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Video: Solving Compound Linear Inequalities

Tim Burnham

We solve a series of inequalities that have more than one inequality symbol (compound inequalities). We then present our answers using simplified inequalities, number line diagrams, intervals, and set notation.

15:53

Video Transcript

In this video, we’re gonna solve some compound linear inequalities and give our answers in a variety of formats, using inequality symbols, number line diagrams, intervals, and set notation. We’ll start with simpler positive integer solutions and move on to harder examples involving negative and rational, or fractional, solutions.

First so, let’s just quickly recap the notation. So we’ve got 𝑥 is greater than three, this is an inequality. The value of 𝑥 is greater than three, but it cannot be equal to three. And we can represent this on a number line diagram, or a graph, by using the open circle above the three because the 𝑥 can’t be equal to three. But 𝑥 can take any value greater than three, so it’s this region here.

We can also represent that using interval notation. So the curved parentheses represents a closed interval. So that interval goes down to, but not including three. At the upper end, we go to infinity, but we still closed that with a parenthesis, or round bracket, rather than a square bracket. But we’ll come onto square brackets in a moment.

And we can also represent it using set notation. So we’ve got the set of 𝑥, such that 𝑥 is a member of the real numbers where 𝑥 has a value greater than three. Now you may have seen other methods of representing set notation, but this is the one that we’re gonna be using in this video.

Okay, let’s look at another inequality, 𝑥 is less than or equal to five. So the valid values of 𝑥, 𝑥 could be equal to five, or it could be anything smaller than that. So it could be four point nine recurring, or it could be minus infinity. So on a number line, we put a solid dot above the five to say that five is included. It can be equal to five and/or it can be anything less than that, so we have a big arrow pointing to the left. And in interval format, we now use the square bracket to the right of the five and that denotes the fact the five is included in that interval. And because we’re going down to negative infinity, the infinity is always enclosed by the round parentheses. And in set notation, it’s the set of 𝑥 such that 𝑥 is real and less than or equal to five.

So now let’s consider this situation where we’ve got this compound linear inequality. Basically, we’re saying that 𝑥 could be greater than three but it also has to be less than or equal to five. So we’re combining the two above, and-and we can represent that in this way. So three is less than 𝑥 is less than or equal to five. So 𝑥 can be equal to five but it still gotta be greater than three all the time. So greater than three, less than or equal to five, that means we allow it to be equal to five, so put a-a solid dot above the five. It’s an open dot above the three because it’s not allowed to be equal to three. But it can be anything in between those two, so we join them up with a line here.

Using interval notation, three and five are at the either end of the interval. But the five is included in the interval, so we put a square bracket around that. And the three isn’t included in the interval, so we put a round parenthesis around that. And in set notation, we’ve got the set of 𝑥 such that 𝑥 is real and is between three but- and less than or equal to five.

Okay. So that’s the notation we’re gonna be using in our questions. Let’s go ahead and look at some questions. So we’ve gotta find all the values of 𝑥 that satisfy the compound inequality forty-five minus 𝑥 is greater than 𝑥 minus five is greater than or equal to nine minus 𝑥. So this is telling us that this bit is bigger than this bit, and this bit is, this bit here, is bigger than or equal to this bit. So, so long if you do the same thing to each of them, that kind of step of, you know, big, slightly smaller, slightly smaller, will be maintained. Now we do have to be careful though, if we multiply or divide everything through by some negative number, then we’d have to reverse the signs. We’re not gonna look at that for now. But let’s just go through and try to get 𝑥 on its own in the middle and try to eliminate 𝑥 from the outsides.

Now first, let’s see I’ve got an 𝑥- a negative 𝑥 here, a negative 𝑥 here, and I’ve got an 𝑥 here as well. So if I add 𝑥 to all of them, then the inequalities will be maintained. And it will eliminate the negative 𝑥 from here and here, so negative 𝑥 plus 𝑥 is gonna be zero. So I’m gonna add 𝑥 to each of those sections. And forty-five minus 𝑥 plus 𝑥 just leaves us with forty-five. So that’s greater than. So I’ve got 𝑥 minus five, add another 𝑥, that’s gonna make that two 𝑥 minus five. And on the last section, I’ve got nine minus 𝑥, add an 𝑥, that’s just gonna leave me with nine.

