Question:
$\frac{x-1}{3}+4<\frac{x-5}{5}-2$
Solution:
$\frac{x-1}{3}+4<\frac{x-5}{5}-2$
$\Rightarrow \frac{x-1}{3}-\frac{x-5}{5}<-2-4 \quad\left[\right.$ Transposing 4 to the RHS and $\frac{\mathrm{x}-5}{5}$ to the LHS $]$
$\Rightarrow \frac{5(x-1)-3(x-5)}{15}<-6$
$\Rightarrow \frac{5 x-5-3 x+15}{15}<-6$
$\Rightarrow \frac{2 x+10}{15}<-6$
$\Rightarrow 2 x+10<-90$
$\Rightarrow 2 x<-90-10$ [Transposing 10 to the RHS]
$\Rightarrow 2 x<-100$
$\Rightarrow x<-\frac{100}{2}$ [Dividing both the sides by 2 ]
$\Rightarrow x<-50$
Hence, the solution of the given inequation is $(-\infty,-50)$.