Question 1144798: Email - aurora.0723@outlook.com
Question - when trying to determine the x value of an inequality, why do we sometime flip the inequality even though it is not being divided the zero. For example,
Solve [1/(x-1)] < 5
Case 1 = {x-1<0 x<1
{ (1/x-1)<5 (1/x-1)>5 x<(6/5)
Then you do the second case. However, in the first case, why does (1/x-1)<5 turn to (1/x-1)>5? Why does the inequality flip?
Thanks for the help
Answer by greenestamps(13334)   (Show Source):
You can put this solution on YOUR website!
Solution strategy: multiply both sides by (x-1) to clear fractions.
Case 1: x-1 < 0 (i.e., x < 1)
If x < 1, (x-1) is negative, so multiplying by (x-1) reverses the direction of the inequality.
So if x < 1, an "additional" restriction is that x < 6/5. But ALL values of x that are less than 1 are less than 6/5. So all numbers less than 1 are part of the solution set.
Case 2: x-1 > 0 (i.e., x > 1)
If x > 1, (x-1) is positive, so multiplying by (x-1) does not change the direction of the inequality.
So if x > 1, the additional restriction is that x > 6/5. So the other part of the solution set is all the number greater than 6/5 = 1.2.
Final solution set: (-infinity, 1) union (1.2, infinity)
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