Easy inequality going wrong 
Question to solve: $$\frac{3}{x+1} + \frac{7}{x+2} \leq \frac{6}{x-1}$$

My method:
$$\implies \frac{10x + 13}{(x+1)(x+2)} - \frac{6}{x-1} \leq 0$$
$$\implies \frac{4x^2 -15x-25}{(x-1)(x+1)(x+2)} \leq 0$$
$$\implies (x-5)(4x+5)(x-1)(x+1)(x+2) \leq 0$$
Using method of intervals, I get:
For $x\leq-2, -5/4\leq x \leq -1$ and $-1\leq x\leq1$, x is less than or equal to zero.
But, this range i got is incorrect. Where did I go wrong then?

Your only mistake is right at the end. Your polynomial changes sign at the points $-2,-\frac54, -1, 1, 5$, meaning that it is:


*

*Negative on $(-\infty, -2)$

*Positive on $(-2,-\frac54)$

*Negative on $(\frac54, -1)$

*Positive on $(-1,1)$

*Negative on $(1, 5)$

*Positive on $(5,\infty)$


As for the precise values in which the sign changes, you must be careful, because $$(x-5)(4x+5)(x-1)(x+1)(x+2)$$ may be defined on the whole real line, but the fraction $$\frac{(4x+5)(x-5)}{(x+1)(x-1)(x+2)}$$
is NOT defined on $x=1,-1,-2$, so the inequality cannot hold for these three numbers. It does hold for $x=-\frac54$ and $x=5$, however, so your whole answer is
$$x<-2\text{ or }-\frac54 \leq x < -1\text{ or } 1<x\leq 5$$
