question_answer

Solve for x: \[\frac{1}{x+1}+\frac{3}{5x+1}=\frac{5}{x+4},x\ne -1,-\frac{1}{5},-4\]

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 \[x=\frac{-11}{17}\]