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in Linear Equations by (15.2k points)
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Find the solution set of the in equation \(\frac{{\text{x}} + 1}{{\text{x}} + 2}\) <1

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by (15.7k points)
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We have to find values of x for which \(\frac{-1}{x + 2} < 0 \) that is \(\frac{-1}{x + 2}\) is negative

The numerator of \(\frac{-1}{x + 2}\)  is -1 which is negative hence for \(\frac{-1}{x + 2}\) to be negative x + 2 must be positive (otherwise if x + 2 is negative then negative upon negative will be positive)

That is x + 2 should be greater than 0

⇒ x + 2 > 0 

⇒ x > -2

Hence x should be greater than -2 for \(\frac{-1}{x + 2}\) < 0

x > -2 means x can take values from -2 to ∞ hence x ∈ (-2, ∞)

Hence the solution set for \(\frac{x+1}{x+ 2} < 0\) is (-2, ∞)

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