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Question
\[\frac{x - 1}{x + 3} > 2\]
Solution
\[\text{ We have }, \frac{x - 1}{x + 3} > 2\]
\[ \Rightarrow \frac{x - 1}{x + 3} - 2 > 0\]
\[ \Rightarrow \frac{x - 1 - 2\left( x + 3 \right)}{x + 3} > 0\]
\[ \Rightarrow \frac{x - 1 - 2x - 6}{x + 3} > 0\]
\[ \Rightarrow \frac{- x - 7}{x + 3} > 0 \]
\[ \Rightarrow \frac{x + 7}{x + 3} < 0\]
∴ \[x \in \left( - 7, - 3 \right)\]
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