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# Video: Solving Equations with Absolute Values of Unknowns on Both Sides of the Equation

Tim Burnham

This video explains how to solve equations that involve finding the absolute value of a variable or the simple linear expression involving that variable with such absolute value expressions on both sides of the equation (e.g., |𝑥 + 3| = |2𝑥 − 6|).

15:38

### Video Transcript

In this video, we’re gonna be solving equations with absolute values of unknowns on both sides of the equation. Now remember taking the absolute value means just taking the positive version of the number. So for example, if you evaluate an expression and it comes up with a negative number, you take the negative of that negative number to turn it into a positive number.

So we’re gonna have a good look at this question that you can see in front of you now, and we’re gonna try and solve it in two different ways. Firstly, we’re gonna solve algebraically and we’re gonna consider all the different possible values and different ranges of 𝑥-values that we can consider this for. And then we’re gonna look at the graphs of it and solve it graphically as well, and hopefully the two will tally.

So the question is solve the absolute value of 𝑥 plus three is equal to the absolute value of two 𝑥 minus six. So we’ve got to find all of the different 𝑥-values that satisfy this equation, taking into account the absolute values that are going on.

So let’s look at the left-hand side first. There are two things that can happen: either the contents of the absolute value, the 𝑥 plus three, could naturally give us a positive number — in which case, well we’ve got a positive number; we’ve got nothing more to do — or 𝑥 plus three could yield a negative number, and we need to take the negative of that negative number to turn it into a positive number, so two different situations.

So either 𝑥 plus three is greater than or equal to zero or 𝑥 plus three is less than zero. Now technically 𝑥 plus three should be equal to zero on that right-hand channel there because if we take the negative of zero is still zero. But let’s just leave it like this for the moment. Well if 𝑥 plus three is greater than or equal to zero, it means the 𝑥 is greater than or equal to negative three.

And if 𝑥 plus three is less than zero, it means the 𝑥 must be less than negative three. So if 𝑥 plus three is positive, as we said we’ve got nothing to do. We can just use the expression 𝑥 plus three to evaluate the value of the absolute value of 𝑥 plus three. But if the 𝑥 plus three gave us a negative answer, as I said before we need to take the negative of that.

And now we can do the same sort of analysis over on the right-hand side. So if we plug the value of 𝑥 into two 𝑥 minus six and it gives us an answer which is zero or bigger, positive, then that’s fine. We don’t need to do anything; the absolute value of a positive number is just the same positive number.

So that two 𝑥 minus six expression is absolutely fine, and now that happens when rearranging that inequality up here, two 𝑥 is greater than or equal to six or 𝑥 is greater than or equal to three. But when 𝑥 is less than three, the value of two 𝑥 minus six is gonna generate a negative answer. So we- to take the absolute value, we need to take the negative of that.

Now to evaluate each side of that equation here, we need to know what value of 𝑥 is going in there. And depending on what value of 𝑥 we use, we may have to use one of these different expressions down here on each side of the equation. So let’s summarise that all in a sort of number line format.

Now the critical values of 𝑥 were negative three and positive three so I’ve marked those onto the number line. And let’s look at the left-hand side first. So if 𝑥 is greater than or equal to negative three, so if it’s greater than or equal to negative three, then we’re gonna be using the expression 𝑥 plus three for the left-hand side.

So that’s here and here. But if 𝑥 is less than negative three, on this side, then we’re gonna be using the negative of that. Now for the right-hand side, if 𝑥 is greater than or equal to three, then we’re gonna use two 𝑥 minus six. But if it’s less than three, then we’re gonna use the negative of that.

So for practical purposes, the exact equation that we’re gonna use will depend very much on the value of 𝑥. So if 𝑥 is less than three, for the left-hand side we’re going to use this and for the right hand side we’re going to use this; if 𝑥 is between positive three and negative three, the left-hand side is gonna be that and the right hand side is gonna be that; and if 𝑥 is greater than tree, the left-hand side is gonna be that and the right hand side is gonna be that.

So let’s make some space up to the- at the top of the screen and go ahead and solve each three of- each of those three equations So on the left first, negative 𝑥 plus three is equal to negative of two 𝑥 minus six. So using the distributive property to put those minus signs, remember that’s minus one times each of the parentheses there, we get negative 𝑥 minus three is equal to negative two 𝑥 plus six.

Then just adding two 𝑥 to each side of that equation gives us 𝑥 minus three is equal to six and then adding three to each side gives us 𝑥 equals nine.

