Solving a Rational InequalityDate: 11/15/2004 at 19:33:39 From: Christina Subject: Special Rules for Inequalities? Interesting inequality... I thought I knew how to solve inequalities, but I found this one to be interesting and challenging: (x-1)/(x+3) < 3 When I solve this inequality, I get the answer to be x > -5. However, if I substitute in -5 for x, the inequality doesn't check. However, -2 works, and all the numbers greater than it work too. Also, -6, -7, and all the numbers smaller than those work too. Why is this happening? Is there a systematic way to figure out this inequality? Date: 11/15/2004 at 22:46:54 From: Doctor Peterson Subject: Re: Special Rules for Inequalities? Interesting inequality... Hi, Christina. I wish you'd shown how you got your answer, so I could see where you made a mistake, but I think I know what you did. A common mistake people make in solving an inequality like this is to multiply both sides by x+3: x - 1 < 3(x + 3) You would then get x - 1 < 3x + 9 -1 < 2x + 9 -10 < 2x -5 < x That looks good, until you remember that multiplying an inequality by a negative number reverses the direction of the inequality. Since you don't know ahead of time whether x+3 is positive or negative, you don't know whether the result should be what I wrote above, or x - 1 > 3(x + 3) It is possible to treat the problem in two parts, first considering the case where x+3 is positive, and then separately considering when x+3 is negative, and then combining them. But there's an easier way. If we can make an inequality with 0 on one side, we will have a neat way to solve it. So let's subtract 3 from both sides: x - 1 ----- - 3 < 0 x + 3 Now simplify the left side: x - 1 3x + 9 ----- - ------ < 0 x + 3 x + 3 -2x - 10 -------- < 0 x + 3 Let's multiply by -1/2 to clear out the common factor of -2 in the numerator: x + 5 ----- > 0 x + 3 (I had to reverse the inequality there, since I multiplied by a negative number.) Now, when will this be positive? The quotient of two numbers is positive if both are positive, or if both are negative. So either x+5 and x+3 are both positive: x > -5 and x > -3, which is true whenever x > -3 or x+5 and x+3 are both negative: x < -5 and x < -3, which is true whenever x < -5 So the solution is really the set of all x that are EITHER greater than -3, OR less than -5. That fits what you found, that -4 doesn't work, but anything from -2 up or from -6 down does. You'd also find that -2.99 and -5.01 work! If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 11/16/2004 at 13:19:17 From: Christina Subject: Thank you (Special Rules for Inequalities? Interesting inequality...) Thanks so much for your assistance! I've referred to the Dr. Math website many times before, and found everything to be explained clearly. This is also evident in the response to the question I submitted. Thanks again! Everything makes so much sense now :) |
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