jmc

Moving all the terms to the left-hand side, we have $$\frac{1}{x+1}+\frac{3}{x+7}-\frac{2}{3}\ge 0.$$To solve this inequality, we find a common denominator: $$\frac{3(x+7)+3\cdot 3(x+1)-2(x+1)(x+7)}{3(x+1)(x+7)}\ge 0,$$which simplifies to $$-\frac{2(x+4)(x-2)}{3(x+1)(x+7)}\ge 0.$$Therefore, we want the values of $x$ such that $$f(x)=\frac{(x+4)(x-2)}{(x+1)(x+7)}\le 0.$$To do this, we make the following sign table: \begin{array}{c|cccc|c} &$x+4$ &$x-2$ &$x+1$ &$x+7$ &$f(x)$ \\ \hline$x<-7$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$+$\\ [.1cm]$-7<x<-4$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$+%%DISP_0%%amp;$-$\\ [.1cm]$-4<x<-1$ &$+%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]$-1<x<2$ &$+%%DISP_0%%amp;$-%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$-$\\ [.1cm]$x>2$ &$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]\end{array}Because the inequality $f(x)\le 0$ is nonstrict, we must also include the values of $x$ such that $f(x)=0,$ which are $x=-4$ and $x=2.$ Putting it all together, the solutions to the inequality are $$x\in \boxed{(-7,-4]\cup (-1,2]}.$$