Well 𝑥 equals nine sounds like a fine answer. But remember, we’re in the region where 𝑥 has to be less than negative three. So 𝑥 equals nine isn’t in the region 𝑥 is less than three. So that’s actually not a proper answer; we can’t use that answer.

So when you’re doing this sort of question, you always have to check that the answers you’ll get match the region that you’re in at the time you doing the algebra. So let’s move on to the next region between 𝑥 is negative three and 𝑥 is three. And in this region, the left-hand side is 𝑥 plus three and the right-hand side is negative of two 𝑥 minus six. So just evaluating the right-hand side, so distributing that minus sign through the parentheses, then we’ll add two 𝑥 to both sides then subtract three from both sides and finally divide by three.

And this gives us a solution of 𝑥 is equal to one; and yes one is between negative three and positive three, so it’s in the region. So that looks like it’s gonna be a valid solution. Now moving into the final region where 𝑥 is greater than three, the left-hand side in this case will be 𝑥 plus three and the right-hand side will be two 𝑥 minus six.

So this time I’m gonna subtract 𝑥 from each side of the equation to eliminate 𝑥 from the left-hand side. And now I’m gonna add six to both sides to leave 𝑥 on its own on the right-hand side. And this is giving us a solution of 𝑥 equals nine. Well 𝑥 has gotta be greater than three to be in this region, so 𝑥 equals nine is a valid answer in this region.

So after all of that working out, we’ve concluded that there are two possible values of 𝑥 which work in that equation, and that is when 𝑥 is equal to one and when 𝑥 is equal to nine. Okay, let’s go on and look at this graphically and see how this looks in graph shape.

So again, let’s consider the left-hand side first. So we’re gonna try to look- to sketch the graph of 𝑦 equals the absolute value of 𝑥 plus three. Now before we do that, let’s just sketch the graph of 𝑦 equals 𝑥 plus three. And just by looking at that, it’s a straight line with a slope of one and it cuts the 𝑦-axis at three. So when the 𝑦-coordinate is zero, it cuts the 𝑥-axis. So we can see that it cuts the 𝑥-axis at negative three, it cuts the 𝑦-axis at three, and it’s a straight line with slope one. So we can now sketch that.

So that’s the line 𝑦 equals 𝑥 plus three. But of course, that’s not actually the line we’re looking for. We’re looking for 𝑦 equals the absolute value of 𝑥 plus three. Well all this means is that any point on the line which is below the 𝑥-axis which has a 𝑦-coordinate of less than zero, we’re gonna take the positive version of that. So any point down here, we’re going to take the negative of those 𝑦-coordinates and reflect them in the 𝑥-axis to their corresponding positive 𝑦-coordinates up here. So we’re gonna generate a line which goes sort of up in this direction. Where can I draw it properly? There we go.

So the line 𝑦 equals the absolute value of 𝑥 plus three is gonna look like this blue line here, gonna go there, and it’s gonna to go up here like this. And to the right of this point here, negative three, we’re gonna be using the equation 𝑦 equals 𝑥 plus three. And to the left of that, over here, the equation of that line, well we’ve taken the negative of those 𝑦-coordinates so we’re gonna be using 𝑦 is equal to the negative of 𝑥 plus three.

Okay, let’s go through a similar exercise for the right-hand side. We want to sketch 𝑦 is equal to the absolute value of two 𝑥 minus six. But first of all, we’re just gonna sketch 𝑦 equals two 𝑥 minus six and then we’ll turn that into the absolute value afterwards.

And that is gonna be a straight line with a slope of two, so it’s gonna be steeper than the line we’ve just drawn and it’s gonna cut the 𝑦-axis at negative six and it’s gonna cut the 𝑥-axis when the 𝑦-coordinate is zero so putting 𝑦 equals zero into that equation, zero equals two 𝑥 minus six. So adding six to both sides, six is equal to two 𝑥. Dividing both sides by two, we’ve got three is equal to 𝑥. So it cuts the 𝑥-axis when 𝑥 is equal to three and it cuts the 𝑦-axis at negative six, And it’s a bit- and it’s steeper than the previous line that we’ve just drawn.

