Answer :
Given:
The equation is:
[tex]\dfrac{x-3}{x+2}+\dfrac{3x-3}{(x+1)(x+2)}=\dfrac{x}{x+1}-\dfrac{x-2}{x+2}[/tex]
To find:
The solution for the given equation.
Solution:
We have,
[tex]\dfrac{x-3}{x+2}+\dfrac{3x-3}{(x+1)(x+2)}=\dfrac{x}{x+1}-\dfrac{x-2}{x+2}[/tex]
Taking LCM on both sides, we get
[tex]\dfrac{(x-3)(x+1)+3x-3}{(x+1)(x+2)}=\dfrac{x(x+2)-(x+1)(x-2)}{(x+1)(x+2)}[/tex]
[tex](x-3)(x+1)+3x-3=x(x+2)-(x+1)(x-2)[/tex]
[tex]x^2+x-3x-3+3x-3=x^2+2x-(x^2-2x+x-2)[/tex]
[tex]x^2+x-6=x^2+2x-x^2+2x-x+2[/tex]
On further simplification, we get
[tex]x^2+x-6=3x+2[/tex]
[tex]x^2+x-6-3x-2=0[/tex]
[tex]x^2-2x-8=0[/tex]
Splitting the middle term, we get
[tex]x^2-4x+2x-8=0[/tex]
[tex]x(x-4)+2(x-4)=0[/tex]
[tex](x+2)(x-4)=0[/tex]
[tex]x=-2,4[/tex]
But [tex]x\neq -2[/tex] because for [tex]x=-2[/tex] the given equation is not defined.
Therefore, the only solution of the given equation is [tex]x=4[/tex].