Solve for x : [latex]\frac{x}{x+1} + \frac{2}{x-1} = \frac{3x+2}{(x+1)(x-1)}[/latex]

Check the restrictions: x ≠ -1 and 1

Combine the two fractions on the left side by finding a common denominator

[latex]\frac{x}{x+1} + \frac{2}{x-1}[/latex]

[latex]\frac{x}{x+1}\left(\frac{x\ -\ 1}{x\ -\ 1}\right)+\frac{2}{x\ -\ 1}\left(\frac{x+1}{x+1}\right)[/latex]

[latex]\frac{x\ (x\ -\ 1)}{(x+1)(x\ -\ 1)}+\frac{2\ (x\ +1)}{(x+1)(x\ -\ 1)}[/latex]

[latex]\frac{x^2\ -\ x}{(x+1)(x\ -\ 1)}+\frac{2\ x\ +2}{(x+1)(x\ -\ 1)}[/latex]

[latex]\frac{x^2\ +x+2}{(x+1)(x\ -\ 1)}[/latex]

There a multiple ways to solve this, you can move the equation on the right over and set it equal to zero, or you can cross multiply.

[latex](x^2\ +x+2)(x+1)(x\ -\ 1)=(3x+2)(x+1)(x\ -\ 1)[/latex]

Since both sides contain the terms (x + 1) and (x – 1), we can divide that out

[latex]\frac{{(x}^2\ +x+2)\cancel{(x+1)(x\ -\ 1)}}{\cancel{(x+1)(x\ -\ 1)}}=\frac{(3x+2)\cancel{(x+1)(x\ -\ 1)}}{\cancel{(x+1)(x\ -\ 1)}}[/latex]

Cancel out terms and set the new values equal to each other. Remember in order to solve for x, we need to have the equation equal zero, so we will move the terms from the right side over

[latex]x^2\ +x+2\ =\ 3x+2[/latex]

[latex]x^2\ -\ 2x\ =\ 0[/latex]

Factor and solve

[latex]x(x\ -\ 2)\ =\ 0[/latex]

So our roots are x = 0 and 2.

Check to make sure that those values are not within the restrictions. Since the restrictions are -1 and 1, both of these solutions are valid