The presence of an absolute value in the equation makes for an interesting case. Let’s see now how we deal with such equations. There will be more than one main type. The first of them is absolute f equals h. How do we solve such an equation? Well, we do it by considering two cases. The first case will be within the domain where f has positive values. So we define D1 that way. You can see that when x is in that domain, then absolute f is the same as f. We can just remove the absolute value signs, and our original equation becomes f equals h. In principle, we can solve that equation. We find its solution set. We call it S1.
The second case is to work within the domain where f is negative. Then the equation becomes minus f equals h. And we find a solution set, call it S2. So the solution set of our original equation is the union of these two solution sets from each of the two cases. Let’s take a look at this approach for the simple example, absolute x equals 2x minus 1. The natural domain of this equation is the whole real line. Case one. If x is positive, then the solution, the equation rather, is equivalent to x equals 2x minus 1, or x equals 1, which is in fact greater or equal 0 as required by case one.
So we retain this possibility, and the solution set, S1, is the singleton 1. The second case, x less than or equal 0, leads to x equals 1/3. However, 1/3 is not less than or equal 0, which was the underlying assumption of case two. So we have to reject this x, and therefore the solution set S2 of this case is the empty set. Conclusion, our overall solution set is the singleton 1. Now, beware it is a common error to think that the equation absolute f equals g is the same as f equals plus or minus g. That’s simply not so.
The second equation we’ll introduce in general solutions where g of x is negative, that can’t possibly work in the first equation. Now let’s look at the second type of equation where absolute values intervene. There can be a sum of absolute values or a difference. Let’s treat the case of a sum. The other one is quite similar. To handle this kind of equation, we’re going to have to consider four cases. Why four? Well, because in general and independently f can be positive or negative and g can be positive or negative. Hence four cases in all. In the first case, we define the domain to be, D1 to be those points where f and g are both positive.
We simply remove the absolute value signs, and we have the equation f plus g equals h. We solve it in the domain D1. We find its solution set S2. Similarly, the second case f positive g negative, that introduces a minus g in the equation rather than a plus g. We find the set S2 of solutions, S3 and S4 for each case. And then the overall answer is the union of these four solution sets, all the solutions coming from the various cases. Let’s illustrate this with the following example. How many cases will there be in solving this equation? Well, there are two absolute values, so there will be four cases in all.
The first case will be when x minus 1 is positive and x plus 3 is positive. That amounts to saying x is greater or equal to 1. When we remove all the absolute values, we get a simple linear equation whose solution is minus 7/8. However, minus 7/8 is inconsistent with x greater or equal 1, so we reject it. Solution set from case one is empty. Now case two, that’s x minus 1 positive and x plus 3 negative. That amounts to saying that x is greater than 1 and less than minus 3. Well, of course, that’s impossible, so immediately we see the solution set S2 is empty.
Similarly, when we look at the third case, we’ll get an empty solution set because the x that arises is not within the domain for the third case. Now, the fourth case, our last hope to find a solution, it amounts to asking that x minus 1 be negative and x plus 3 positive. That is, x has to be between minus 3 and 1. The equation becomes a linear equation, which gives us minus 1/2. Is minus 1/2 consistent with that interval? Yes, it is. We can keep this solution. So S4 is the singleton set minus 1/2. The overall solution set, then, the union of these four is minus 1/2.
Now, do you remember there was a cowboy approach to solving equations with radicals? We didn’t worry about the domain. We just went full speed ahead, thinking that we’ll just check whatever we get at the end to see what we should keep. Well, we can sometimes solve absolute value equations that way too. We may wish to simply consider all the possible cases that arise, not even think about the domains, and we’ll check later. The fact is, when absolute values are concerned, this may or may not work, unlike the case when we only had radicals. Let’s look at a case where it does work.
This is exactly the same example we saw a moment ago, but now I’m going to solve it in cowboy mode. I write out the four equations that could arise from removing the absolute values. I solve each of these four linear equations. They give me in each case a value of x. Are these values solutions? I don’t know. I’m going to go back and check. I take minus 7/8. It’s not a solution. How do I know? I plugged it into the original equation, and it didn’t work. Similarly, the second x, the third, not solutions. The fourth one is. Therefore, the solution set is minus 1/2. Now, that was fast, wasn’t it? The cowboy got there before the surgeon.
The fact is, though, that there are cases when the cowboy will never get there, and here’s an example of how this approach may not work. If we want to solve this equation and we do it without thinking about the domain, the case one that we consider is usually the one where we just remove the absolute value signs. So x minus 3 is x minus 2 minus 1. When you simplify that and it comes out to x equals x, well, that’s always true. It’s true for all x. So later on, when we have to go back and check the candidates we have accumulated to see if they work in the original equation, we’re back where we started.
We need to find the x’s that work in the original equation. So no progress whatever has been done. Now, a surgeon could solve this because if you do the solution in a way that identifies the domain, you’ll get the correct solution set 3 infinity. So there are an infinite number of x’s that satisfy the equation. We now look at the last type of equation. It’s absolute f equals absolute g. You might think it considers four cases because there are two absolute value signs, but, no, it’s easy to see in this case that the equation is equivalent to f equals plus or minus g in this case because of the extra absolute value. Here’s an example.
To solve this equation, I write 2x minus 5 equals plus or minus 1 minus x. That leads very simply to two possible values of x. That gives you our solution set. And, of course, we do check the answers because we’re not androids.