[Algebra]
Problem: Determine all values of $x$ for which $0<\frac{x^{2}-11}{x+1}<7$.

Solution: We consider two cases: $x>-1$ (that is, $x+1>0$ ) and $x<-1$ (that is, $x+1<0$ ). Note that $x \neq-1$.

Case 1: $x>-1$

We take the given inequality $0<\frac{x^{2}-11}{x+1}<7$ and multiply through by $x+1$, which is positive, to obtain $0<x^{2}-11<7 x+7$.

Thus, $x^{2}-11>0$ and $x^{2}-11<7 x+7$.

From the first, we obtain $x^{2}>11$ and so $x>\sqrt{11}$ or $x<-\sqrt{11}$.

Since $x>-1$, then $x>\sqrt{11}$. (Note that $-\sqrt{11}<-1$.)

From the second, we obtain $x^{2}-7 x-18<0$ or $(x-9)(x+2)<0$. Thus, $-2<x<9$. (Since $y=x^{2}-7 x-18$ represents a parabola opening upwards, its $y$-values are negative between its $x$-intercepts.)

Since $x>-1$ and $-2<x<9$, then $-1<x<9$.

Since $x>\sqrt{11}$ and $-1<x<9$, then the solution in this case is $\sqrt{11}<x<9$.

Case 2: $x<-1$

We take the given inequality $0<\frac{x^{2}-11}{x+1}<7$ and multiply through by $x+1$, which is negative, to obtain $0>x^{2}-11>7 x+7$.

Thus, $x^{2}-11<0$ and $x^{2}-11>7 x+7$.

From the first, we obtain $x^{2}<11$ and so $-\sqrt{11}<x<\sqrt{11}$.

Since $x<-1$ and $-\sqrt{11}<x<\sqrt{11}$, then $-\sqrt{11}<x<-1$.

From the second, we obtain $x^{2}-7 x-18>0$ or $(x-9)(x+2)>0$. Thus, $x<-2$ or $x>9$. (Since $y=x^{2}-7 x-18$ represents a parabola opening upwards, its $y$-values are positive outside its $x$-intercepts.)

Since $x<-1$, we obtain $x<-2$.

Since $-\sqrt{11}<x<-1$ and $x<-2$, then the solution in this case is $-\sqrt{11}<x<-2$.

In summary, the values of $x$ for which $0<\frac{x^{2}-11}{x+1}<7$ those $x$ with $-\sqrt{11}<x<-2$ and those $x$ with $\sqrt{11}<x<9$.

Final answer: $(-\sqrt{11},-2)\cup (\sqrt{11},9)$