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Q. Solution of $ \left|1+ \frac{3}{x}\right| > 2$ is

Linear Inequalities

Solution:

$ \left|1+ \frac{3}{x}\right| > 2$
Case $I$ : $1+\frac{3}{x} > 2 \Rightarrow \frac{3}{x} > 1$ (Clearly $x > 0$)
$\Rightarrow \quad3 > x$ or $x < 3$.
Case $II$ : $1 + \frac{3}{x} < - 2 \Rightarrow \frac{3}{x} < - 3$ (Clearly $x < 0$)
$\Rightarrow \quad 3 < - 3x \Rightarrow -1 < x$ or $x > -1$
Hence, either $0 < x < 3$ or $- 1< x < 0$.