So to turn that into 𝑦 is equal to the absolute value of two 𝑥 minus six wherever the 𝑦-coordinate is negative, we need to reflect in the 𝑥-axis to its corresponding positive version. So negative six here is gonna go up to positive six here. And if we draw that in, we’re gonna get the green lines here. And to the right of 𝑥 equals three, we’re just using the equation 𝑦 is equal to two 𝑥 minus six to generate those 𝑦-coordinates. But to the left of that, we’re using the negative of those 𝑦-coordinates so we’re using 𝑦 is equal to the negative of two 𝑥 minus six.

Okay, so let’s merge together those two graphs onto the same pair of axes. And when we do that, we can see that they intersect in two places here and here. So what we’d really like to know is what is the 𝑥-coordinate here and what is the 𝑥-coordinate here. So solving the equation just means when is the 𝑦-coordinate at this line equal to the 𝑦-coordinate at this line and that obviously is the points at which they intersect. So that’s why we’re interested in these two 𝑥-coordinates here. So let’s call this solution 𝑎, and we’ll call this solution 𝑏. And solution 𝑎 is gonna be the intersect of this part of the blue line, which would have an equation of 𝑦 equals 𝑥 plus three, and this part of the green line, which would have an equation of 𝑦 equals negative of two 𝑥 minus six.

Now I’m not gonna go through the algebra in any great detail because we’ve already done this calculation once, so let’s put it all down here. And we’ve got an answer of 𝑥 equals one. So this 𝑥-coordinate here is one. And to work out the solution for 𝑏, we’ve got this part of the blue line up here, which has this equation, and we’ve got this part of the green line here, which has this equation, so we’re gonna equate those two things 𝑥 plus three equals two 𝑥 minus six, which as we said before gives a solution of 𝑥 is equal to nine.

So the 𝑥-coordinate for solution 𝑏 is nine. So I think there’s an advantage to doing the graphs because you can really see the lines intersecting. You can see whether they’re valid solutions or not. So just before we move on and do the summary, let’s take a look at that original solution, the spurious solution of 𝑥 equals nine that we’ve got when we were analysing 𝑥 is less than three in our algebraic solution

So when we’re in this region over to the left, the equation that we were using for the blue line was 𝑦 is the negative of 𝑥 plus three and the equation that we were using for the green line is 𝑦 is the negative of two 𝑥 minus six. So essentially because this green line here is steeper than this blue line here, if we follow them off in this direction, they are never gonna meet. They’re sort of diverging. That’s moving more sort of that way and the blue one is moving more sort of that way, so they’re never gonna meet in that direction. But if you extend those both back in the other direction over here, then down here somewhere — that’s difficult — maybe just off the screen here, but down here somewhere they’re going to cross over when 𝑥 equals nine. So when we got that 𝑥 equals nine solution for the left-hand region, that’s what it was finding; it was finding this point over here, which of course isn’t in the region 𝑥 is less than negative three.

Now it just so happened in this particular question with these particular numbers that we ended up getting an answer of 𝑥 equals later on. That doesn’t always happen, so you’ve gotta be very careful. These extra erroneous solutions that crop up, you’ve gotta be careful. Are they in the region that we’re actually looking for or are they just coming out as a result of extending the lines in the other direction and finding where they cross somewhere else on the graph?

So let’s go through a quick summary then of the stages that we went through. First we had to consider the positive and the negative versions of each expression. So for example on the left-hand side, if 𝑥 plus three was greater than or equal to zero, then in terms of absolute value we wouldn’t have any work to do. We could just use the expression 𝑥 plus three. But if 𝑥 plus three gave us a negative value, then we’d have to take the negative of that negative value to turn it into a positive value because that’s what the absolute value function is doing.

And of course in our case, we had to do that for both sides of the equation. And then we have to consider the regions in which we would need to use each expression. So that means analysing like this to work out whether 𝑥 is gonna be greater than or less than various different values. And then we have to go through region by region solving the equations, and then we need to check the solutions that we’ve get for each we can to make sure that they’re valid for that region.

And in our case, when we went for the first region, equating negative of 𝑥 plus three with the negative of two 𝑥 minus six gave us an answer of 𝑥 equals nine. But of course nine isn’t less than negative three so that was an invalid solution in that region. The second gave us an answer of 𝑥 equals one, which was valid for that region, and the third gave us an answer of 𝑥 equals nine which was valid for that region.

And lastly doing the graph gave a bit more insight into why we had this spurious solution of 𝑥 equals in this particular region. So having done all this practice, hopefully you’ll now be able to solve equations with absolute values of unknowns on both sides of the equation.