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5.14

## Precalculus

Skip to 0 minutes and 10 seconds Inequalities in which feature absolute values are very common in mathematics. Let’s see how to deal with them. As in the case of equations, the method of cases, paying due attention to the domains of each case will always work in principle. You’ll be able to remove the absolute value signs and solve each case, and at the end, find the overall solution set. Now we have seen examples though that indicate the prudence that we must use. If you don’t take careful account of the domain as you’re doing this, false solutions could be generated. Let’s illustrate that with the following inequality, which we wish to solve. And as you can see, it features two absolute values.

Skip to 0 minutes and 57 seconds Because there are two absolute values to be removed, so to speak, that will result in four different cases, depending upon whether the expressions and the absolute values are positive or negative– four possibilities. The first possibility is when x minus 1 is positive and x plus 3 is positive. That amounts to requiring that x be greater than 1. So we carefully note this. Now we can remove the two absolute values from the inequality, and it becomes a simple linear inequality. It leads to the solution set x greater than minus 7/8.

Skip to 1 minute and 35 seconds Now we can’t keep all those x’s, though– we only keep the ones that are consistent with the restrictions that defined as case 1 that we’re looking at– namely x greater than 1. So that leads to this solution set S1 of this first case, the interval 1 infinity. We now look at case 2. That’s where the first expression in absolute values is positive and the second negative. That turns out to require that x be simultaneously greater than 1 and less than minus 3, not possible. Therefore, the solution set from case 2 is empty. Case 3 turns out to require that x be less than minus 3, and the resulting linear inequality gives x greater than 7 over 6.

Skip to 2 minutes and 25 seconds These two are contradictory, therefore, no solutions from case 3. And finally case 4. Case 4 requires that x lie in the interval minus 3, 1. And the resulting linear inequality says that x is greater than minus 1/2. So you take the intersection of these two conditions, and you find the interval S4 minus 1/2 to 1. You now have treated all the possible cases carefully. You take the union of the four resulting solution sets, and that will be your overall solution set. In this case, we see that we get minus 1/2 to infinity. Now notice how this example was solved. We remained calm, we were thorough, we found our inner surgeon.

Skip to 3 minutes and 12 seconds We treated each case carefully, we dissected them, and then we found the answer at the end. In order to better understand absolute value inequalities, we’re now going to look at two special types that occur frequently and try and understand the distinction between them. The first type will be absolute f less than g. And the second type will be absolute f greater than g. So we start with absolute value of f less than g. It will be helpful to recall something that we saw in week 1 about absolute values, namely to say that the absolute value of a real number a is less than r, is equivalent to saying that a is in the interval minus r, r.

Skip to 3 minutes and 59 seconds To put that another way, it’s equivalent to two inequalities– a is greater than minus r, and a is less than r. Of course here r is a positive number. If we apply this observation from the past to our inequality, absolute f less than g, we see that it is equivalent to simultaneously two inequalities in which there are no absolute values– f greater than minus g and f less than g. Consider now the other type of inequality– the reverse– absolute f greater than g of x. In order to better understand that one, we recall another fact about absolute values.

Skip to 4 minutes and 40 seconds To say that the absolute value of a is greater than r is the same as saying that either a is less than minus r or a is greater than r. Notice the either or. Now this is always true this last conclusion when r is negative. Applying this observation to our inequality– absolute f greater than g– we see that it is equivalent to require that either f is less than minus g or f is greater than g. So you notice the distinction between the word “and” in the first type of inequality and the word “or” in this second type of inequality. This is the distinction that we wish to focus on. Look at the pattern.

Skip to 5 minutes and 26 seconds When you have absolute f less than g, that requires simultaneously that two conditions be satisfied– intersection. When you have the inequality absolute f greater than g, that’s an either/or situation. Either one is true or the other– it’s a sort of union. To put this yet another way, absolute f being less than something is a strong condition. Fewer points will satisfy it. It leads to an intersection. Whereas absolute f greater than something is a weaker type of condition. More points will satisfy it– the union of two conditions. Let’s apply this reasoning to a simple inequality of the type we often see– absolute value of 2x minus 4 greater than 3.

Skip to 6 minutes and 16 seconds According to what we’ve said, that is equivalent to asking either that 2x minus 4 be greater than 3 or that 2x minus 4 be less than minus 3. These are both simple linear inequalities that we easily solve, and we find that either x is greater than 7/2 or x is less than 1/2. And therefore, we have our solution set. The union of these two open intervals. So this way of thinking about absolute value inequalities is an aid to memory in many cases.

# Inequalities with absolute values

Analysis by cases, two special inequalities