<s>[INST] Can someone explain how to solve linear inequality? 
Can someone explain how to solve this linear inequality?
 [/INST] 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 $>0$
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