SOLUTION: Hi! I got stuck with this problem...how should I go about solving this? 2(is less than or equal to) {{{-2/(x-1)}}} Thank you!
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Question 1104052: Hi! I got stuck with this problem...how should I go about solving this? 2(is less than or equal to) Thank you!
Found 3 solutions by Alan3354, greenestamps, ikleyn:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Multiply by x-1
Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!
The other tutor's solution is wrong....
The given inequality is
The other tutor simply multiplied both sides of the inequality by (x-1). That won't give you the full solution set; and it might give you a solution set which is NOT right. The problem with that method is that (x-1) might be negative; if it is, the direction of the inequality changes.
You can solve the problem by multiplying both sides of the inequality by (x-1); but you need to look at two cases -- when (x-1) is positive and when it is negative.
Case 1: If (x-1) is positive (that is, x>1), then
That is the "solution" that the other tutor gave; but it is not a solution at all. For this case 1, we are only considering solutions with x>1; "x <= 0" does not give any solutions with x>1.
So this first case gives us NO solutions to the inequality.
Case 2: If (x-1) is negative (that is, x<1), then
[we multiplied by a negative; the direction of the inequality changes]
Since case 2 only considers values of x less than 1, the solution set for this case is 0 <= x < 1.
So the complete solution set is 0 <= x < 1; or, in interval notation, [0,1).
As an alternative to the above method of solution where we consider two different cases, we can modify the given inequality so that one side of the inequality is 0. Following is a solution to the problem using that strategy.
For x values less than 0, both numerator and denominator are negative, making the fraction positive; so x<0 is NOT part of the solution set.
For x values between 0 and 1, the numerator is positive and the denominator is negative, making the fraction negative; so x between 0 and 1 IS part of the solution set. Note the endpoint of that interval x=0 is part of the solution set, because it makes the numerator 0; the endpoint x=1 is NOT part of the solution set, because it makes the denominator 0.
For x values greater than 1, both numerator and denominator are positive, making the fraction positive; so x>1 is NOT part of the solution set.
And so the solution set -- as before, of course -- is [0,1)
Answer by ikleyn(52765) (Show Source): You can put this solution on YOUR website!
.
Many of similar problems are solved with detailed explanations in the lesson
- Solving inequalities for rational functions with non-zero right side
in this site.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lesson is the part of this online textbook under the topic "Inequalities".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.
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