Absolute Value InequalitiesDate: 06/23/2004 at 11:43:53 From: Cristyn Subject: which inequality sign is "and" and which sign is "or" I have to solve |x| + 3 < 5 and I don't remember if it's "and" or "or". I think it's "or" so I think that I set the problem up like this: 5 < x + 3 > -5 but I don't think that's right. Date: 06/23/2004 at 12:23:53 From: Doctor Peterson Subject: Re: which inequality sign is "and" and which sign is "or" Hi, Cristyn. There is no "and" or "or" implicit in an inequality like this. It comes out of the meaning of the inequality as a whole. That means you don't have to memorize which is "or" and which is "and"; instead, think about what it is saying, and you will discover the right way to state the answer. One handy way to think about absolute value inequalities is to use the idea that the distance between two numbers on a number line can be calculated as the absolute value of the difference of the two numbers. Thus, the distance between -2 and 5 on the number line is |(-2) - (5)| = |-7| = 7 or |(5) - (-2)| = |7| = 7 Applying that thinking, we can generalize that |x - a| is the distance between x and a on the number line. So, if you see the inequality |x - 3| < 5 the meaning is that the distance from x to 3 is less than 5; this means in turn that x is between 3 - 5 and 3 + 5: -5 < x - 3 < 5 -2 < x < 8 The fact that x is BETWEEN -2 and 8, and that we write that as a single statement with two inequalities, can be expressed using "and": x > -2 AND x < 8 But nothing in the original inequality said that directly. On the other hand, if you have |x - 3| > 5 it means that x is MORE than 5 units away from 3, which can be in EITHER direction, either 5 units more than 3 or 5 units less than 3: x - 3 > 5 ==> x > 8 OR x - 3 < -5 ==> x < -2 So this inequality can be expressed using "or". You can see these two solutions visually by drawing the number line: <--|---|---|---|---|---|---|---|---|---|---|---|---|--> -3 -2 -1 0 1 2 3 4 5 6 7 8 9 For |x - 3| > 5, find all points whose distance from 3 is greater than 5: <*****O | O******> <--|---|---|---|---|---|---|---|---|---|---|---|---|--> -3 -2 -1 0 1 2 3 4 5 6 7 8 9 Thus, x < -2 OR x > 8 For |x - 3| < 5, find all points whose distance from 3 is less than 5: O*******************|*******************O <--|---|---|---|---|---|---|---|---|---|---|---|---|--> -3 -2 -1 0 1 2 3 4 5 6 7 8 9 Thus, x > -2 AND x < 8, which can be written -2 < x < 8. The basic ideas on which these solutions are built are that |x| < a means -a < x < a and |x| > a means x > a or x < -a Now, a teacher might oversimplify the process by saying that if an absolute value is less than something, you will end up with "and", and if it is greater, you get "or". But that is a very risky "rule" to try to apply to any but the simplest problems, which take the form above. It's essential to think through the meaning of the inequality carefully. Your problem does not quite look like my examples, and its answer is a little different: |x| + 3 < 5 Here you want to FIRST get the absolute value alone on the left, by subtracting 3 from both sides. Only then can you do the kind of thinking I demonstrated. The most orderly way to attack an absolute value inequality is to break it down into cases, based on the definition of the absolute value: |x| = x if x >= 0 = -x if x < 0 So for my first example, |x - 3| < 5 We have two cases: if x - 3 >= 0, it means (x - 3) < 5 if x - 3 < 0, it means -(x - 3) < 5 This can be stated using BOTH "and" and "or": either x - 3 >= 0 AND x - 3 < 5 OR x - 3 < 0 AND 3 - x < 5 This can be simplified to either x >= 3 AND x < 8 OR x < 3 AND x > -2 Thus the solution set consists of two parts (two intervals on the number line), but since those parts are contiguous, meeting at x = 0, they can be combined into x < 8 AND x > -2 If the inequality is reversed (giving my second example), then we end up with two separate intervals that have to be combined using "or"; the "and" disappears because any number greater than 8 is also greater than 3: either x >= 3 AND x > 8 OR x < 3 AND x < -2 either x > 8 OR x < -2 Try doing the same kind of thinking with your problem, and see what you get. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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