In every absolute-value equation we've seen so far, there has been one absolute-value expression, and it could be "isolated"; that is, we could get it by itself on one side of the "equals" sign. What if there are two absolute-value expressions? Can we use the same method? Yes, but only if there are exactly just the two absolute values, so that we can "isolate" each of them, one on either side of the equation.
Let's consider the following equation:
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Either the arguments of the two absolute values are both "plus" (so nothing changes when I drop the bars), or else they're both "minus" (so they both get a "minus", which can be divided off, so nothing changes), or else they have opposite signs (in which case one of them changes sign when I drop the bars, and the other doesn't). So I can deal with all three cases by dropping the bars on either side, and considering a "plus" and a "minus" case for the right-hand side. (I could have done the "plus" and the "minus" on the left-hand side, but I'm a creature of habit.) I'll do the "minus" case first:
x + 2 = –(3 – x)
x + 2 = –3 + x
2 = –3
Clearly, this case has no solution. So now I'll try the "plus" case:
x + 2 = 3 – x
2x = 1
x = ½
This case does work. So my answer is:
x = ½
(If you're not sure of that solution, graph the two associated absolute-value functions, and confirm that the two lines intersect at x = –½. No graphing calculator handy? Try here.)
Well, the equation above solved nicely. But it had exactly two absolute-value expressions, and nothing else, so the equation could accommodate the isolation of each of the two absolute values. In other words, that equation was the one and only "nice" case of having two or more absolute values.
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But what happens if there are three (or more) absolute-value expressions, or if there are two such expressions and they also have loose numbers or variables with them, so it is simply not possible to isolate the expressions to get the absolute values by themselves on one side (or both sides) of the equation? To solve such an equation, we will need a different solution method. To see why, let's consider the following example:
This equation looks similar to what we've seen before; it doesn't look particularly much more complicated than the others.
But it is a very different case, so I'm going to discuss it a bit, before showing the necessary solution method.
If I split the original equation above into two cases for the argument on the left-hand side, move the 1 from the right-hand side to the the left, and split each of the results into another two cases, I'll get four solutions: –3, –2, 0, and ½.
For instance, just working down the "plus" branches, and starting on the left-hand side of the equation, my work would look like this:
| x – 3 | = | 3x + 2 | – 1
x – 3 = | 3x + 2 | – 1
x – 3 = | 3x + 2 | – 1
x – 2 = | 3x + 2 |
x – 2 = 3x + 2
–4 = 2x
–2 = x
But of the four solutions listed at the beginning (namely, –3, –2, 0, and ½), only two are actually correct. As we can see in the graph below, the solution I just "proved" above is very clearly wrong; the two lines do not in fact intersect at x = –2:
I got too many answers from using the previous method. That method does not work for equations of this particular type. The previous method allowed us to avoid some very nasty algebra, but for an equation with two (or more) un-nested absolute values, and where there is also a loose number (or some other variable, etc), we have no choice but to get technical.
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Why? Because every time we consider a "plus" or a "minus" case when taking the bars off an absolute value, we're making an assumption about what we're doing; in particular, we're making an implicit assumption about the portion(s) of the number line for which the argument is one sign or another. But when we try to make assumptions about two separate arguments (and thus two probably-different sets of intervals) at the same time (as one must, in the case of the current equation), then we might be finding "solutions" in intervals that don't actually even exist. That's why I got a completely wrong answer in my working above. The previous method works only if we can "isolate" the absolute value (that is, if we can get the absolute value all by itself), with one entity on the other side of the "equals" sign. But we can't do that with the current equation.
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To get around this failure of the regular solution method, we must make explicit what previously had been implicit; we must explicitly consider the different intervals created by the breakpoints of the absolute values' arguments. (A "breakpoint" is where the argument changes sign, or where, on a graph of the associated absolute-value function, we get that "V" shape.) And then we must consider each interval separately. Returning to that equation from above, here's how the new method works:
The first absolute-value expression, in the left-hand side of the equation, is positive when the argument is positive. I'll solve to find that interval:
x – 3 > 0
x > 3
The argument of this absolute value will be negative before the breakpoint (at x = 3) and positive after. (It's equal to zero at the breakpoint.)
The second absolute-value expression, in the right-hand side of the equation, is positive for:
3x + 2 > 0
3x > –2
The argument of this absolute value will be negative before the breakpoint, and positive after. But this argument's breakpoint is at , which does not match the breakpoint for the previous argument.
These computations give me the breakpoints of each of the two absolute-value expressions. These breakpoints are the endpoints of my intervals, and are at . These endpoints split up the number line into the following intervals:
On the first interval, , I'm below the left-most breakpoint, so I know that the arguments for each of the absolute values is negative. This means that I'll have to change the sign on each of them when I drop the absolute-value bars. Then I can solve:
| x – 3 | = | 3x + 2 | – 1
▬(x – 3) = ▬(3x + 2) – 1
–x + 3 = –3x – 2 – 1
2x = –6
x = –3
Since this solution value fits within the current interval, , this solution is valid.
On the second interval, , the argument for the absolute value on the left-hand side of the equation is still negative (because I'm below x = 3), so I'll have to flip the sign on that expression when I drop the bars. Since the other argument is positive on this interval (because I'm above ), I can just drop the bars and proceed.
| x – 3 | = | 3x + 2 | – 1
▬(x – 3) = 3x + 2 – 1
–x + 3 = 3x + 1
2 = 4x
½ = x
Because this value is within the current interval, , this solution is valid.
On the third and final interval, (3, +∞), each of the two arguments is positive, so I can drop the bars to solve:
| x – 3 | = | 3x + 2 | – 1
x – 3 = 3x + 2 – 1
–4 = 2x
–2 = x
And here I see why I need to be careful about my intervals. This solution value does not fit within the targetted interval of (3, +∞). So this value cannot actually be a valid solution to the original equation.
But the other two values were valid, so my final answer is:
x = –3, ½
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You should expect to see nested absolute-value equations, and equations where the arguments are other than simply linear (such as the quadratic example that we did on the previous page). If your book doesn't cover absolute-value equations where the absolute values cannot be isolated (and doesn't explain the method of finding intervals and then solving on each of the intervals), then you may not need this page's method until you reach trigonometry or calculus.
However, your instructor in that later math class may assume that your algebra class did cover this other solution method. So keep this other method in the back of your head, for in case you need it later. You can always return here and refresh, when and if it becomes necessary.
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