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Video: Solving Absolute Value Linear Inequalities

Tim Burnham

A detailed and careful exploration of linear inequalities that involve absolute values. We give step-by-step instructions as to how the graph of an absolute value linear function is constructed and apply this knowledge to help us solve the inequalities.

14:04

Video Transcript

In this video, we’re gonna look at how to solve absolute value linear inequalities; these involve expressions that we evaluate and take the absolute value of. So if the result was negative, then we just use the positive of that number. We’ll look at a graphical problem and an algebraic one, and we’ll talk about how to approach them both. We’ll also give you some answers in different formats, so inequality format, and also using interval notation. Okay, let’s go ahead and look at our first problem. So in this question, we’re given a graph which shows two functions: we’ve got 𝑓 of 𝑥 which is the absolute value or the modulus of 𝑥 plus two and another straight line, horizontal line, a function 𝑔 of 𝑥, which is equal to five. And we’ve gotta find the values of 𝑥 that satisfy the inequality 𝑓 of 𝑥 is less than 𝑔 of 𝑥. So this is a strict inequality. So we’re looking for 𝑥-coordinates, where the corresponding 𝑦-coordinate on the 𝑓 function is lower than the 𝑔 function. So we’re looking for what areas of the graph is the green line below the black line. And when we’re given a graph like this, it’s just a matter of really doing a bit of highlighting on the graph and reading some values of the graph. And so we’re gonna do that, but we’ll also go one stage further and do something the question doesn’t ask us to do and solve it algebraically. And look- and that’ll just hopefully provide the link to understanding how the algebraic solution works.

So in trying to understand where the 𝑓 of 𝑥 values generated, the 𝑦-coordinates, are less than the 𝑔 of 𝑥 values created, so the 𝑦-coordinates for that function, we need to really start off by thinking, “Well where are they equal?” And then we sort of think about do we- do we go to the left or the right of that in each case. So we can see that the lines intersect here and here: so 𝑓 of 𝑥 is equal to 𝑔 of 𝑥 when 𝑥 is negative seven, and 𝑓 of 𝑥 is equal to 𝑔 of 𝑥 when 𝑥 is equal to three. So 𝑓 of 𝑥 is less than 𝑔 of 𝑥 as we said when the green line is lower than the black line on here, and that is all points in between 𝑥 equals negative seven and 𝑥 equals three. So negative seven is not included, three is not included cause the functions are equal there. But everything in between the two, the green line is below the black line. So writing that as an inequality, we say that negative seven is at the left-hand end. And that is always less than 𝑥 because if 𝑥 is equal to negative seven, then the two functions will be equal. And 𝑥 is always smaller than three because if 𝑥 was three or larger, then the 𝑓 of 𝑥 values generated will be greater than the 𝑔 of 𝑥 value.

So that’s our answer. Or we could write it as an interval. Now the critical values are negative seven and three at either end of the interval. Now negative seven is not included so we use the round parenthesis, and three is not included so we use the round parenthesis again. So that would be another way of presenting that solution. In fact, we can do it in set notation here. So we’ve got the set of 𝑥 such that 𝑥 is a real number where negative seven is less than 𝑥 is less than three, so you’ve got three different ways of presenting your answer.

Okay, so that was pretty straightforward. What we’re gonna do though is we’re just gonna take another look at it, sort of solving it algebraically, but still looking at the graph as we do it. So 𝑓 of 𝑥 is the absolute value of 𝑥 plus two. But let’s- before we think about, that let’s consider the graph of 𝑦 equals 𝑥 plus two, so ignoring the absolute value. Now 𝑥 plus two would basically be a line which has a slope of positive one and it would cut the 𝑦-axis at positive two. And that’s basically this line here, but of course it would carry on going down here forever. So that has, as we said, it cuts the 𝑦-axis at two, and the slope of one means that every time I increase my 𝑥-coordinate by one then the corresponding 𝑦-coordinate goes up by one as well. So I’ve taken two points on the line here, so to go from here to the next point my 𝑥-coordinate increases by one and my corresponding 𝑦-coordinate would also increase by one. And that’s the same wherever you are on that line. So that’s all well and good; that’s 𝑦 equals 𝑥 plus two. But the function that we’re looking at is represented by 𝑦 equals absolute value of 𝑥 plus two. Now every 𝑦-coordinate that we generate, if that turns out to be negative, we’ve got to turn it into the positive. So that doesn’t really sort of apply to the right of this line.