Okay. So I’ve now only got 𝑥s in the middle section, so that’s a bit of progress. But I’ve got two 𝑥 and I’ve got minus five. So I’m gonna add five to each section next, to eliminate the minus five in the middle bit. So forty-five plus five gives us fifty, two 𝑥 minus five plus five just leaves us with two 𝑥, and nine plus five gives us fourteen. So now I want to just have one 𝑥 in the middle, so I can half everything. If I took away 𝑥 from each of them, then, well here I’d be undoing the work that I did over here, but I’d be ending up with 𝑥s here and here as well. So I’m just gonna half all the way through. And half of fifty is twenty-five, and half of two 𝑥 is 𝑥, and half of fourteen is seven. So that definitely works, because let’s imagine I had two 𝑥 and I had fourteen. We’re saying that two 𝑥 is greater than or equal to fourteen, so we’re saying that this is bigger, or it might be equal to this. Now if I chop that in half and I chop that in half, because we’ve got half of the bigger amount and half of the smaller amount, that ratio of which one is bigger and which one’s smaller will still be maintained.

So that’s kind of our first solution then, twenty-five is greater than 𝑥 is greater than or equal to seven. Now we quite often will tend to actually write that the other way around. We tend to start with the-the smaller numbers on the left and kind of work our way upwards, cause then that lines up with the graphs that we’re gonna draw. So I’m just gonna turn that around. So seven is less than or equal to 𝑥 is less than twenty-five. So we’ve turned the whole thing around and we’ve turned each in-each individual sign around as well. So let’s represent that on the number line. So the critical values are seven and twenty-five. So we’ve put those on that number line and now we’ve got to try to work our 𝑥s. Well 𝑥 is greater than or equal to seven. So it’s allowed to be equal to, so we’ve got a solid dot. And then it’s greater than, so it’s off to the right of that. Now 𝑥 can’t be equal to twenty-five, so we’re gonna put a- an open dot above that. But it’s got to be less than that, so we’re going to the left of that. So this line is gonna join up with this line here, and that’s our interval.

Now in interval format, seven and twenty-five are the ends of the interval. Now it’s not allowed to be equal to twenty-five, so we use a round parenthesis for that. And it is allowed to be equal to seven, so we use a square bracket to represent that.

And in set notation, 𝑥 is the set where 𝑥 is real and it’s between seven and twenty-five.

So for number two, find all the values of 𝑥 that satisfy two 𝑥 is less than five 𝑥 plus nine is less than two 𝑥 plus thirty-nine. So we’ve got another compound linear inequality to solve. So we’re gonna take the same approach as before, try to get 𝑥 on its own in the middle. Now what we can see is, we’ve got two 𝑥s here, we’ve got five 𝑥s here, we’ve got two 𝑥s here. So if we took away two 𝑥 from each of those, and we’re taking away the same thing from each so the inequalities are still gonna match, then that’s gonna eliminate 𝑥s from the left and from the right. And so on the left, we’ve got two 𝑥 take away two 𝑥, well that’s nothing. And that’s less than. But five 𝑥 take away two 𝑥 is three 𝑥, and we still got our plus nine. And two 𝑥 take away two 𝑥 is nothing, so that still leaves us with our thirty-nine. So zero is less than three 𝑥 plus nine is less than thirty-nine. Now we want to get rid — well we wanna try and get the 𝑥 on its own, so we’ve gotta get rid of the plus nine and we’ve gotta get rid of the times three. So the first thing I’m gonna do, is subtract nine from everything. Now zero take away nine is negative nine, three 𝑥 ta- take aw- plus nine take away nine is just three 𝑥 and then thirty-nine take away nine is just thirty. And then lastly, we’ve got three times 𝑥, so we want to divide everything by three, so we’ve just got one times 𝑥. And a third of negative nine is negative three, a third of three 𝑥 is just one 𝑥, and a third of thirty is ten. So minus three is less than 𝑥 is less than ten. So that’s our first answer.

Now we’re gonna take that inequality format, and draw our number line, or graph. So we’re looking at values of 𝑥 and the critical values are negative three and ten. I’m just gonna put zero’s in the middle there somewhere as well. Now 𝑥 is gotta be bigger than negative three, the bigger end of the sign, of the inequality sign, is against the 𝑥 and the smaller end is against the negative three. So 𝑥 is to the right of negative three, it can’t be equal to it, so we have an open circle above the negative three. 𝑥 is less than ten, can’t be equal to ten, so it’s an open circle above the ten. But it is allowed to be any value in between those. So that’s what our number line solution would look like.

As an interval, well the critical values again are negative three at one end and ten at the other. And in both cases, we’re not allowed to be equal to those, so we’re gonna enclose them with these round parentheses.

And then in set notation, we’ve got the set of 𝑥 such that 𝑥 is real and it’s between negative three and ten, not including negative three and ten. So these intervals can go into the negative region as well.

