Solve the following


$\frac{2 x+3}{4}-3<\frac{x-4}{3}-2$


$\frac{2 x+3}{4}-3<\frac{x-4}{3}-2$

$\Rightarrow \frac{2 x+3}{4}-\frac{x-4}{3}<-2+3 \quad$ (Transposing $\frac{x-4}{3}$ to the LHS and $-3$ to the RHS)

$\Rightarrow \frac{3(2 \mathrm{x}+3)-4(\mathrm{x}-4)}{12}<1$

$\Rightarrow 3(2 x+3)-4(x-4)<12 \quad$ (Multiplying both the sides by 12$)$

$\Rightarrow 6 x+9-4 x+16<12$

$\Rightarrow 2 x+25<12$

$\Rightarrow 2 x<12-25$

$\Rightarrow 2 x<-13$

$\Rightarrow x<-\frac{13}{2} \quad$ (Dividing both the sides by 2 )

Hence, the solution of the given inequation is $\left(-\infty,-\frac{13}{2}\right)$.

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