Question 1104052
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The other tutor's solution is wrong....<br>
The given inequality is<br>
{{{2 <= -2/(x-1)}}}<br>
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.<br>
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.<br>
Case 1: If (x-1) is positive (that is, x>1), then
{{{2(x-1) <= -2}}}
{{{2x-2 <= -2}}}
{{{2x <= 0}}}
{{{x <= 0}}}<br>
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.<br>
So this first case gives us NO solutions to the inequality.<br>
Case 2: If (x-1) is negative (that is, x<1), then
{{{2(x-1) >= -2}}}  [we multiplied by a negative; the direction of the inequality changes]
{{{2x-2 >= -2}}}
{{{2x >= 0}}}
{{{x >= 0}}}<br>
Since case 2 only considers values of x less than 1, the solution set for this case is 0 <= x < 1.<br>
So the complete solution set is 0 <= x < 1; or, in interval notation, [0,1).<br>
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.<br>
{{{2 <= -2/(x-1)}}}
{{{2 + 2/(x-1) <= 0}}}
{{{(2(x-1))/(x-1) + 2/(x-1) <= 0}}}
{{{(2x-2+2)/(x-1) <= 0}}}
{{{2x/(x-1) <= 0}}}<br>
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.<br>
And so the solution set -- as before, of course -- is [0,1)