Now let’s move to the world of fractions. And so we’ve got to find all the values of 𝑥 that satisfy two and two-thirds minus 𝑥 is less than or equal to two is less or equal to three and a half minus 𝑥. So we’ve got 𝑥 here and here, but we want 𝑥 in here, ideally. So I’m going to eliminate the 𝑥s from the outer two sections. I’m just gonna add 𝑥 to each section. And two and two-thirds minus 𝑥 plus 𝑥 is just two and two thirds, and that’s less than or equal to two. If I add 𝑥, let’s call that two plus 𝑥, and if I add 𝑥 to the last section, three and a half minus 𝑥 plus 𝑥 is just three and a half. So now I’ve got rid of my 𝑥s from the outside and I’ve got my 𝑥 in the middle. All I need to do is subtract this two from each of them and I’ll just have 𝑥 on its own in the middle. So two and two-thirds take away two is just two-thirds, and two plus 𝑥 take away two is just 𝑥, and three and a half take away two is one and a half.

So there’s our simplified compound linear inequality and it involves fractions this time. So drawing the graph, we’re thinking about our 𝑥-values and the critical values are two-thirds and one and a half. And we could put zero on there, if we want as well. And we could even put sort of one in here as well, if we wanted to, don’t need to do that. We’re allowed to be equal to two-thirds, so that’s gonna be a solid dot. We’re allowed to be equal to one and a half, so that’s gonna be a solid dot. And 𝑥 can take any value in between the two, so that’s what our graph is gonna look like.

In interval format, two-thirds and one and a half are the either end of that interval. Both of those values are included, so we use the square brackets to denote that we’re including those.

And then in set notation, we’ve got the set of 𝑥 such that 𝑥 is real and 𝑥 is between two-thirds and one and a half. And it’s allowed to be equal to two-thirds and one and a half.

So just one more example then, find all values of 𝑥 that satisfy the inequalities three 𝑥 plus five is less than eleven or two 𝑥 minus three is greater than or equal to seven. So we’ve actually got two distinct regions here and when we-when we go through and we simplify them, we’ll actually see that in practice. So when you’ve got two inequalities like this, we need to simplify each one individually. Sometimes you can fit them back into one compound inequality and sometimes you can’t. So let’s go ahead and do this. We’ll do the one on the left first. We’ve got three 𝑥 plus five is less than eleven. So if I took away five from both sides, I’d be ending up with just three 𝑥 on the left-hand side. So three 𝑥 is less than six. Now if I divide everything by three, I know that 𝑥 has got to be less than two. So let’s look at the other inequality now. So the first thing we’re gonna do is add three to both sides, to get rid of that negative three from the left-hand side. Now remember, we can do this, we can add three instead of taking away five because this is a completely separate inequality. So adding three to both sides gives us two 𝑥 is greater than or equal to ten. Now dividing by two tells me that 𝑥 is greater than or equal to five. So 𝑥 can be less than two or it can be greater than or equal to five. So two and five are the critical values. 𝑥 can be greater than or equal to five, so it can be equal to five or it can be greater than; so it can be in this region here. And it can’t be equal to two but it can be less than two, so we’ve got this region here. So this is a non-continuous region, so we have to represent this as two separate inequalities. So acceptable values for 𝑥 are less than two or greater than or equal to five.

Now in this case, in interval notation, we’ve actually got to represent two different intervals and then we’re gonna represent the union of those two intervals. So the one on the left is everything from minus infinity up to two. And the one on the right is everything from five up to positive infinity. But we’ve gotta be careful about those, whether the two and the five are included or not. So I’ve written down my critical values and we’re going to union them together. The infinities are always enclosed there by round parentheses. Now the two is not included, so we have to include the round parenthesis there. The five is included, so we write the square parenthesis, or the square bracket, around the five. So that’s how we’d represent this as the union of two intervals.

Now in set notation, we’ve got the set of 𝑥 such that 𝑥 is real and 𝑥 is less than two or 𝑥 is greater than or equal to five.

So let’s just summarise what we’ve done with compound linear inequalities then. So we can add or subtract the same thing from each. In this case we added 𝑥 to each, but we could’ve added five to each. So long as we add or subtract the same, then those inequalities are gonna be maintained. And we also saw that we can double or halve or triple; we can multiply or divide by the same for each, and again maintain those inequalities. The thing to watch out for though is, if you multiply or divide by a negative number, then you’re gonna have to swap the signs around so they point in a different direction. And all that adding, subtracting, multiplying, dividing, is aiming to get a single 𝑥 on its own in the middle part of that lin- of that compound inequality. And lastly, you can use interval, or graph, or number line, or set, or inequality notation, to present your answer as a choice of different ways to present that answer.