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Topic: Help with inequality
Replies: 1   Last Post: Dec 21, 2012 9:30 AM

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Posts: 132
Registered: 11/27/12
Re: Help with inequality
Posted: Dec 21, 2012 9:30 AM
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I take that when you said you "attempted was to solve it the same as though it were an equation", you mean that you multiplied both sides by x- 1. The problem with that is that multiplying by a NEGATIVE number will reverse the direction of the inequality: 1< 3 but -1> -3. Here, we want to multiply by x- 1 but because we don't know x, we don't know whether x- 1 is positive or negative.

So one way to do this is to do it as two cases:
Assume x- 1> 0. Then multiplying both sides by x- 1 does NOT change the inequality: 2>= -(x- 1)= -x+ 1. Adding x to both sides and subtracting 2 from both sides gives x>= -1. But, of course, that is true for all positive numbers and this was assuming x- 1> 0 which means x> 1.

Assume x- 1< 0. Then multiplying by x- 1 changes the direction of the inequality: 2<= -(x- 1)= -x+ 1. Adding x to both sides and subtracting 2 from both sides now gives x<= -1. Of course, any x less than -1 also satisfies x- 1< 0 so we have x<= -1.

Putting those together, the inequality is satisfied for all x< -1 and all x> 1.

Another way to do this is to solve the associated equation: 2/(x- 1)= -1 is (for all x other than 1) the same as 2= -x+ 1 or x= -1. The point of that is that we can change from "<" to ">", or vice versa, only where we have "=" or the function is not continuous. We have seen that we have equality where x= -1 and a fraction is not continuous where the denominator is 0, that is, at x= 1. Those two points Are the only places where the inequality can change. Check one point in each interval, x< -1, -1< x< 1, x> 1 to determine whether or not the inequality is satisfied by all points in that interval.

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