[tex] \dfrac{x}{x - 2} + \dfrac{x - 1}{x + 1} = -1 [/tex]
[tex] (x - 2)(x + 1) \times \dfrac{x}{x - 2} + (x - 2)(x + 1) \times \dfrac{x - 1}{x + 1} = -1(x - 2)(x + 1) [/tex]
[tex] (x + 1)x + (x - 2)(x - 1) = -1(x - 2)(x + 1) [/tex]
[tex] x^2 + x + x^2 - 3x + 2 = -x^2 + x + 2 [/tex]
[tex] 3x^2 - 3x = 0 [/tex]
[tex] 3x(x - 1) = 0 [/tex]
[tex] x = 0~~~or~~~x - 1 = 0 [/tex]
[tex] x = 0 ~~~ or ~~~ x = 1 [/tex]
Since neither 0 nor 1 cause a zero in a denominator in the original equation, 0 and 1 are acceptable as solutions.
Answer: x = 0 or x = 1