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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].