\(2x+13=0\\
2x=-13\\
x=-6.5\)
The function is linear where \(a\ \textgreater \ 0\), so it's negative for \(x\ \textless \ -6.5\) and positive for \(x\ \textgreater \ -6.5\)
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\(D:x\in\mathbb{R}\setminus\{-1,1,0\}\\\\
\dfrac{a+3}{a^2-1} - \dfrac{1}{a^2+a}=\\
\dfrac{a+3}{(a-1)(a+1)} - \dfrac{1}{a(a+1)}=\\
\dfrac{a(a+3)}{a(a-1)(a+1)} - \dfrac{a-1}{a(a-1)(a+1)}=\\
\dfrac{a^2+3a-a+1}{a(a-1)(a+1)}=\\
\dfrac{a^2+2a+1}{a(a-1)(a+1)}=\\
\dfrac{(a+1)^2}{a(a-1)(a+1)}=\\
\dfrac{a+1}{a(a-1)}\)
for \(a\in(-\infty,-1)\) it's negative
for \(a\in(-1,0)\) it's positive
for \(a\in(0,1)\) it's negative
for \(a\in(1,\infty)\) it's positive
