Answer:
[tex]\displaystyle x = 0 \text{ and } x = 1[/tex]
Step-by-step explanation:
We are given and asked to solve the equation:
[tex]\displaystyle \frac{x}{x-2} + \frac{x-1}{x+1} = -1[/tex]
We can first remove the rational expressions by multiplying both sides of the equation by its LCM (in this case, by (x - 2)(x + 1)):
[tex]\displaystyle (x-2)(x+1) \left( \displaystyle \frac{x}{x-2} + \frac{x-1}{x+1}\right) = -1(x-2)(x+1)[/tex]
Simplify:
[tex]\displaystyle x(x+1)+(x-1)(x-2) = -(x-2)(x+1)[/tex]
Expand and isolate:
[tex]\displaystyle \begin{aligned} (x^2 + x) + (x^2 - 3x + 2) & = -(x^2 -x -2) \\ \\ 2x^2 - 2x + 2& = -x^2 + x + 2 \\ \\ 3x^2 - 3x & = 0 \end{aligned}[/tex]
Hence solve for x:
[tex]\displaystyle \begin{aligned} 3x(x - 1) & = 0 \\ \\ x & = 0\text{ or } x = 1\end{aligned}[/tex]
In conclusion, the solutions of the equation are x = 0 and x = 1.
