Question:
x >= 6/(x-1)
I tried to solve it, but I keep getting fractional answers (which I know are incorrect). If someone could provide a worked solution, that'd be great. Cheers.
Hi Glitch,
Here is one way. We have
EDIT: The following contains errors. Please see other posts below. I grayed everything out, except that the LaTeX still renders in black.
$\displaystyle x \ge \frac{6}{x-1}$
$\displaystyle x(x-1) \ge 6 $
We can solve by finding out where
$\displaystyle x(x-1) = 6$
So we find that
$\displaystyle x^2-x-6 = 0$
$\displaystyle (x-3)(x+2) = 0$
So we have three intervals, $\displaystyle (-\infty,-2), (-2,3), (3,\infty)$ and we can try out values from each interval.
$\displaystyle x = 0 \Rightarrow x(x-1) = 0 \not \ge 6$ so the interval $\displaystyle (-2,3)$ does not satisfy the inequality.
$\displaystyle x = -3 \Rightarrow -3(-3-1) = 12 \ge 6$ so $\displaystyle (-\infty,-2)$ satisfies the inequality.
$\displaystyle x = 4 \Rightarrow 4(4-1) = 12 \ge 6$ so $\displaystyle (3, \infty)$ satisfies the inequality.
So the answer is: $\displaystyle x \le -2$ OR $\displaystyle x \ge 3$.
You're right! I did something illegal, which led to a wrong answer.
I would rectify it by saying off the bat that we can solve
$\displaystyle x \ge \frac{6}{x-1}$
by finding where
$\displaystyle x = \frac{6}{x-1}$
and then testing the intervals as before.
EDIT: Disregard the grayed out part, which happens to contain more than one silly error. See below post.
So the first two intervals fail and only the third interval passes, giving the answer: $\displaystyle x \ge 3$.
Sorry about that!
Ah, I'm sorry to have to post again due to carelessness. Maybe I shouldn't be doing this at 1:43 AM.
We have to consider x = 1 as a critical point too. So we have the four intervals $\displaystyle (-\infty,-2), (-2,1), (1,3), (3,\infty)$
The second and fourth intervals pass, leading to the answer: $\displaystyle -2 \le x < 1$ OR $\displaystyle x \ge 3$.
Should be mistake-free now.. finally..
Here comes a slightly different approach:
$\displaystyle x \geq \frac6{x-1}~\implies~x - \frac6{x-1}\geq 0~\implies~\frac{x^2-x-6}{x-1}\geq 0$
A quotient is positive (or zero) that means greater than zero (or zero) if the signs of the numerator and the denominator are equal:
$\displaystyle x^2-x-6\geq 0 \wedge x-1 > 0~\vee~x^2-x-6\leq 0 \wedge x-1 < 0$
After moving some stuff around this "chain" of inequalities simplifies to:
$\displaystyle x \geq 3 ~\vee~ -2\leq x < 1$
By the way, if you're allowed to use a graphing calculator, then you might find you have a better understanding of this problem by looking at the graphs of $\displaystyle y=x$ and $\displaystyle y=\frac{6}{x-1}$ and seeing where the first graph is above or intersects with the second graph.
Will do. I don't have a graphing calculator, but Wolfram|Alpha—Computational Knowledge Engine works just as well.
@undefined the earboth's method should be use in solving this kind of inequalities.
Your method is not good at all, because it is not correct.
For example.
NOT correct.
(x-1)/x>1
(x-1)>x
-1>0 which leads to contradiction.
CORRECT:
(x-1)/x -1 >0
(x-1-x)/x >0
-1/x >0
-1<0
so x<0