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This video explains how to solve equations that involve finding the absolute value of a variable or the simple linear expression involving that variable and also include that variable on the other side of the equation (e.g., |𝑥 + 5| = −5𝑥 + 2).
In this video, we’re gonna look at how to solve equations that involve an unknown in two places, one requiring an absolute value to be taken and one not. Now remember, taking the absolute value just means taking the positive version of a number, so ignoring the negative sign. We’ll look at two ways to solve the problems: one purely algebraically considering different possible values and the other by looking at the graphs and considering the effect of taking the absolute value. Okay, let’s look at our first example.
Well we’ve got to solve the absolute value of 𝑥 plus two is equal to 𝑥 plus two. And we’ve got two different situations to consider: if 𝑥 plus two is greater than or equal to zero, then it’s already positive so we don’t need to do anything fancy; if 𝑥 plus two is less than zero, then we need to take the negative of that value in order to turn it into a positive value. That’s what the absolute value function is doing. So the values of 𝑥 down each of those branches are, well subtracting two from either side of the inequality on the left there, is 𝑥 is greater than or equal to negative two, and doing the same for the other inequality, 𝑥 is less than negative two. So if the value of 𝑥 is less than negative two, we’re gonna come down the right hand-branch; and if the value of 𝑥 is greater than or equal to negative two, we’re gonna come down to the left-hand branch. And if we come down the left-hand branch, that means that the value of 𝑥 plus two is positive so we don’t need to worry. So we’ll just use the expression 𝑥 plus two. So what we’re saying is 𝑥 plus two on the left-hand side is equal to 𝑥 plus two on the right-hand side, and that’s true for all values of 𝑥. But remember, although it’s all values of 𝑥, we were stretching ourselves to only values of 𝑥 which were greater than or equal to negative two in order to get in here. So it’s all values of 𝑥 that are greater than or equal to negative two. Now going down the right channel, we know that 𝑥 is less than negative two, and we know that 𝑥 plus two is giving us an answer which is negative, so we’ve gotta take the negative of that negative to turn it into a positive. So that means the negative of 𝑥 plus two is equal to 𝑥 plus two or negative 𝑥 take away two is equal to 𝑥 plus two. Well if I add 𝑥 to both sides of that equation, I get negative two is equal to two 𝑥 plus two; then if I subtract two from both sides, I get negative four is equal to two 𝑥; and then if I divide both sides by two, I get the answer 𝑥 is equal to negative two. Now if we’re being really strict about that, we said that 𝑥 was less than negative two to come down this route in the first place, so that is not a valid solution. But in fact, strictly speaking, 𝑥 could be equal to negative two to come down this route because the negative of zero is still zero. So taking the absolute value of that would in fact work. In fact, it doesn’t matter in either case because we have already covered that solution on the other side over here. So the answer is 𝑥 is greater than or equal to negative two, which we can also write in interval notation like this. negative two is included, so we’ve got the square bracket and it goes all the way up to positive infinity, which is- always has the parentheses or the round bracket.
Now we also promised to have a look at this graphically. So if I draw the graph of 𝑦 equals 𝑥 plus two, straight away from the equation 𝑦 equals 𝑥 plus two, we can see the slope is one. So every time I increase my 𝑥-coordinate by one my 𝑦-coordinate also goes up by one, and the 𝑦-intercept is two. And it cuts the 𝑥-axis when 𝑦 is equal to zero, so zero is equal to 𝑥 plus two. Subtracting two from both sides, 𝑥 is equal to negative two. So it cuts the 𝑥-axis, 𝑥 is negative two, cuts the 𝑦-axis at two, and has a slope of one. Now let’s think about the graph of 𝑦 equals the absolute value of 𝑥 plus two. Well, this is gonna be just like the graph of 𝑦 equals 𝑥 plus two when it happens to be above the 𝑥-axis, when the 𝑦-coordinates are positive. But when the 𝑦-coordinates go negative, we’re reflecting that in the 𝑥-axis so that all the negative 𝑦-coordinates map onto their positive equivalents. So the orange line there is the line 𝑦 equals the absolute value of 𝑥 plus two. And to the right of where the 𝑦-coordinate is zero at 𝑥 equals negative two and in fact even including when 𝑥 equals negative two. So I put a little solid dot in here. That region there, we’re just using the equation 𝑦 equals 𝑥 plus two to generate the 𝑦-coordinates for that orange line from the 𝑥-coordinates. Now to the left of that, that equation would generate negative 𝑦-coordinates, so we’re taking the negative of the negative 𝑦-coordinates to generate these positive coordinates up here. So to the left of 𝑥 equals negative two we’re using the equation that the 𝑦-coordinate is equal to the negative of the coordinate generated by 𝑥 plus two. In other words, 𝑦 equals negative 𝑥 take away two So the question was, when are those two things equal? When does the line 𝑦 equals the absolute value of 𝑥 plus two intersect with the line 𝑦 equals 𝑥 plus two? And it’s all of these values here, including when 𝑥 is equal to negative two, and all of these off into infinity in that direction here. So the set of 𝑥-values that match that is when 𝑥 is equal to negative two and everything to the right of that. So we’ve got the same answers, by either doing it algebraically or using the graph, but using the graph gives you a bit more insight into why that’s the answer and what’s really going on.
