用代数方法求满足下列不等式的 $x$ 的取值集合：

绝对值不等式要拆成两个同时成立的条件，分别解分式不等式后取交集。通分时先整理分子，再根据各因子的正负判断，不直接乘以符号未定的分母。注意 $x=-2$ 不在定义域内。

For $x\ne-2$, let

The modulus inequality is equivalent to both

First solve $F(x)<7-x$. On putting $F(x)-(7-x)$ over the denominator $x+2$, let its numerator be $N(x)$. Then

Hence the first inequality becomes

The numerator is positive for $x<-1$ or $x>2$, and negative for $-1<x<2$. The denominator is negative for $x<-2$ and positive for $x>-2$. The fraction is negative when the signs differ, giving

Next solve $F(x)>-(7-x)$. On putting $F(x)+(7-x)$ over the denominator $x+2$, its numerator is

Thus, dividing out the positive constant $8$,

The numerator and denominator have the same sign when

Taking the intersection of the two solution sets, the required values are