question_answer

Solve for x:

\[\frac{3}{x+1}+\frac{4}{x-1}=\frac{29}{4x-1};x\ne 1,-1,\frac{1}{4}\]

Answer:

We have,

\[\frac{3}{x+1}+\frac{4}{x-1}=\frac{29}{4x-1};x\ne 1,-1,\frac{1}{4}\]

\[\Rightarrow \,3(x-1)(4x-1)+4(x+1)(4x-1)=29(x+1)(x-1)\]

\[\Rightarrow \,3(4{{x}^{2}}-4x-x+1)+4(4{{x}^{2}}+4x-x-1)=29({{x}^{2}}-1)\]

\[\Rightarrow \,12{{x}^{2}}-15x+3+16{{x}^{2}}+12x-4=29{{x}^{2}}-29\]

\[\Rightarrow 28{{x}^{2}}-3x-1=29{{x}^{2}}-29\]

\[\Rightarrow {{x}^{2}}+3x-28=0\]

\[\Rightarrow {{x}^{2}}+7x-4x-28=0\]

\[\Rightarrow x(x+7)-4(x+7)=0\]

\[\Rightarrow (x-4)(x+7)=0\]

\[\Rightarrow x=-7\] or \[4\]