What should $a$ be for the system of inequalities to have one solution?

$$(1) \frac{3}{x-a}\geq 1 (2)\left | x-2a-2 \right | \leq 1$$

For them to have one solution, the solution must be an endpoint of an interval. (1) gives me $$x\leq a+3$$, but I don't know how to proceed from here. Any hint would be appreciated.


The first it's $$\frac{3}{x-a}-1\geq0$$ or $$\frac{3+a-x}{x-a}\geq0$$ or $$a<x\leq 3+a.$$ The second it's $$2a+1\leq x\leq3+2a.$$ Now, we see that an unique possibility it's $$3+a=2a+1.$$ Can you end it now?

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  • $\begingroup$ Can you elaborate on the whole argument? $\endgroup$ – Aleksandr Aug 3 '19 at 4:02
  • $\begingroup$ @Aleksandr Ask please a concrete question. $\endgroup$ – Michael Rozenberg Aug 3 '19 at 4:03
  • $\begingroup$ Where does the first line come from? $\endgroup$ – Aleksandr Aug 3 '19 at 4:06
  • 1
    $\begingroup$ @Aleksandr The first it's the intervals method. By the definition of $|\cdot|$ we have $|x|\leq a$ it's $-a\leq x\leq a$. Thus, the second it's $-1\leq x-2a-2\leq1$ or $2a+1\leq x\leq2a+3.$ $\endgroup$ – Michael Rozenberg Aug 3 '19 at 4:08
  • $\begingroup$ Why can't $x-a$ be smaller than $0$? $\endgroup$ – Aleksandr Aug 3 '19 at 4:31

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