Answer:
Given, \[\frac{1}{x+1}+\frac{3}{5x+1}=\frac{5}{x+4}\] \[\Rightarrow \] \[\frac{1}{x+1}-\frac{5}{x+4}=\frac{-3}{5x+1}\] \[\Rightarrow \] \[\frac{(x+4)-5(x+1)}{(x+1)(x+4)}=\frac{-3}{5x+1}\] \[\Rightarrow \] \[\frac{x+4-5x-5}{{{x}^{2}}+5x+4}=\frac{-3}{5x+1}\] \[\Rightarrow \] \[\frac{(-4x-1)}{{{x}^{2}}+5x+4}=\frac{-3}{5x+1}\] \[\Rightarrow \] \[(4x+1)(5x+1)=3({{x}^{2}}+5x+4)\] \[\Rightarrow \] \[20{{x}^{2}}+4x+5x+1=3{{x}^{2}}+15x+12\] \[\Rightarrow \] \[17{{x}^{2}}-6x-11=0\] \[\Rightarrow \] \[17{{x}^{2}}-17x+11x-11=0\] \[\Rightarrow \] \[17x(x-1)+11(x-1)=0\] \[\Rightarrow \] \[(x-1)(17x+11)=0\] \[\Rightarrow \] Either \[x=1\] or \[x=\frac{-11}{17}\]
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