question_answer

Solve for x:

\[\frac{2}{x+1}+\frac{3}{2(x-2)}=\frac{23}{5x},x\ne 0,-1,2\]

Answer:

We have, \[\frac{2}{x+1}+\frac{3}{2(x-2)}=\frac{23}{5x},x\ne 0,-1,2\]

\[\Rightarrow 2(10x)(x-2)+3(5x)(x+1)=23(2)(x+1)(x-2)\]

\[\Rightarrow 20x(x-2)+15x(x+1)=46(x+1)(x-2)\]

\[\Rightarrow 20{{x}^{2}}-40x+15{{x}^{2}}+15x=46({{x}^{2}}+x-2x-2)\]

\[\Rightarrow 20{{x}^{2}}-40x+15{{x}^{2}}+15x=46{{x}^{2}}-46x-92\]

\[\Rightarrow 11{{x}^{2}}-21x-92=0\]

\[\Rightarrow x=\frac{21\pm \sqrt{441+4048}}{22}\]

\[\Rightarrow x=\frac{21\pm \sqrt{4489}}{22}\]

\[\Rightarrow x=\frac{21\pm 67}{22}\]

\[\Rightarrow x=\frac{21+67}{22}or\,\frac{21-67}{22}\]

\[\Rightarrow x=\frac{88}{22}or-\,\frac{46}{22}\]

\[\therefore x=4\] or \[-\frac{23}{11}\]