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posted by  Anonymous on 10/6/2008 3:06:51 AM  |  status: Closed  

Solving inequalities

Course Textbook Chapter Problem
Precalculus N/A N/A N/A
Question Details:
Can you show me how to solve this?

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posted by Handyman on 10/6/2008 3:33:20 AM  |  status: Live
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Anonymous's comment:
"The first sentence of the response is not true and at the bottom it states that every number greater than 2 is correct, but the solution given is (1,2)."
Response Details:
First, make sure that the expression is a single rational expression on the left and zero on the right.
-which is true for this expression.



Determine the changing points.


and


Use the changing points to break up the number line into intervals.



Select three test values and substitute each one into the inequality.
- A number less than 1: x=0

*(this means that every number left of 1 will be false for this expression)

-A number between 1 and 2:  x=1.5


*(This means that every number between 1 and 2 is true for this expression)

-A number larger than 2:  x=3


*(This means that every number larger than 2 is correct for this expression)

So for this expression only the interval between 1 and 2 result in a true statement yielding the interval notation.

(1,2)
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posted by Anonymous on 10/6/2008 3:45:32 AM  |  status: Live
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Response Details:
Does it matter that it's 1 on the right and not 0?
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posted by Whackpro on 10/6/2008 5:42:38 AM  |  status: Live
Asker's Rating: Helpful   
Response Details:
 
 
 
 
 
  (x - 2)/(x - 1)
 
 = (x - 1 - 1)/(x - 1)
 
 = 1 - 1/(x-1)
 
  now, for (x-2)/(x-1) < 1,
 
  1 - 1/(x-1) < 1
 
  -1/(x-1) < 0
 
  => 1/(x - 1) > 0
 
  x - 1 > 0
 
  =>   1 < x < ∞
 
 
 
  now, as for your follow up question,
 
  if,
 
  (x-2)/(x-1) < 0
 
  x - 2 < 0 and  x - 1 > 0,
 
=>  1 < x < 2
 
 
 or,
 
  x - 2 > 0  and x - 1 < 0
 
  x > 2 and x < 1, which is impossible.
 
 
so, the solution in this case is  1 < x < 2
 
  
did my answer help you ? if it was wrong or you didn't quite get it kindly send me a private message giving the details.
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posted by Mea on 10/6/2008 6:16:43 AM  |  status: Live
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Anonymous's comment:
"This was most helpful. But when I checked the intervals, only x>1 satisfied the inequality."
Response Details:
Hey.... i am going to try and solve this inequality.. i will try to go about it step-by-step so that you understand what i am doing.
 
Ok.. the first thing you have to do is to get a 0 in the place of the 1 on the right-hand-side. This you can do by subtracting 1 from both the sides.
Your ineqaulity becomes .
 
Now, you have to make the left-hand side as a single fraction.
 
The denominator will becoem zero if x = 1. So, you get one boundary point on a number line. This boundary point, 1, divides the number line into two intervals .
 
Take a test point in each interval and check it in .
 
Your answer is the interval to which the point satisfying the inequality belongs to. I found that both intervals satisfy the inequality... you can check and confirm.
 
Hope i helped....
 
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