Question

(x)/(x-2)+(x-1)/(x+1)=-1

Answer 1

Answer:

$x=0$ and $x=1$

Step-by-step explanation:

$\frac{x}{x-2}+\frac{x-1}{x+1}=-1$

$\frac{x(x+1)}{(x-2)(x+1)}+\frac{(x-2)(x-1)}{(x-2)(x+1)}=-1$

$\frac{x(x+1)+(x-2)(x-1)}{(x-2)(x+1)}=-1$

$x(x+1)+(x-2)(x-1)=-(x-2)(x+1)$

$x^2+x+x^2-3x+2=-(x^2-x-2)$

$2x^2-2x+2=-x^2+x+2$

$3x^2-2x+2=x+2$

$3x^2-3x=0$

$3x(x-1)=0$

$x=0$ and $x=1$

Answer 2

Answer:

x = 0 or x = 1

Step-by-step explanation:

(x)/(x - 2) + (x - 1)/(x + 1) = -1

The LCD is (x - 2)(x + 1).

Multiply both sides of the equation by the LCD.

(x - 2)(x + 1) × (x)/(x - 2) + (x - 2)(x + 1) × (x - 1)/(x + 1) = (x - 2)(x + 1) × (-1)

The factors in bold cancel out of each product.

x(x + 1) + (x - 2)(x - 1) = -(x - 2)(x + 1)

x² + x + x² - x - 2x + 2 = -(x² + x - 2x - 2)

2x² - 2x + 2 = -x² + x + 2

3x² - 3x = 0

3x(x - 1) = 0

x = 0 or x = 1

The restrictions on the domain are:

x - 2 = 0

x = 2

x + 1 = 0

x = -1

Since our solutions do not include the excluded values, x = 2 and x = -1, both solutions x = 0 and x = 1 are valid.

Answer: x = 0 or x = 1