So over here, for all these 𝑥-coordinates, just thinking about 𝑦 equals 𝑥 plus two will always generate zero or a positive 𝑦-coordinate anyway. So if we’d limited the 𝑥-coordinates we could use to negative two or above, then just using the function represented by 𝑦 equals 𝑥 plus two would be fine. But to the left of that line, when we’ve got 𝑥-coordinates less than negative two, every time we generate these negative 𝑦-coordinates, what we’re doing is we’re reflecting them in the 𝑥-axis and taking the positive version. So what we were effectively doing is we’re taking a negative value then taking the negative of that negative value, which is giving us the positive value. So to the left of negative two, the 𝑦-value is always the negative of 𝑥 plus two, the negative of the function.

So although 𝑓 is equal to the absolute value of 𝑥 plus two, the practical way of interpreting that is if the 𝑥-coordinate is negative two or above, then we’re just gonna use this equation to work out our 𝑦-coordinates; but if the 𝑥-coordinate is less than negative two, then we’re gonna use this equation to work out our 𝑦-coordinates.

So if we hadn’t been given the graph but we had been given the functions absolute 𝑥 plus two and five, then we could’ve now sketched the graph. And if we’d done that, we’d have a graph that looks something like this. So now the process will be similar-ish to what we did before; we’d still be interested in where do those two lines intersect, and then we’ll be looking for where is the 𝑓 of 𝑥 line lower on the graph than the 𝑔 of 𝑥 line. And that’s obviously between those two points. So we’re quite interested in working out what that 𝑥-coordinate is and we’re trying- are quite interested in working out what that 𝑥-coordinate is.

Well 𝑔 of 𝑥 is always five. So let’s look at the left-hand point 𝑓 of 𝑥 is equal to 𝑔 of 𝑥. Well in that region, 𝑓 of 𝑥 is equal to minus 𝑥 minus two; and in that region, 𝑔 of 𝑥 is equal to five. Well in every region 𝑔 of 𝑥 is equal to minus five. So we just need to add 𝑥 to both sides, then subtract five from both sides. And we see that the 𝑥-coordinate there is negative seven. We can do the same for the right-hand end. And in that region, the equation for the 𝑓 of 𝑥 function is 𝑦 equals 𝑥 plus two. So 𝑓 of 𝑥 is 𝑥 plus two and 𝑔 of 𝑥 is five.

Now just subtracting two from each side on that tells us that over there, the 𝑥-coordinate is three. So now we’ve got exactly the same information that we were given straight away in the graph last time. We can say, “But when is the blue line below the black line?” Well not when 𝑥 is negative seven, not when 𝑥 is three, but every point in between. So again, we can write out our answer in any of those three formats. Okay let’s have a look at a purely algebraic question from scratch.

Find the values of 𝑥 that satisfy the inequality the absolute value of six minus 𝑥 is greater than or equal to ten. So it’s not a strict inequality; that can be equal to ten. So to start off, we’re just gonna think about the left-hand side. So we’re just gonna consider 𝑦 equals six minus 𝑥 and we’re gonna ignore the absolute function for now. We’ll work that out in a bit. So that’s a linear function. And because the coefficient of 𝑥 is negative one, that means the slope is one; every time my 𝑥-coordinate increases by one then my corresponding 𝑦-coordinate will go down by one. The 𝐶, the intercept, the constant term is six, so it means it cuts the 𝑦-axis at six. So now we just need to think about where would it cut the 𝑥-axis. Well it’ll cut the 𝑥-axis when 𝑦 is zero. And putting 𝑦 equals to zero gives us zero equals six minus 𝑥. So If I just add 𝑥 to both sides, I get 𝑥 equals six. So let’s just sketch that on some axes. So that’s what 𝑦 equals six minus 𝑥 would look like. But of course remember, we wanted the absolute value of 𝑦 equals six minus 𝑥. So anywhere that that line dips below the 𝑥-axis and gives us negative 𝑦-coordinates we’re just gonna take the corresponding positive 𝑦 version of that. So that line is positive all the way up to 𝑥 equals six, and then after that it’s gone below the 𝑥-axis. So we’re gonna have to reflect all of these points in the 𝑥-axis up to here, take the corresponding positive versions of those 𝑦-coordinates. So for the absolute value of six minus 𝑥 we ignore the blue line below the axis there and we just reflect it in the 𝑥-axis up here, so what we’re doing is we’re taking the negative of all the 𝑦-coordinates, that’s the negative of that function 𝑦 equals six minus 𝑥. So to the left of 𝑥 equals six, we’re just using the straight line 𝑦 equals six minus 𝑥, but to the right of it we’ve got 𝑦 equals 𝑥 minus six and negative of the other one.

