• # question_answer Solve for x: $\frac{x+1}{x+1}+\frac{x-2}{x+2}=4-\frac{2x+3}{x-2};x\ne 1,-2,2$

 We have, $\frac{x+1}{x-1}+\frac{x-2}{x+2}=4-\frac{2x+3}{x-2};x\ne 1,-2,2$ $\frac{(x+1)(x+2)+(x-2)(x-1)}{(x-1)(x+2)}=\frac{4(x-2)-(2x+3)}{x-2}$ $(x-2)[{{x}^{2}}+x+2x+2+{{x}^{2}}-2x-x+2]=[4x-8-2x-3]({{x}^{2}}+x-2)$ $(x-2)(2{{x}^{2}}+4)=(2x-11)({{x}^{2}}+x-2)$ $2{{x}^{3}}+4x-4{{x}^{2}}-8=2{{x}^{3}}+2{{x}^{2}}-4x-11{{x}^{2}}-11x+22$ $4x-4{{x}^{2}}-8=-9{{x}^{2}}-15x+22$ $5{{x}^{2}}+19x-30=0$ $5{{x}^{2}}+25x-6x-30=0$ $5x(x+5)-6(x+5)=0$ $(5x-6)(x+5)=0$ $x=-5,\frac{6}{5}$ $\therefore x=-5$ or $x=\frac{6}{5}$