question_answer

Solve for x: \[\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0,x\ne 3,-3/2\]

Answer:

We have, \[\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0,x\ne 3,-3/2\]

\[2x(2x+3)+(x-3)+(3x+9)=0\]

\[4{{x}^{2}}+6x+x-3+3x+9=0\]

\[4{{x}^{2}}+10x+6=0\]

\[2{{x}^{2}}+5x+3=0\]

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

\[2x(x+1)+3(x+1)=0\]

\[(2x+3)(x+1)=0\]

\[x=-1,\frac{-3}{2}\]

\[\therefore x=-1\]                                   [\[\because \] Given \[x\ne -3/2\]]