Rational InequalityDate: 2/15/96 at 19:43:46 From: Ruth Subject: Algebra 1 We are going over old tests to get ready for a math contest. One of the problems requires us to find the sum of all values of x such that 6x is an integer and (x-4)/(x-2) >= 3 This is on an Algebra 1 test. Students understand that as the absolute value of x gets really big the -4 and -2 become unimportant. They also know they can't divide by zero, but know nothing about limits! Any idea how to explain this on their level? Thanks....Ruth Date: 8/16/96 at 15:20:57 From: Doctor James Subject: Re: Algebra 1 Well, first let's start off with 6*x being an integer. That means that there must be an integer n such that x = n/6. So let's plug that into our equation, (x-4)/(x-2) >= 3. We get (n/6 - 4)/(n/6 - 2) >= 3 multiply the left side by 6/6 = 1, and we get (n - 24)/(n - 12) >= 3 To solve this, we need to rewrite it as a comparison of a rational expression to zero: n - 24 ------ - 3 >= 0 n - 12 n - 24 3n - 36 ------ - ------- >= 0 n - 12 n - 12 -2n + 12 -------- >= 0 n - 12 Dividing both sides by -2, this becomes n - 6 ------ <= 0 n - 12 The left side will be negative when either n-6 OR n-12 is negative, but not both. The inequality will be true when n - 6 = 0, but not when n - 12 = 0. So the solution is 6 <= n < 12 That is, the solutions are n = 6, 7, 8, 9, 10, 11, and therefore x = 6/6, 7/6, 8/6, 9/6, 10/6, 11/6, or x = 1, 7/6, 4/3, 3/2, 5/3, 11/6 and the answer to the original question is the sum of these fractions. Hope this helps! -Doctor James, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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