Lesson Video: Solving Systems of Linear Inequalities | Nagwa Lesson Video: Solving Systems of Linear Inequalities | Nagwa

# Lesson Video: Solving Systems of Linear Inequalities

In this video, we will learn how to solve systems of linear inequalities by graphing them and identify the regions representing the solution.

08:37

### Video Transcript

Graph Two-Variable Inequalities

There, given this inequality, we can say this is, 𝑦 is greater than or equal to three 𝑥 minus one. So how we will plot this is, we’ll basically pretend that we’re plotting instead. And if we were plotting the line 𝑦 equals three 𝑥 minus one. First thing we would need is a table of values. Well we need to pick three coordinates. My favourites is negative one, zero, and one. It can be more than three values, but it can never be less. So we’re going to take each of these 𝑥-coordinates that we’ve chosen now and substitute them into the function to get the 𝑦-value.

So for the first one, we put 𝑦 equals three multiplied by negative one minus one. Well three multiplied by negative one is negative three. And subtracting one from that, we get negative four. Then having a go for zero, we know that three multiplied by zero is zero. So then zero minus one is minus one. And then finally, substituting in one on the 𝑥-coordinate, we’ll have three multiplied by one, which is three, and minus one is two.

So now we have some coordinates. Those coordinates are: Negative one, negative four; zero, negative one; and one, two. We’ve got each of these coordinates by obviously looking at the 𝑥-coordinates in the table of values and looking at the 𝑦, and then saying, well that’s the 𝑥 and that’s the 𝑦, to find each of the set of coordinates. Now what we’ll need to do is plot this graph. But this is where we need to care about it being an inequality again.

So plotting each of these points we’ve got negative one, negative four; zero, negative one; and one, two. Now this is an inequality that is an “or equal to” inequality, as we can see here. So what that means is, when we come to plot the graph we must have a straight line, rather than a dotty line. Because for dotty line, it’s just greater than or less than, whereas with a straight line, we also include “or equal to”. And this follows, if you remember putting inequalities on a number line, we put a hollow circle in the case where we had just greater than and or less than, and a coloured-in circle for “or equal to”.

So anyway, going back to the question, we can see ours is an “or equal to”, so it needs to be a straight line. This, that I did earlier. Now the thing with two-variable inequalities, the thing that’s different from when we have simple linear inequalities, is we can’t just say “Oh okay that means I’m going to shade this side”. We need to actually do some testing. So we’re going to test above and test below, and see which fits our inequality. So if we test this point below of five, five and we’ll test negative five, ten above. We can really pick any points. But we just need to see, do these points satisfy the inequality. So we’ve got that 𝑦 is ten, so we put ten is greater than or equal to three multiplied by the 𝑥-value, which is negative five, minus one. Well we know that three multiplied by negative five is negative fifteen, and that negative fifteen minus one is negative sixteen. So ten is greater than or equal to negative sixteen. So above works. But we’re not just going to take that as an answer and say “Well there we go. I’m going to shade the region above, as that satisfies the inequality” because we have to assume that we always might’ve done something wrong. So let’s check our test below, and it should be wrong; it should not satisfy the inequality.

So we’ve got five is greater than or equal to three multiplied by five minus one. Well three multiplied by five is fifteen, and subtracting one from that we get fourteen. Now it’s not true that five is greater than or equal to fourteen. So we know that below doesn’t work, which is exactly what we wanted. So we can shade the region that satisfies the inequality, and that is above. And now we have shaded the correct inequality by plotting the line and testing above and testing below. Now one thing that has to be in place for us to be able to do a table of values is that our inequality has to be in the form 𝑦 equals 𝑚𝑥 plus 𝑐, which we can see that it doesn’t in this question.

So in this question, we need to first get our inequality into the form 𝑦 equals 𝑚𝑥 plus 𝑐.

So what we essentially need to do is, rearrange to get 𝑦 the subject of the function. So what we’re gonna do first, is subtract two 𝑥 from both sides. And that gives us on the left-hand side three 𝑦, which is less than or equal to seven minus two 𝑥 on the right-hand side. We can see that this still isn’t quite in the way that we need it. We need it to be just 𝑦, is it less than or equal to. So we’re going to have to divide both sides by three, giving us 𝑦 is less than or equal to seven minus two 𝑥, all divided by three. And although this isn’t exactly in the form that we want it, we can see that it is some number and some coefficient of 𝑥 on the right-hand side with 𝑦 by itself on the left.

So we can use this to have a table of values and plot the graph. So then substituting in negative one to the function, we get 𝑦 is less than or equal to seven minus two times negative one all divided by three. The negatives on the top become a positive, so we’ve got seven add two, which is nine, divided by three, just gives us an answer of three. And then substituting in zero and one, we get some fractions of seven over three and five over three. So now this gives us some coordinates of negative one, three; zero, seven over three; and one, five over three. These aren’t the nicest coordinates to plot, but we can definitely use them. And when we do plot our graph, we get this.

And then finally, we must test above and test below. So we’ll choose five, five above and negative five, negative five below. So to make it easier for us, let’s use the original function. Well we’ve got three multiplied by five, which is fifteen. Plus two multiplied by five, which is ten. And that must be less than or equal to seven, which is not. So the above doesn’t work. So we need to test below.

Now again, using the original function, we’ve got three multiplied by negative five, which is negative fifteen. Plus two multiplied by negative five, which is negative ten. And that is less than or equal to seven. So below works. So to shade the region that satisfies the inequality, we need to shade below the line.

And now we have it. We have finished the inequalities with two variables. And the most important things are to remember to plot using a table of values, and to check above and below the line, to be able to know what side to shade.

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