Let’s look at number two then. Solve the absolute value of 𝑥 plus five is equal to negative five 𝑥 plus two. So again we’ve got an absolute value, so there are two different scenarios. Either the value inside the absolute function there 𝑥 plus five is already greater than or equal to zero or it’s negative and it needs changing into a positive value. So just thinking about the 𝑥-values that correspond to that, if 𝑥 plus five is greater than or equal to zero, that happens when 𝑥 is greater than or equal to negative five; if 𝑥 plus five is less than zero, then that happens when 𝑥 is less than negative five. So going down the left-hand branch first, if the contents of the absolute value were already positive, then we can just take 𝑥 plus five because it’s already positive and we can solve that equation. So adding five 𝑥 to both sides gives us six 𝑥 plus five equals two. And then subtracting five from both sides gives us a six 𝑥 is equal to negative three. And then dividing both sides by six gives us 𝑥 is equal to negative three over six which is negative a half. So just before we consider that a proper solution, we do need to think, “Is that in the region that would have got us here anyway?” In order to get down this branch, 𝑥 had to be greater than or equal to negative five, and negative a half is greater than or equal to negative five, so we’re fine. That is a good solution. So let’s look at the other side.
On this side, remember 𝑥 plus five gave us a negative answer so we need to convert that to its positive equivalent. So instead of taking 𝑥 plus five, we’re taking the negative of 𝑥 plus five to turn it into a positive coordinate, and that’s equal to negative five 𝑥 plus two. So distributing the negative over the parentheses there, we’ve got negative 𝑥 take away five is equal to negative five 𝑥 plus two, and so adding five 𝑥 to both sides now gives us four 𝑥 minus five on the left-hand side and just two on the right-hand side. so if I now add five to both sides, I’ve got four 𝑥 equals seven. And now finally, just divide both sides by four and I get 𝑥 is equal to seven over four. But wait! We said to get down this branch, 𝑥 needed to be less than negative five; but in doing that, we’ve come up with an answer 𝑥 is equal to positive seven over four. So this is a spurious nonsense solution, so that’s not a correct solution here. So the only answer we’ve got that’s valid is 𝑥 equal to negative a half. Let’s have a look at the graph of that and just see why that’s the case.
So let’s first consider the graph of 𝑦 equals negative five 𝑥 plus two that has a slope of negative five and it cuts the 𝑦-axis at two. So negative five every time I increase my 𝑥-coordinate by one, my 𝑦 coordinate is gonna go down by five. So we’re looking at a very sharp downhill line, and it cuts the 𝑥-axis when 𝑦 equals zero. So putting the 𝑦-coordinate equal to zero, we’ve got zero is equal to negative five 𝑥 plus two, adding five 𝑥 to both sides gives me five 𝑥 is equal to two, and then dividing both sides by five gives me 𝑥 is equal to two-fifths. So it cuts the 𝑥-axis when 𝑥 is two-fifths. So that’s roughly what the line 𝑦 equals minus five 𝑥 plus two looks like. Now let’s consider the line of the absolute value of 𝑥 plus five. Well that’s gonna be just the same as the line 𝑦 equals 𝑥 plus five unless 𝑦-coordinate would’ve been negative, in which case it becomes positive. So 𝑦 equals 𝑥 plus five has a slope of one. It cuts the 𝑦-axis at 𝑦 equals five and it cuts the 𝑥-axis at 𝑥 equals minus five. So that’s the line 𝑦 equals 𝑥 plus five. But remember, we said the absolute value of 𝑥 plus five wherever we’ve got negative 𝑦 coordinates they’re gonna get reflected to their sort of positive equivalents up here. So this part of the line to the left of 𝑥 equals negative five is actually gonna bounce up like this in that direction. Now those two lines cross over in just one place; and in the region where they cross over, we look at the intersection of this line 𝑦 equals negative five 𝑥 plus two with this line 𝑦 equals just 𝑥 plus five. And when we solve that equation, we get an 𝑥-coordinate of negative a half, which is exactly what we got when we just did it algebraically. So why did our algebraic approach give us this false answer of 𝑥 is equal to seven over four? Well what it did was it extended this part of the line here off in this direction; and then way off down here with this line and this line intersected, it calculated the corresponding 𝑥-coordinate up here of seven over four. Now that isn’t a valid solution, so you do have to be very very careful when using the algebraic approach. Using these graphs helps you really to visualise which are valid solutions and which are not valid solutions.
So let’s just summarise the main points of what we’ve learned. The first thing we have to consider is when the absolute value kicks in. And by that, we just mean when do we have to change the answers that we get from our original function. So for example, if we’re looking at the absolute value of 𝑥 plus five, if the value of 𝑥 plus five is greater than or equal to zero anyway, we can just take that value; it’s already positive. But if the value of 𝑥 plus five was less than zero, we’d have to take that negative answer and turn it into a positive answer. And the way that we do that is we take the negative of that negative answer. And remember, when 𝑥 plus five is less than zero, if I take away five from each side, that happens when 𝑥 is less than negative five. So if 𝑥 is less than negative five, I’m going to take the negative of 𝑥 plus five to turn it into a positive value; but if 𝑥 is greater than or equal to negative five, then I can just use the 𝑥 plus five value because it’s already greater than zero or equal to zero. Then we have to do two sets of calculations, one in this situation and one in this situation, and we end up with two different solution sets. And thirdly, you have to check if the solutions that you get match the criteria above them. So for example over here, we know that 𝑥 has got to be less than negative five. So if this comes up with an answer of 𝑥 equals four, then we know that’s not valid because it’s not less than negative five; but if it came up with an answer of negative ten, then that is in the valid region; it is a valid answer. And finally, consider using drawing a graph. The graphical approach is a visual approach which helps you to identify valid solutions, perhaps a bit more easily more obviously. Good luck then with solving equations involving absolute and nonabsolute values of an unknown.
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