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Question
\[\frac{7x - 5}{8x + 3} > 4\]
Solution
We have,
\[\frac{7x - 5}{8x + 3} > 4\]
\[ \Rightarrow \frac{7x - 5}{8x + 3} - 4 > 0\]
\[ \Rightarrow \frac{7x - 5 - 4\left( 8x + 3 \right)}{8x + 3} > 0\]
\[ \Rightarrow \frac{7x - 5 - 32x - 12}{8x + 3} > 0\]
\[ \Rightarrow \frac{- 25x - 17}{8x + 3} > 0\]
\[ \Rightarrow \frac{25x + 17}{8x + 3} < 0 (\text{ Multiplying by - 1 to make the coefficient of x in the LHS hspace{0.167em} positive })\]
\[x \in \left( \frac{- 17}{25}, \frac{- 3}{8} \right)\]
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