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Rational Inequality


Date: 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/   
    
Associated Topics:
High School Basic Algebra

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