Can someone explain how to solve linear inequality? 
Can someone explain how to solve this linear inequality?
 A: Multiply the both sides by $x^2-9=(x-3)(x+3)$. 
But note that you need to have two cases as $x^2-9\gt 0$ or $x^2-9\lt 0$.
(Case 1)
If $x^2-9\gt 0\iff x\lt -3\ \text{or}\ x\gt 3$, then 
$$\frac{6(x^2-9)}{x^2-9}+\frac{(13-x)(x+3)(x-3)}{x+3}\le\frac{3(x+3)(x-3)}{x+3}-\frac{2(x+3)(x-3)}{-(x-3)}$$
i.e.
$$6+(13-x)(x-3)\le 3(x-3)+2(x+3)\iff x\le 5\ \text{or}\ x\ge 6.$$
Hence, in this case, we have
$$x\lt -3\ \ \text{or}\ \ 3\lt x\le 5\ \ \text{or}\ \ x\ge 6.$$
(Case 2)
If $x^2-9\lt 0\iff -3\lt x\lt 3$, then 
$$\frac{6(x^2-9)}{x^2-9}+\frac{(13-x)(x+3)(x-3)}{x+3}\color{red}{\ge}\frac{3(x+3)(x-3)}{x+3}-\frac{2(x+3)(x-3)}{-(x-3)}$$
i.e.
$$6+(13-x)(x-3)\ge 3(x-3)+2(x+3)\iff 5\le x\le 6.$$
Hence, in this case, there is no such $x\in\mathbb R$.
Hence, the answer is
$$x\lt -3\ \ \text{or}\ \ 3\lt x\le 5\ \ \text{or}\ \ x\ge 6.$$
A: We have $$\frac{6+(13-x)(x-3)-3(x-3)-2(x+3)}{x^2-9}\le0$$
$$\iff-\frac{(x-6)(x-5)}{(x-3)(x+3)}\le0$$
For the equality $x=6$  or $5$
For the inequality, 
$$-\frac{(x-6)(x-5)}{(x-3)(x+3)}<0\iff\frac{(x-6)(x-5)}{(x-3)(x+3)}>0$$
$$\iff[x-(-3)](x-3)(x-5)(x-6)>0 (\text{multiplying either sides by }$(x-3)(x+3)^2>0)$$
Check for $-\infty<x<-3,-3<x<3,3<x<5,5<x<6,6<x<\infty$
As the product needs to positive, we need even number of multiplicand $>0$
which is evidently true if $x<-3$ or $x>6$
For $-3<x<3,$ only one is $>0$
For $3<x<5,$ two are $>0$
For $5<x<6,$ three are $>0$
