\(\displaystyle\frac{{{3}–{x}}}{{{x}+{5}}}\ge{1}\)

\(\displaystyle\frac{{{3}–{x}}}{{{x}+{5}}}–{1}\ge{0}\)

\(\displaystyle{\left({3}–{x}\right)}–{\left({x}+{5}\right)}\frac{{)}}{{{x}+{5}}}\ge{0}\)

\(\displaystyle\frac{{-{2}{\left({x}+{1}\right)}}}{{{x}+{5}}}\ge{0}\)

\(\displaystyle\frac{{{x}+{1}}}{{{x}+{5}}}\ge{0}\), so

\(\displaystyle{f{{\left({x}\right)}}}=\frac{{{x}+{1}}}{{{x}+{5}}}\)

\(\displaystyle{x}=-{5},{x}=-{1}\)

\(\displaystyle\frac{{{3}–{x}}}{{{x}+{5}}}–{1}\ge{0}\)

\(\displaystyle{\left({3}–{x}\right)}–{\left({x}+{5}\right)}\frac{{)}}{{{x}+{5}}}\ge{0}\)

\(\displaystyle\frac{{-{2}{\left({x}+{1}\right)}}}{{{x}+{5}}}\ge{0}\)

\(\displaystyle\frac{{{x}+{1}}}{{{x}+{5}}}\ge{0}\), so

\(\displaystyle{f{{\left({x}\right)}}}=\frac{{{x}+{1}}}{{{x}+{5}}}\)

\(\displaystyle{x}=-{5},{x}=-{1}\)