Solve: $$ x \geq \frac{6}{x - 1} $$

I figure that I can't really multiply by both sides by $(x - 1)$ since I'm not sure if it will be positive or negative. So I multiplied by $(x - 1)^2$ to get:

$$ x^3 - 2x^2 -5x \geq -6 $$

but wasn't sure how to proceed to an answer.

  • $\begingroup$ In principle, what you did is correct, but it introduces extra solutions. You should multiply by $(x-1)$, move the 6 to the left-hand side, and factorize the resulting inequality. You will then have two cases. Hope this helps; it is better if you do it by yourself. $\endgroup$ – Manolito Pérez Mar 11 '12 at 7:10

You have two straightforward alternatives. One is to go back to the original inequality and split the problem into two cases. You know that $x$ can’t be $1$, so either $x>1$, or $x<1$. If $x>1$, then $x-1$ is positive, and you can multiply through by it to get $$x(x-1)\ge 6\quad\text{and}\quad x>1\;,$$ or $$x^2-x-6\ge 0\quad\text{and}\quad x>1\;.$$ The quadratic factors, so you have $(x+2)(x-3)\ge 0$, implying that $x\le -2$ or $x\ge 3$. But this case applies only if $x>1$ so in fact you just get $x\ge 3$.

If $x<1$, on the other hand, $x-1$ is negative, and you have $$(x+2)(x-3)\le 0\quad\text{and}\quad x<1\;,$$ which you can solve in the same general fashion to find the other part of the solution.

The other is to factor the cubic $x^3-2x^2-5x+6$. By the rational root test the only possible rational roots are $\pm1,\pm2,\pm3$, and $\pm6$. $1$ is quickly seen to be a root, so $x-1$ is a factor. Dividing it out leaves the quadratic $x^2-x-6$, which is easily factored. Then you merely solve the inequality $(x-1)(x+2)(x-3)\ge 0$, bearing in mind that $x$ cannot be $1$. However, the solution of this inequality involves looking at essentially the same cases as the first solution, so you’ve not really gained anything. I’d argue that the solution by cases is actually easier.

  • 1
    $\begingroup$ +1 for the elegant and simplified explanation. Thank you. $\endgroup$ – NoChance Mar 11 '12 at 8:39

You can also solve it by a case by case analysis instead.

First assume that $(x-1) > 0$. Then we can multiply out by $x-1$ without reversing the inequality.

Hence, we get $x^2-x \geq 6$ i.e. $x^2 - x -6 \geq 0$ i.e. $(x-3)(x+2) \geq 0$. This along with the assumption that $(x-1) > 0$, gives us that $x \geq 3$.

Similarly, assume $(x-1) < 0$ and proceed accordingly.

  • $\begingroup$ And don't forget you will be getting two answers. Will it be that both must be true ("and" logic) or that each one individually is true ("or" logic)? $\endgroup$ – David Lewis Mar 11 '12 at 7:15

$$ x \ge \frac{6}{x-1} $$ Move everything to one side: $$ x-\frac{6}{x-1} \ge 0 $$ The common denominator is $x-1$, so we have: $$ \frac{x(x-1)}{x-1} - \frac{6}{x-1} \ge 0 $$ Simplify: $$ \frac{x^2-x-6}{x-1}\ge 0 $$ Factor: $$ \frac{(x-3)(x+2)}{x-1} \ge 0 $$ This changes signs at $-2$, $1$, and $3$. It is positive if $x>3$, negative if $x$ is between $3$ and $1$, positive if $x$ is between $1$ and $-2$, and negative if $x<-2$. It is zero when the numerator is $0$, i.e. when $x=-2$ or $x=3$.

So the solution is $$ x\ge3\text{ or } -2\le x<1. $$


$x \ge \dfrac{6}{x-1}$

$x - \dfrac{6}{x-1} \ge 0$

$\dfrac{x^2 - x - 6}{x - 1} \ge 0$

$\dfrac{(x + 2)(x - 3)}{x - 1} \ge 0$

(x-1): ---------------------------o+++++++++++++++++++++++
(x+2): ---------●+++++++++++++++++++++++++++++++++++++++++
(x-3): ---------------------------------------●+++++++++++

 soln: ---------●+++++++++++++++++o-----------●+++++++++++

         -3    -2    -1     0     1     2     3     4

solution set $= [-2, 1) \cup $[3, \infty)


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