[tex]2x+13=0\\
2x=-13\\
x=-6.5[/tex]
The function is linear where [tex]a\ \textgreater \ 0[/tex], so it's negative for [tex]x\ \textless \ -6.5[/tex] and positive for [tex]x\ \textgreater \ -6.5[/tex]
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[tex]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)}
[/tex]
for [tex]a\in(-\infty,-1)[/tex] it's negative
for [tex]a\in(-1,0)[/tex] it's positive
for [tex]a\in(0,1)[/tex] it's negative
for [tex]a\in(1,\infty)[/tex] it's positive
