Solve inequality, please: (3 – x)/(x + 5) >= 1

amanf 2021-08-14 Answered
Solve inequality, please: \(\displaystyle\frac{{{3}–{x}}}{{{x}+{5}}}\ge{1}\)

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Usamah Prosser
Answered 2021-08-15 Author has 17306 answers
\(\displaystyle\frac{{{3}–{x}}}{{{x}+{5}}}\ge{1}\)
\(\displaystyle\frac{{{3}–{x}}}{{{x}+{5}}}–{1}\ge{0}\)
\(\displaystyle{\left({3}–{x}\right)}–{\left({x}+{5}\right)}\frac{{)}}{{{x}+{5}}}\ge{0}\)
\(\displaystyle\frac{{-{2}{\left({x}+{1}\right)}}}{{{x}+{5}}}\ge{0}\)
\(\displaystyle\frac{{{x}+{1}}}{{{x}+{5}}}\ge{0}\), so
\(\displaystyle{f{{\left({x}\right)}}}=\frac{{{x}+{1}}}{{{x}+{5}}}\)
\(\displaystyle{x}=-{5},{x}=-{1}\)
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