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Question
\[\frac{2x + 3}{4} - 3 < \frac{x - 4}{3} - 2\]
Solution
\[\frac{2x + 3}{4} - 3 < \frac{x - 4}{3} - 2\]
\[ \Rightarrow \frac{2x + 3}{4} - \frac{x - 4}{3} < - 2 + 3 (\text{ Transposing } \frac{x - 4}{3} \text{ to the LHS }\hspace{0.167em} \text{ and - 3 to the RHS })\]
\[ \Rightarrow \frac{3\left( 2x + 3 \right) - 4\left( x - 4 \right)}{12} < 1\]
\[ \Rightarrow 3\left( 2x + 3 \right) - 4\left( x - 4 \right) < 12 (\text{ Multiplying both the sides by } 12)\]
\[ \Rightarrow 6x + 9 - 4x + 16 < 12\]
\[ \Rightarrow 2x + 25 < 12\]
\[ \Rightarrow 2x < 12 - 25\]
\[ \Rightarrow 2x < - 13\]
\[ \Rightarrow x < - \frac{13}{2} (\text{ Dividing both the sides by 2 })\]
\[\text{ Hence, the solution of the given inequation is } \left( - \infty , - \frac{13}{2} \right) .\]
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