Where is the mistake while solving $\frac{|x-3|}{x^2-5x+6}\ge2$

Question

I have split it into $2$: a)$$-\Biggl(\frac{x-3}{x^2-5x+6}\Biggr)\le2,$$ $$\frac{2x^2-9x+9}{(x-2)(x-3)}\ge0.$$ b) $$\frac{x-3}{x^2-5x+6}\ge2,$$ $$\frac{2x^2-11x+15}{(x-2)(x-3)}\le0.$$ The answers for the inequality containing the absolute value are in the interval $x\in [2;2.5)$ or

But it is $[3/2;2)$ in the book.

I have redone the problem several times, nonetheless I can't discover my mistake.

Answer 1

Hint: for $x \ge 3$ we have $\frac{|x-3|}{x^2-5x+6}=\frac{1}{x-2}$

and

for $x < 3$ and $x \ne 2$ we have $\frac{|x-3|}{x^2-5x+6}=-\frac{1}{x-2}.$

Answer 2

Your first case is wrong it should be $$\frac{1}{x-2} \le -2$$ Or $$\frac{-1}{x-2} \ge 2$$

by definition the $|x-3|$ is $-(x-3)$ for $ (x-3) \le 0$

What I think your doing wrong in the first case is changing the inequality sign without dividing by negative 1.

Answer 3

After the help from the community and research I finally found my mistake. Firstly, the split points must be found: $x_1\ne2, \ x_2\ne3,$ which forms the intervals $(-\infty;2)\cup(2;3)\cup(3;\infty).$

Case 1: $$x\lt2 \Rightarrow \frac{-(x-3)}{(x-2)(x-3)}\ge2 \Rightarrow \frac{3-2x}{x-4}\ge0.$$ The interval that satisfies the last inequality is $$x\in[3/2;2),$$ which also satisfies $x\lt2.$ So this is the answer of the modulus inequality above.

Case 2: $$2\lt x\lt3.$$ This again gives us $$\frac{3-2x}{x-2}\ge2 \rightarrow x\in[3/2;2).$$ Notwithstanding, it isn't the true answer because it doesn't satisfy $2\lt x\lt3.$

Case 3: $$x\gt3 \Rightarrow \frac{x-3}{(x-2)(x-3)}\ge2 \Rightarrow \frac{5-2x}{x-2}\ge0 \rightarrow x\in(2;5/2],$$ which, obviously, again isn't the correct answer.

The final answer for the inequality involving the absolute value sign is $x\in[3/2;2).$

P.S: Cancelling out common terms (in the fractions in this example) are crucial in solving this problem as you wouldn't get the right answer, otherwisely.