draw that. It's going to go-- I'll draw up one more attempt. Maybe if I start from the bottom it'll be easier. That was a better attempt. Let me bold that in a little bit, so you can make sure you can see it. So that line represents this right over here. Now if we're talking about the less than area, what would that imply? So if we think about it, when g is equal to 0, 0.89s is less than 25. So when g is equal to 0, if we really wanted the less than there, we could think of it this way. It's less than instead of just doing less than or equal to. So s is less than 28.08. So it'll be the region below. When s is 0, g-- so if we think s is 0, if we use this original equation, 1.99g will be less than or equal to. I use this just to plot the graph, but if we actually care about the actual inequality, we get 1.99g is less than 25. g would be less than or equal to 12.56. So when s is equal to 0, g is less than 12.56. So the area that satisfies this second constraint is everything below this graph. Now we want the region that satisfies both constraints. So it's going to be the overlap of the regions that satisfy one of the two. So the overlap is going to be this region right here. Below the orange graph and above the blue graph, including both of them. So if you pick any combination-- so if he buys 4 games and 14 songs, that would work. Or if he bought 2 games and 16 songs, that would work. So you can kind of get the idea. Anything in that region-- and he can only buy integer values-- would satisfy