So now we know what the absolute value of six minus 𝑥 looks like. Let’s just sketch on our graph 𝑦 equals ten, and we want to know when this is above this. So again, let’s look at those critical points where they intersect. So looking at the left hand of those two first, absolute values of six minus 𝑥 is represented just by 𝑦 equals six minus 𝑥 in that region, so we need to put that equal to ten.

And then I’m gonna add 𝑥 to both sides and then subtract ten. So the 𝑥-coordinate where they intersect is negative four And over on the right-hand side in that region, 𝑦 equals 𝑥 minus six tells us about that straight line, and we need to put that equal to ten. So I’m just gonna add six to both sides to work out what 𝑥 is. And 𝑥 is equal to sixteen in that case, so I can write sixteen on my graph on my axes.

So we know where those two things are equal; we now need to look back at the question that’s asking for when the absolute value of six minus 𝑥 is greater than ten. So this graph has to be above this graph on our axes. So at the left-hand point here where they intersect, they’re equal; and anywhere to the left of that, the absolute value of six minus 𝑥 is above 𝑦 equals ten. And over on the right-hand side here is where they’re equal, and anything to the right of that is where modulus or the absolute value of six minus 𝑥 is greater than 𝑦 equals ten.

So thinking about the 𝑥-values that correspond with those regions of the graph, when 𝑥 is negative four, that’s great; the two things are equal. And anywhere to the left of that, as we’ve illustrated on here, is good region; that’s where our inequality is satisfied. And then over on the right-hand side, when 𝑥 is sixteen, that’s in the region we’re looking for; the inequality is satisfied. And anywhere to the right of that is satisfied as well. So we’ve ended up with a noncontinuous region: so anywhere to the including or to the left of negative four, anywhere including or to the right of sixteen. So we can write that as a pair of inequalities like this.

And when it comes to writing that in interval format, we’ve got to write that as the union of two separate intervals. So we’re going all the way from negative infinity up to negative four and then we’re going from sixteen up to positive infinity. Well the infinities always have a round parenthesis by them. And negative four is included in that region, so we put a square bracket around that one. And sixteen is included so we put a square bracket around that one. And remembering that both of those two are valid regions, we’re going to union them together. So that’s one way of representing it.

Now another way of representing that is the fact that, remember we’re sort of thinking about the 𝑥-axis here, all of those are the real numbers on the number line. So we could say it’s all real numbers minus this region here in between negative four and sixteen. But remember, we don’t wanna take away negative four because negative four is part of the valid region. So I’m gonna put a round parenthesis to say this isn’t included in the bit that we’re taking away. And likewise for sixteen, round parenthesis says that sixteen is not in the bit that we’re taking away. And again I could use set notation: we’ve got the set of 𝑥, where 𝑥 is real such that 𝑥 is less than or equal to negative four or 𝑥 is greater than or equal to sixteen.

So to solve absolute value linear inequalities, really you need to think about what the line would look like without the absolute signs then adjust for the absolute sign, and then just start looking at where regions of graph overlap or above or below other regions. And then you’ve got a variety of different ways that you can present your answer. So top tip is always to do these sketches and do these graphs because I think that really helps to aid your thinking and communicate your thinking, and also it helps you to avoid making some silly mistakes. Okay, good luck with your linear inequalities with absolute functions.